Class 12 Medical Physics - Alternating Current

In a series LCR circuit, obtain an expression for the resonant frequency.

inductive reactance

$={ X }_{ L }=2\pi fL=\omega L$

Capacitive reactance

${ X }_{ C }=\dfrac { 1 }{ 2\pi fC } =\dfrac { 1 }{ \omega C }$

total circuit reactance

$={ X }_{ T }={ X }_{ L }-{ X }_{ C }$ or

${ X }_{ C }-{ X }_{ L }$

total circuit impendance

$=Z=\sqrt { { R }^{ 2 }+{ { X }_{ T } }^{ 2 } }$

for minimum impendance

${ X }_{ T }$ should be equals to

$0$

$\Rightarrow { X }_{ T }={ X }_{ C }-{ X }_{ L }=\dfrac { 1 }{ \omega C } -\omega L=0$

$\Rightarrow \omega =\sqrt { LC } =2\pi f$

$\Rightarrow f=\dfrac { \sqrt { LC } }{ 2\pi }Hertz$

When a series combination of inductance and resistance are connected with a $10V.50Hz$ a.c source, a current of $1A$ flows in the circuit.The voltage leads the current by phase angle of $\cfrac { \pi }{ 3 }$ radian. Calculate the values of resistance and inductive reactance.

Phase difference between current and voltage

$\phi = \dfrac{\pi}{3}$
Rms value of source voltage

$V_{rms} = 10 \ V$
Rms value of current

$I_{rms} = 1 \ A$
Thus impedance of circuit

$Z = \dfrac{V_{rms}}{I_{rms}} = \dfrac{10}{1} = 10\Omega$
Using

$\cos\phi = \dfrac{R}{Z}$
We get

$\cos\dfrac{\pi}{3} = \dfrac{R}{10}$

$\implies \ R= 5\Omega$
Also using

$\tan\phi = \dfrac{X_L}{R}$

$\therefore$

$\tan\dfrac{\pi}{3} = \dfrac{X_L}{5}$

$\implies \ X_L = 5\sqrt{3}\Omega$

Calculate the impedance of a condenser in order to run a bulb rated at 10 watt and 60 volt when connected in series to an A.C.Source of 100 volt.

Power of the bulb

$P_b = 10 \$ Watt
Operating voltage of bulb

$V_b = 60 \ V$
Impedance of bulb

$Z= \dfrac{V_b^2}{P_b} = \dfrac{60^2}{10} = 360 \ \Omega$

The current in the shown circuit is found to be $4 \sin\left(314t-\dfrac{\pi}{4}\right)A$ . Find the value of inductance.

Solution:-

In L-R circuit, we have given

$V_0\sin\left(314t\right) \; I=4\sin\left(314t-\cfrac{\pi}{4}\right)$

$\phi=\cfrac{\pi}{4}$

For calculating

$L$ , we use,

$\tan\phi=\cfrac{\omega L}{R} \Rightarrow \;L=\cfrac{R\tan\phi}{\omega}$

$\cfrac{314.\tan\cfrac{\pi}{4}}{314}$

L = 1H

An alternating voltage $E=200 \,sin \,(300t)$ volt is applied across a series combination of $R=10 \Omega$ and inductance of $800\ mH$ . Calculate the impedance of the circuit.

Given :

$L = 800 \ mH$

$R = 10\Omega$
Alternating source voltage

$E = 200 \ \sin(300 t)$
Comparing with

$E = E_o\ sin wt$
we get

$w = 300$
Inductive reactance

$X_L = wL = 300\times 800\times 10^{-3} = 240 \ \Omega$
Impedance of the circuit

$Z = \sqrt{R^2+X_L^2}$

$\implies \ Z = \sqrt{10^2+240^2} = 240.21\Omega$

A coil of reactance $100 \Omega$ and resistance $100 \Omega$ is connected to a 240 V, 50 Hz a.c. supply. Find the impedance of the circuit.

Reactance of coil,

$X = 100\Omega$

Resistance of circuit,

$R = 100\Omega$

So impedance of circuit

$Z = \sqrt{R^2+X^2}$

$\implies \ Z = \sqrt{100^2 + 100^2} = 100\sqrt{2}\Omega = 141.4\Omega$

Draw the graph of $I_{rms} \rightarrow \omega$ for an A.C., L-C-R series circuit and hence explain Q-factor.

The graph shows variation of current $I_{rms}$ with angular frequency $\omega$

The Q factor is a dimensionless parameter that indicates the energy losses within a resonant element which could be anything from a mechanical pendulum, an element in a mechanical structure or within electronic circuit such as resonant circuit.

The current in the shown circuit is found to be $4\sin { \left( 314t-\cfrac { \pi }{ 4 } \right) A }$ . Find the value of inductance.

Current=

$4\sin\left(314t-\cfrac {\pi} {4}\right)$

$\omega=314$

Phase angle

$=\cfrac {\omega L} {R}=\cfrac{\pi} 4$

$=\cfrac {314L} R=\cfrac {\pi} 4$

$L=\cfrac {\pi} 4 \cfrac {R} {314}= \cfrac{ \pi \times 314 }{ 4\times 314 }$

$L=\cfrac{\pi} 4=0.785H$

A $15.0\ \mu\ F$ capacitor is connected to a $220\ V, 50\ Hz$ source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled , what happens to the capacitive reactance and the current ?

Given :

$C = 15 \ \mu F$

$f = 50 \ Hz$

$V_{rms} = 220 \ V$
Capacitive reactance

$X_c = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi\times 50\times 15\times 10^{-6}} = 212.3 \ \Omega$
Rms current in the circuit

$I_{rms} = \dfrac{V_{rms}}{X_c} = \dfrac{220}{212.3} = 1.04 \ A$
Peak current

$I_o = \sqrt{2}I_{rms} = 1.414\times 1.04 = 1.47 \ A$
Since

$X_c\propto \dfrac{1}{f}$
So capacitive reactance will get halved if the frequency is doubled.
Also,

$I_{rms}\propto \dfrac{1}{X_c}\propto f$
Thus current will get doubled if frequency is doubled.

A capacitor of capacitance $0.5\ \mu F$ is connected to a source of alternating electromagnetic force of frequency $100\ Hz$ . What is the capacitive reactance? ( $\pi = 3.142$ )

In alternating current

$AC$ circuit the opposition to the flow of current is known as impedance. In

$DC$ it is only resistance.

The impedance offered by capacitor or inductor is known as reactance.

Capacitive reactance is given by

$X_C=\cfrac1{2\pi fC}=$

$C=0.5\times10^{-6}F$

$f=100Hz$

$X_C=?$

$\therefore X_C=(2\times3.142\times100\times0.5\times10^{-6})^{-1}$

$\cfrac{10^4}{3.142}=3182.686$

$Ohms$

An AC rms voltage of $2V$ having a frequency of $50\ KHz$ is applied to a condenser of the capacity of $10\mu F$ . The maximum value of the magnetic field between the plates of the condenser if the radius of a plate is $10 cm$ is _____.

Magnetic field

$\overrightarrow B=\cfrac{\mu_oI}{2A}\overrightarrow r$

$r=$ radius of plate

$A=$ area of plate

$I=\cfrac{2}{X_c}=20\times \omega C=20\times 5\times 10^{3}\times 10\times 10^{-6}=1$

$A$

Now,

$B=\cfrac{\mu_oIr}{2A}=\cfrac{4\pi\times 10^{-7}\times 10\times 10^{-2}}{2\times pi\times (10\times 10^{-2})^2}=\cfrac{2\times 10^{-7}}{0.1}=2$

$\mu {wb/m^2}=2\times 10^{-6}$

${wb/m^2}$

Calculate Impedance of the circuit at the given frequency.

Given :

$C = 16\times 10^{-6} \ F$

$L = 1 \ H$
Resistance of the resistor

$R = 100 \ \Omega$
Given frequency

$w = 100\pi$
Capacitive reactance

$X_c = \dfrac{1}{wC} = \dfrac{1}{100\pi\times 16\times 10^{-6}} = 200\Omega$
Inductive reactance

$X_L = wL = 100\pi\times 1 = 314\Omega$
Impedance of the circuit

$Z = \sqrt{R^2+(X_L-X_c)^2}$

$\implies = \sqrt{(100)^2+(314-200)^2} = 151.6\Omega$

Calculate inductive reactance of coil in the following figure.

Voltage supply is given as

$V =10\ \sin (1000t)$
Comparing it with

$V = V_o \ \sin wt$
We get

$w = 1000$
Inductance of coil is given as

$L = 20 \ mH = 20\times 10^{-3} \ H$
Thus inductive reactance of coil

$X_L = wL$

$\implies \ X_L = 1000\times 20\times 10^{-3} = 20 \ \Omega$

Using the expressions for charge and current for L-C oscillator, explain L-C oscillations.

Refer IMAGE

$01$

Using KVL,

$L\cfrac{di}{dt}+\cfrac{q}{c}=V$

$L\cfrac{d^2i}{dt^2}+\cfrac{1}{C}\cfrac{dq}{dt}=0$

$\cfrac{di}{dt}=-\left(\cfrac{1}{CL}\right)I\quad$ (equation

$1$ )

This will gives sin or cosine function.

$i=l_0 \sin(\Omega t+\phi)$

$\cfrac{d^2i}{dt^2}=-\omega L\quad$ (equation

$2$ )

Comparing

$1$ and

$2$

$\omega =\cfrac{1}{\sqrt{LC}}$

This is frequency of oscillation.

Refer IMAGE

$02$

In the circuit diagram shown, $X_C=100 \Omega,X_L=200 \omega$ and $R=100 \omega$ . Find the effective current through the source.

Admittance

$=Y$

Given,

$X_C=100\Omega, X_L=200\Omega$ and

$R=X_L-X_C=100\Omega$

$Y=\cfrac{1}{R}+j\cfrac{1}{(X_L-X_C)}$

$\Rightarrow |Y|=\sqrt{\left(\cfrac{1}{R}\right)^2+\left(\cfrac{1}{100}\right)^2}=\sqrt { \left( \cfrac { 1 }{ 100 } \right) ^{ 2 }+\left( \cfrac { 1 }{ 100 } \right) ^{ 2 } }=\cfrac{\sqrt{2}}{100}$

Current through source,

$I=VY=200\times\cfrac{\sqrt{2}}{100}=2\sqrt{2}A$

What do you mean by resonance in $LCR$ series circuit? Write the formula for resonant frequency.

The resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase.

The sharp minimum in impedance which occurs is useful in tuning applications.

Now, the formula of resonant frequency is

${{f}_{r}}=\sqrt{\dfrac{1}{4{{\pi }^{2}}LC}}$

An alternating voltage $E = 200\sin 300t$ is applied across a series combination of $R = 10\ ohm$ and an inductor of $800\ mH$ calculate.
(i) impedance of the circuit.
(ii) Peak value of current in the circuit.

$(i)$ Impedance

$Z=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } }$

$=\sqrt { { R }^{ 2 }+(\omega { L) }^{ 2 } }$

$=\sqrt { { 10 }^{ 2 }+({ 300\times 800\times { 10 }^{ -3 }) }^{ 2 } }$

$Z=240.2 \Omega$

$(ii)$

$I_0=\cfrac {V_0}{Z}=\cfrac{200}{240.2}=0.832A$

What is wattles current?

The current in

$AC$ circuit is said to be wattless current when the average power consumed in such circuit correspond to zero such current is also called idle current.

A series $LCR$ circuit has an inductance of $100\ mH$ , a capacitor $0.1\mu F$ and a resistance of $200\Omega$ . Find the impedance at resonance and the resonant frequency.

At resonance ,

Impedance

$Z=R=200\Omega$

Resonance Frequency

$f_r=\cfrac { 2\pi }{ \sqrt { LC } }$

$=\cfrac { 2\pi }{ \sqrt { 100\times { 10 }^{ -3 }\times 0.1\times { 10 }^{ -6 } } }$

$=2\pi \times 10^4 Hz$

An inductor $2/\pi$ F and a resistor $100/\pi$ are connected in series across a source of emf v= $10$ sin $100$ [ $\pi$ t]. Here t is in second. (a) find the impedance of the circuit find the energy dissipated in the circuit in $20$ minutes.

We know that, impedance of a LR circuit is

$Z=\sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } }$ ---- (1)

emf, V=

$10sin100\pi t$

$\omega =2\pi f=100\pi \\ \quad f=50Hz$

Therefore,

$\omega L=2\pi fL=2\times 50\times \pi \times \dfrac { 2 }{ \pi } \\ \quad \quad \omega L=2\times 50\times 2\\ \quad \quad =\quad 200$

(1)=.>

$Z=\sqrt { { (\dfrac { 100 }{ \pi } ) }^{ 2 }+{ 200 }^{ 2 } }$

=202 ohm

Hence impedance of the circuit, Z= 202 ohm

A voltage V = 200 sin 377 t volt is applied across an inductance L having a resistance of $1.0 \Omega$ The maximum current is found to be 10 A. Find the value of L.

1343607_1111521_ans_4dcac30fb6ed4379941a01505376449a.jpg

Calculate the total reactance if two inductor of 10mH and 50 mH are connected in series with 10 kHz AC.

Total Reactance is given by

$X=X_1+X_2=\omega(L_1+L_2)$

$\implies X=2\pi f(L_1+L_2)$

$\implies X=2\times 10^4\times \pi (60\times 10^{-3})$

$\implies X=3768\Omega=3.768 K\Omega$

A sinusodial voltage of peak value 283 V and frequency 50 Hz i applied across a LCR circuit in which $R = 3 \Omega$ $L = 25.48\space mH$ $C = 796 \space \mu F$ . Find the impedance of the circuit.

$v=283 v\ \ f=50 Hz$

$\omega =2 \pi f=2 \times \pi \times 50$

$\omega =10\ pi rad/s$

$R=3 \Omega\ \ L=25.48 m/s\ \ \ C=796 \mu F$

$X_L=\omega L$

$=100 \times \pi \times 25.48 \times 10^{-3}$

$X_L=8 \Omega$

$X_c=\dfrac{1}{\omega c}=\dfrac{1}{100 \pi \times 796 \times 10^{-6}}=$

$X_c=4 \Omega$

Now

$100$ perdance

$z=\sqrt{R^2+(X_L-X_C)^2}$

$z=\sqrt{3^2+(8-4)}=\sqrt{3^2+4^2}$

$z=5 \Omega$

Explain the production of the electrical oscillation in LC circuit. Under what conditions are the oscillation produce in the circuit.

1311233_1138286_ans_d80c0d93b9e94540b05ad60aea2f4893.jpg

A $20\Omega$ resistor, $1.5$ H inductor and $35\mu F$ capacitors are connected in series with $200$ V, $50$ Hz ac supply. Calculate the impedance of the circuit and also find the current through the circuit?

Impedence equal to resistance

$z=R\quad \phi ={ 0 }^{ o }$

$z=20\Omega$

${ I }_{ v }=\dfrac { { E }_{ v } }{ Z } =\dfrac { 200 }{ 20 } =10A$

Average power transferred

$P={ E }_{ v }{ I }_{ v }\cos { { \theta }^{ o } }$

$200\times 20\times 1$

$=200watt$

If an LC circuit has $C= 1\mu F$ & $L = 0.25$ henry. neglecting any resistance in the circuit, what is the frequency of oscillation ?

Solution

In LC circuit,

$R = 0\Omega$ ,

Net reactance = 0

$\Rightarrow Lw -\dfrac{1}{Lw} = 0$

$\Rightarrow w = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\dfrac{1}{4}\times 10^{-6}} = 2\times 10^{8} rad/s$

1152180_1174999_ans_fa885fba19ec465ba3bd50def52a2306.jpg

The figure shows the graphical variation of the reactance of a capacitor with frequency of ac source.
Find the capacitance of the capacitor.

Let the capacitance

$u$ of the capacitor is

$C$

we know

Reactant

$X_c = \dfrac{1}{\omega C}$

$\omega = 2\pi f$

$X_c = \dfrac{1}{2\pi fC}$

From the graph

$X_c = 2 \Omega, f = 300Hz$

$2 = \dfrac{1}{2\pi \times 300 \times C}$

$C = \dfrac{1}{2\pi \times 2 \times 300} = \dfrac{1}{1200\pi} = 0.0002652 = 265.2\mu F$

1497149_1700258_ans_7177d763a73e4e8e991b35597a24d944.png

An ac source of emf $V = V_0 \, \sin \omega t$ is connected to a capacitor of capacitance C. Deduce the expression for the current (I) flowing in it. Plot the graph of (i) V vs $\omega t$ , and (ii) I vs. $\Omega t$ .

If a capacitor of capacitance 'C' is connected across the alternating source, the instantaneous charge on the capacitor is :-

$q = C V_c = C V_o \sin \omega t$

and the instantaneous current i paring through it is given by :

$i = \dfrac{dq}{dt} = CV_0 \omega \cos \omega t$

$= \dfrac{V_0}{1/\omega c} \sin (\omega t + N_2)$

$\Rightarrow i = i_0 \sin (\omega t + N_2)$

1497110_1700246_ans_3677420fcd194c2692f4f492618821e4.png

Show that in an $AC$ circuit containing a pure inductor, the voltage is ahead of current by $\pi /2$ in phase

AC voltage applied to an inductor

Source,

$v=v_{m}\sin \omega t$

Using Kirchhoff's loop rule,

$\sum{\varepsilon (t)}=0$

$v-L\dfrac{di}{dt}=0$

$\dfrac{di}{dt} = \dfrac{v}{L} = \dfrac{v_{m}\sin \omega t}{L}$

Integrating

$\dfrac{di}{dt}$ with respect to time,

$\int{\dfrac{di}{dt}}dt=\dfrac{v_{m}}{L}\int{\sin(\omega)dt}$

$i=-\dfrac{v_{m}}{\omega}L\cos(\omega t)+constant$

$-\cos \omega t=\sin \left(\omega t-\dfrac{\pi}{2}\right)$

$\therefore i=i_{m}\sin \left(\omega t-\dfrac{\pi}{2}\right)$

$\because$ Where,

$i_{m}=\dfrac{v_{m}}{\omega L}$ is the amplitude of current

1766108_1842193_ans_5c269f0b386f4165a4b1daf748b163bc.png

Answer in brief:
What is the natural frequency of the LC circuit? What is the reactance of this circuit at this frequency?

$X_{L}=-X_{C}$

$\Rightarrow \omega L=\dfrac{1}{\omega C}$

$\omega^{2}LC=1$

$\omega=\dfrac{1}{\sqrt{2C}}$

$\omega=2\pi f\Rightarrow f=\dfrac{\omega}{2}\pi =\dfrac{1}{2\pi \sqrt{LC}}$

Total reactance

$=X_{L}+X_{C}$

$=-X_{C}+X_{C}$

$=0$

$V=\dfrac{1}{2\pi \sqrt{LC}}$
Reactance of the circuit at this frequency is zero.

Alternating $emf$ of $e = 220\sin 100 \pi t$ is applied to a circuit containing an inductance of $(1/\pi)henry$ . Write an equation for instantaneous current through the circuit. What will be the reading of the $AC$ galvanometer connected in the circuit?

From

$e=220\sin 100\pi t$

$e_{0}=220V$ ,

$\omega = 100\pi$ ,

$L=(1/\pi)H$

$X_{I}=\omega L=100\pi \times \dfrac{1}{\pi}=100\Omega$

$\therefore I_{0}=\dfrac{e_{0}}{X_{L}}=\dfrac{220}{100}=2.2A$

As current lags behind the e.m.f by a phase angle

$\pi /2$

$\therefore I=I_{0}\sin (\omega t-\pi/2)$

$I=2.2\sin (100\pi t-\pi/2)$

Reading of a.c. galvanometer,

$I_{v}=\dfrac{I_{0}}{\sqrt{2}}=\dfrac{2.2}{\sqrt{2}}$

$I_{v}=\dfrac{2.2}{\sqrt{2}}\dfrac{\sqrt{2}}{\sqrt{2}}= 1.1\times 1.414=1.555A$

How soon does the current amplitude in an oscillating circuit with quality factor $Q=5000$ decrease $\eta =2.0$ times if the oscillation frequency is $\nu=2.2MHz$ ?

$Q=\dfrac{\pi}{\beta T}$ or

$\beta =\dfrac{\pi}{QT}$
Now,

$\beta t=In\eta$ so

$t=\dfrac{In\eta}{\pi}QT$

$=\dfrac{QIn\eta}{\pi \nu}=0.5ms$

Conceptual Questions
(a) Why does a capacitor act as a short circuit at high frequencies?
(b) Why does a capacitor act as an open circuit at low frequencies?

(a) The capacitive reactance is proportional to the inverse of the frequency. At higher and higher frequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire.
(b) As the frequency goes to zero, the capacitive reactance approaches infinity-the resistance of an open circuit.

Box may have any series combination of L, C and R. Column-I represents source current and column-II represents possible statements.

For option (A) solution : Current and voltage are in phase

$\Rightarrow$ only R or RLC in resonance.
For option (B) solution : Current leads voltage by

$\displaystyle \frac {\pi} {2} \Rightarrow$ only C

$\Rightarrow$ power factor = 0
For option (C) solution : Current leads voltage

$\Rightarrow$ RC or RLC with

$X_C > X_L$
For option (D) solution : Current leads voltage

$\Rightarrow$ RL or RLC with

$X_L > X_C$

it is observed that they resonate at frequency of 325 $\times$ s Hz. Find the value of s.

$f = \displaystyle\frac{1}{2\pi \sqrt{LC}}=$ resonance frequency

$=\displaystyle\frac{1}{2\pi \sqrt{0.03\times 2 \times 10^{-6}}}$

$=650Hz$
At resonance,

$X_L=X_C$ and

$Z=R$

$\displaystyle \therefore \cos \phi=\frac{R}{Z}=1$ or

$\phi=0^o$

An electric lamp which runs at $100V$ dc and consumes $10A$ current is connected to ac mains at $150V, 50 Hz$ cycles with a choke coil in series. The inductance of voltage across the choke is given as $x\times 10^{-3} H$ .Find $x$ Neglect the resistance of choke.

$V_L=\sqrt{V^2-V^2_R}$

$=\sqrt{(150)^2-(100)^2}$

$=111.8V$

$\displaystyle V_L=IX_L=I(2\pi fL)$

$\displaystyle \therefore L=\frac{V_L}{2\pi fI}=\frac{111.8}{2\pi \times 50\times 10}$

$=0.036H$

What is the reactance in ohms of a $2.00H$ inductor at a frequency of $50.0Hz$ ?

$X_L=2\pi fL$

$\text{But}\ L=2\ \text{H},\ f=50\ \text{Hz}$

$\Rightarrow X_L=2\pi\times 50 \times 2 = 628\ \Omega$

Match the following two columns:

(a)

$B=\cfrac{F}{il}$

$[B]=\quad \left[ \cfrac { ML{ T }^{ -2 } }{ AL } \right] =\left[ M{ A }^{ -1 }{ T }^{ -2 } \right]$
(b)

$U=\cfrac { 1 }{ 2 } L{ i }^{ 2 }$

$\therefore \quad \left[ L \right] =\left[ \cfrac { U }{ { i }^{ 2 } } \right] =\left[ \cfrac { M{ L }^{ 2 }{ T }^{ -2 } }{ { A }^{ 2 } } \right]$

$=\left[ M{ { L }^{ 2 }A }^{ -2 }{ T }^{ -2 } \right]$
(c)

$\quad \omega =\cfrac { 1 }{ \sqrt { LC } }$

$\quad \therefore \quad \left[ LC \right] \left[ \cfrac { L }{ { \omega }^{ 2 } } \right] =\left[ { T }^{ 2 } \right]$
(d)

$\left[ \phi \right] =\left[ BS \right] =\left[ M{ { L }^{ 2 }A }^{ -1 }{ T }^{ -2 } \right]$

A series LCR circuit with R = 20 $\Omega$ , L = 1.5 H and C = 35 $\mu$ F is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

At resonance, the frequency of supply power equals to naturan frequency of LCR circuit.

Impedence,

$Z=\sqrt{R^2+(\omega L-1/\omega C)}$

$\omega_L=\dfrac{1}{\omega_C}$

so,

$Z=R$

$I=V/R=200/20=10\ A$

Power,

$W=V\times I=200\times 10=2000\ W$

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

We know that

$X_{L}=\omega L=2\pi fL=2 \times 3.14 \times 50 \times 44 \times 10^{-3}=13.82\Omega$

Therefore,

$I_{rms}=\dfrac{E}{X_{L}}=\dfrac{220}{13.82}=15.92A$

A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 $\mu$ H, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]

Lower tuning,

$f_1=800\times 10^3\ Hz$

Upper tuning,

$f_2=1200\times 10^3\ Hz$

$L=200\times 10^{-6}\ H$

Capacitance of the variable capacitor for

$f_1$ is,

$C_1=1/(\omega^2_1L)$

where,

$\omega_1=2\pi f_1$

so,

$C_1=198.1\ pF$

similarly,

$C_2=88.04\ pF$

$\implies C \in [88.04\ pF, 198.1\ pF]$

What will be the effect on Inductive reactance $X_{L}$ and capacitive reactance $X_{C}$ if frequency of ac source is increased?

The inductive reactance

$X_{L} =\omega L = 2\pi v L$ will increase with the increase of frequency

$v$ while the capacitive reactance

$X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{2\pi v C}$ will decrease with the increase of frequency

$v$ .

What is meant by wattless component of the current?

The current in an AC circuit is said to be Wattless Current when the average power consumed in such circuit corresponds to Zero.Such current is also called as Idle Current

In the given circuit, the potential difference across the inductor $L$ and resistor $R$ are $200 V$ and $150 V$ respectively and the rms value of current is $5 A$ . Calculate the impedance of the circuit.

Voltage applied

$V=\sqrt{V_{L}^{2}+V_{C}^{2}} =\sqrt{(200)^{2}+(150)^2} = 250V$

Impedance of circuit,

$Z=\dfrac{V}{I}=\dfrac{250}{5}=50\Omega$

Calculate the resonant frequency and Q-factor (Quality factor) of a series L-C-R circuit containing a pure inductor of inductance 4 H, capacitor of capacitance $27\mu F$ and resistor of resistance $8\cdot 4\Omega$

Resonant frequency

$w_o = \dfrac{1}{\sqrt{LC}}$

$\therefore$

$w_o = \dfrac{1}{\sqrt{4\times 27\times 10^{-6}}}$

$\implies$

$w_o = 96.22 Hz$

Quality factor

$Q = \dfrac{L w_o}{R}$

$\therefore$

$Q = \dfrac{4\times 96.22}{8.4}$

$\implies$

$Q = 45.82$

For an A.C. circuit containing an inductance L, a capacitance C and a resistance R connected in series, establish the formula for impedance of the circuit and write the relationship between the alternating e.m.f. and currents in each of the following case when :
(i) $w L>\displaystyle\frac{1}{\omega C}$ ,
(ii) $w L<\displaystyle\frac{1}{\omega C}$ ,
(iii) $w L=\displaystyle\frac{1}{\omega C}$ .

Resonant circuit: In L-C-R circuit, when the value of impidence is least, then the current becomes maximum this phenomenon is called resonance.

In this state the frequency of the applied E.M.F. is equal to the natural frequency of the L-C circuit called resonante frequency and this circuit is called resonant electric circuit.

In the figure an inductance L, a capacitance C and a resistance R are connected in series with an alternating voltage.

At any instant alternating voltage is given as

$V=V_osin\omega t$ .........(1)

Where

$V_o$ is the peak value of alternating voltage at any instant, if the current in the circuit be I, then. The potential difference across the inductance L is.

$V_L=X_LI$ ........(2)

Similarly potential difference across the capacitance C is

$V_C$

$V_C=X_CI$ ...........(3)

and the potential difference across the resistance R is

$V_R$

$V_R=RI$ .......(4)

Now

$V_R$ and I are in the same phase while

$V_L$ leading ahead I by

$90^o$ and

$V_C$ logging behind I by

$90^o$ hence the angle between

$V_L$ and VC is

$180^o$ i.e.,

$V_L$ and

$V_C$ are in opposite phase. Now the resultant of

$V_L$ and

$V_C$ is

$V_L-V_C$ then the angle between

$V_L-V_C$ and

$V_R$ is

$90^o$ .

If the resultant of

$(V_L-V_C)$ and

$V_R$ is V.

Then according to the figure (2) by pythagoras theorem.

$V^2=(V_L-V_C)^2+V^2_R$ .......(5)

from equation (2), (3), (4) and (5)

$V^2=(X_LI-X_CI)^2+(RI)^2$

$V^2=I^2(X_L-X_C)^2+I^2R$

$\displaystyle\frac{V^2}{I^2}=(X_L-X_C)2+R2$

$\displaystyle\frac{V}{I}=\sqrt{(X_L-X_C)^2+R^2}$ ..........(6)

On comparing equation (6) with ohms law, the right hand side of equation (6) shows that the effective resistance of this circuit which is called impedance of this circuit and denoted by

$Z_{L-C-R}$

then,

$Z_{L-C-R}=\sqrt{(X_L-X_C)^2+R^2}$

but,

$X_L=WL$ and

$X_C=\displaystyle\dfrac{1}{\omega_C}$

$Z_{L-C-R}=\sqrt{\left(WL-\dfrac{1}{\omega c}\right)^2+R^2}$

equation (7) and (8) are the required expression for the impidence of the this circuit.

Phase difference between V and I-According to the above the circuit from (2) and current I in the leggs behind the potential difference V by phase angle

$\Phi$ then figure.

$tan\Phi =\displaystyle\frac{V_L-V_C}{V_R}=P$

$\Phi =tan^{-1}\left(\displaystyle\frac{V_L-V_C}{V_R}\right)$

From equation (2), (3), (4) and (5)

$\Phi =tan^{-1}\left(\displaystyle\frac{X_LI-XC.I}{RI}\right)$

$\Phi = tan^{-1}\left(\displaystyle\frac{X_L-X_C}{RI}\right)$

but

$X_L=WL, \ \ X_C=\displaystyle\frac{1}{WC}$

then,

$\Phi =tan^{-1}\left(\displaystyle\frac{WL-\displaystyle\frac{1}{WC}}{R}\right)$

equation (9), (10), (11) are required expression for phase difference for current. Thus in the L-C-R circuit the current can be expressed by the following equation.

$I=I_osin(wt-\Phi)$

Where

$\Phi =tan^{-1}\left(\displaystyle\frac{V_L-V_C}{V_R}\right)$

and

$I_o=\displaystyle\frac{V_o}{Z_{L-C-R}}$

here

$Z_{L-R-C}=\sqrt{\left(wL-\displaystyle\frac{1}{wC}\right)^2+R^2}$

Resonance frequency: For resonance condition inductive reactance

$X_L$ and capacitative reactance X equal in this circuit so the resistance of this circuit will be least then the current becomes minimum i.e., in resonance condition.

$X_L=X_C$

but

$X_L=w.L$

$X_C=\displaystyle\frac{1}{wC}$

$w^2=\displaystyle\frac{1}{LC}$

$w=\displaystyle\frac{1}{\sqrt{LC}}$

but

$2\pi f=w$

$2\pi f=\displaystyle\frac{1}{\sqrt{LC}}$

$f=\displaystyle\frac{1}{2\pi \sqrt{L.C}}$

(where f is resonance frequency of this circuit)

This is required expression for resonant frequency.

Calculate resonant frequency and Q-factor of a series L-C-R circuit containing a pure inductor of inductance 3 H, Capacitor of capacitance $27\mu F$ and resistor of resistance $7.4\Omega$ .

Given :

$L = 3H$

$R = 7.4\Omega$

$C = 27\times 10^{-6} F$

Resonant frequency

$w_o = \dfrac{1}{\sqrt{LC}}$

$\therefore$

$w_o = \dfrac{1}{\sqrt{3\times 27\times 10^{-6}}}$

$\implies$

$w_o = 111.11$

Quality factor

$Q = \dfrac{w_o L}{R}$

$\therefore$

$Q = \dfrac{111.11 \times 3}{7.4}$

$\implies$

$Q = 45.04$

Calculate Impedance of the given ac circuit.

Potential difference across capacitance,

$V_{C} = X_{C}I$

$\therefore$ Capacitive reactance,

$X_{C} = \dfrac{V_{C}}{I} = \dfrac{40}{2} = 20\Omega$

Resistance,

$R = \dfrac{V_{R}}{I} = \dfrac{30}{2} = 15\Omega$

Impedance,

$Z = \sqrt{R^{2}+X_{C}^{2}} = \sqrt{(15)^{2}+(20)^{2}}=\sqrt{225+400}=\sqrt{625}\Omega=25\Omega$

What is the power dissipated by an ideal inductor in ac circuit? Explain.

The power

$P = V_{rms}I_{rms} cos \varphi$

An ideal inductor is the one whose resistive component is zero.

where

$cos \varphi = \dfrac{R}{Z}$ ;

For ideal inductor

$R = 0$ .

$\therefore cos\varphi = \dfrac{R}{Z} =0$

$\therefore P = V_{rms}I_{rms} cos \varphi =0$ , i.e., power dissipated by an ideal inductor in ac circuit is zero.

Distinguish between the terms effective value and peak value of alternating current.

Alternating current changes in magnitude as well as direction. The maximum value of the alternating current is called the peak value. It is denoted by

$I_{0}$ . The square root of mean square value of current is called the ‘effective value’ or ‘rms value’ of current. The two are related by

$\text{Effective value, }I_{eff} = \dfrac{I_{0}}{\sqrt{2}}$

You are given many resistances, capacitors and inductors. These are connected to a variable $DC$ voltage source (the first two circuits) or an $AC$ voltage source of $50\ Hz$ frequency (the next three circuits) in different ways as shown in List $2$ . When a current I (steady state for $DC$ or rms for $AC$ ) flows through the circuit, the corresponding voltage $V_{1}$ and $V_{1}$ . (indicated in circuits) are related as shown in List $1$ .Match the two

$X_L$ =

$2\pi(50)(6 \times 10^{-3})$ =

$6\pi \times 10^{-1}$

$X_C$ =

$\displaystyle \frac{1}{2\pi(50) \times 3 \times 10^{-6}}$ =

$\displaystyle \frac{10^4}{3\pi}$

$X_C> X_L$

Answer the following:
What is the impedance of a capacitor of capacitance C in an AC circuit using source of frequency n Hz?

Given, Capacitance of capacitor

$= C$

frequency

$= n \ Hz$

$\omega = 2 \pi n$

$X_C = \dfrac{1}{\omega C}$

$X_C = \dfrac{1}{2 \pi n C}$

Impedance,

$Z = X_C = \dfrac{1}{2 \pi nC}$

When a circuit element $X$ is connected across an a.c. source, a current of $\sqrt{2} A$ flows through it and this current is in phase with the applied voltage. When another element $Y$ is connected across the same a.c. source, the same current flows in the circuit but it leads the voltage by $\dfrac{\pi}{2}$ radians. Name the circuit element $X$ and $Y$ .

When circuit elementis

$X$ , the current is in phase with the applied emf, this implies that

$X$ is

$\text{pure resistance}$

When circuit element

$Y$ is connected, the current leads the voltage by

$\dfrac{\pi}{2}$ so

$Y$ is

$\text{pure capacitance}$

An alternating current of frequency $\omega =314s^{-1}$ is fed to a circuit consisting of a capacitor of capacitance $C=73\mu F$ and an active resistance $R=100\Omega$ connected in parallel. Find the impedance of the circuit.

$\dfrac{1}{z}=\dfrac{1}{R}+\dfrac{1}{\dfrac{1}{i\omega C}}=\dfrac{1}{R}+i\omega C=\dfrac{1+i\omega RC}{R}$

$|z|=\dfrac{r}{\sqrt{1+(\omega RC)^2}}=40 \Omega$

An oscillating circuit consist of a capacitor with capacitance $C=1.2nF$ and a coil with inductance $L=6.0\mu H$ and active resistance $R=0.50\Omega$ . What mean power should be fed to the circuit to maintain undamped harmonic oscillations with voltage amplitude across the capacitor being equal to $V_m=10V$ ?

$=\dfrac{RCV_m^2}{2L}$ as in

$(4.112)$ . we get

$=5\ mW.$

If a circuit made up of a resistance $1 \Omega$ and inductance 0.01H, and alternating emf 200 V at 50 H is connected, then the phase difference between the current and the emf in the circuit is :

$R=1\Omega$

$X_L=\omega L=(2\pi f)L=\pi \Omega$

Phase difference -

$\cos\phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^2+X_L^2}}$ and the current lags the emf as this is an inductive circuit

$\implies \cos\phi=\dfrac{R}{Z}=\dfrac{1}{\sqrt{1+\pi^2}}=0.303$

$\implies \phi=72.34^o$

The current is given as $\dfrac{x}{10} A$ . Find $x$

$\displaystyle Z_1=\frac{115}{3}=38.33\Omega$

$\displaystyle \cos\phi_1=\frac{R_1}{Z_1}$

$\displaystyle \Rightarrow R_1=Z_1\cos\phi_1$

$=\displaystyle (38.33)(0.6)=23\Omega$

$\displaystyle X_L=\sqrt{Z^2_1-R^2_1}=30.67\Omega$

$\displaystyle Z_2=\frac{115}{5}=23\Omega$

$\displaystyle R_2=Z_2\cos\phi_2=(23)(0.707)$

$=16.26\Omega$
When connected in series

$\displaystyle R=R_1+R_2=39.26\Omega$

$\displaystyle X_L-X_C=14.41\Omega$

$\therefore \displaystyle Z=\sqrt{R^2+(X_L-X_C)^2}$

$=41.82\Omega$

$\displaystyle i=\frac{V}{Z}=\frac{230}{41.82}=5.5A$

An $L-C-R$ series circuit with $L=0.120H, R=240\Omega$ , and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$ . What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor?

$R=240\Omega, C=7.3\mu F, L=0.12H,\omega=2\pi\times400=2512$

RMS current is

$i_r=0.45A$

The rate of thermal energy dissipated in Inductor is zero.

As instantaneous power

$p_L=i_msin(\omega t-\pi/2)\times v_msin\omega t=-v_mi_msin2\omega t/2$

Now average power in Inductor is

$P_L=0$ as average of

$sin2\omega t$ is

$0$ .

An $L-C-R$ series circuit with $L=0.120H, R=240\Omega$ , and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$ . The average rate at which electrical energy is converted to thermal energy in the resistor is given as $\dfrac{x}{10} W$ . Find $x$

$R=240\Omega, C=7.3\mu F, L=0.12H,\omega=2\pi\times400=2512$

RMS current is

$i_r=0.45A$

The rate of thermal energy dissipated in resistor is given as

$P=i_r^2 R=0.45^2\times240=48.6W$

$\Rightarrow x/10=48.6\Rightarrow x=486$

Power losses in case of ac (in W) is

Power loss =

${I}_{rms}^{2} R=10^2 \times 5 =500W$

A charged 30 $\mu$ F capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Given: The inductance in the circuit is

$27\ mH$ .

The capacitance in the circuit is

$30\mu F$ .

To find: The angular frequency of free oscillations.

The resonant frequency is given by

$\omega _{r}=\dfrac{1}{\sqrt{LC}}\\=\dfrac{1}{\sqrt{27\times10^{-3} \times 30 \times 10^{-6}}}$

$\Rightarrow1111.11rad/s$ or

$\Rightarrow1.11\times10^3\ rad/s$

The reactance of a $2.00\mu F$ capacitor at a frequency of $50.0Hz$ is $\dfrac{x}{10}k\Omega$ . Find $x$

$reactance\quad =\quad \frac { 1 }{ 2\pi fC } =\frac { 1 }{ 2\pi \times 50\times 2\times { 10 }^{ -6 } } =\frac { x }{ 10 } \\ \Rightarrow x=16$

The capacitance of a capacitor whose reactance is $2.00\Omega$ at $50.0Hz$ is $\dfrac{x}{10}mF$ Find $x$

We have the formula

$C=\frac { 1 }{ 2\pi f{ X }_{ C } }$

$=\frac { 1 }{ 2\pi \times 50\times 2 } \quad =1.6mF$

A circuit operating at $\displaystyle \frac{360}{2\pi}Hz$ contains a $1\mu F$ capacitor and a $20\Omega$ resistor. How large an inductor must be added in series to make the phase angle for the circuit zero? Calculate the current (in A) in the circuit if the applied voltage is $120V$ .
Write value of $\dfrac{L}{1.1}+I$ .

Given :

$f = \dfrac{360}{2\pi}$ Hz,

$C = 10^{-6} F$ and

$R = 20\Omega$

For phase angle

$\phi = 0$ , we must get

$X_L=X_C$
Or

$2\pi f L=\dfrac{1}{2\pi f C}$
Thus inductance

$L=\displaystyle \frac{1}{(2\pi f)^2}C$

$\therefore$

$L=\displaystyle \frac{1}{(360)^2\times 10^{-6}}$

$=7.7H$
Since

$X_L=X_C$ , we get impedance of the circuit

$Z=R = 20\Omega$

Thus current in the circuit

$\displaystyle I=\frac{V}{Z}=\frac{V}{R}$

$\implies I = \dfrac{120}{20}=6A$

The inductance of an inductor whose reactance is $2.00\Omega$ at $50.0Hz$ is $\dfrac{x}{100} mH$ . Find $x$

We have

${ X }_{ l }=\omega L$

${ X }_{ l }=2\pi fL$

$\\ 2=2\times 3.14\times 50\times L\\ L=6.37mH$

Define capacitor reactance. Write its S.I. units

Capacitive reactance is defined as the total opposition to the current due to a capacitor , and is given by ,

$X_{c}=1/\omega C=1/2\pi fC$ ,

where

$\omega=$ angular frequency of circuit and

$f=$ frequency of circuit ,

for DC ,

$f=0$ , so

$X_{c}=\infty$ , capacitor doesn't allow the DC to pass through it .

Its S.I. unit is ohm(

$\Omega$ ).

What is the phase difference between AC e.m.f and current in the following?
Pure resistor and pure inductor.

In a pure resistor, the phase difference between AC Emf and AC current is zero.
In a pure inductor, the AC Emf is ahead of AC current by a phase of exactly

$90^{\circ}$ .

Alternating current flowing through a certain electrical device leads over the potential difference across it by $90^{\circ}$ . State whether this device is a resistor, capacitor or an inductor

In a capacitor, the current leads the voltage by

$\pi/2$ .

In modern days incoming frequency of radio receiver is superposed with a locally produced frequency produce intermediate frequency which is always constant. This makes tuning of the receiver very simple. This is used in superheterodyne oscillators.

A superheterodyne receiver, often shortened to superhet, is a type of radio receiver that uses frequency mixing to convert a received signal to a fixed intermediate frequency (IF) which can be more conveniently processed than the original carrier frequency.

A light bulb is rated at $100W$ for a $220V$ ac supply. Calculate the resistance of the bulb.

In case of

$AC$

$P_{av} = \dfrac{V_{rms}^{2}}{R} =\dfrac{V_{eff}^{2}}{R}$

$R = \dfrac{V_{rms}^{2}}{P} = \dfrac{220 \times 220}{100} = 484\Omega$

When 10V, DC is applies across a coil current through it is 2.5 A, if 10V, 50Hz A. C. is applied current reduces to 2 A. Calculate reluctance of the coil.

$so,I = 2.5$

${V \over R} = I$

$R = {{10} \over {2.5}} = 4\Omega$

$I = {V \over {\rm Z}}$

${\rm Z} = (R + j\omega L)$

$= \sqrt {{R^2} + {\omega ^2}{L^2}}$

$2 = {{10} \over {\sqrt {16 + {\omega ^2}{L^2}} }}$

$16 + {\omega ^2}{L^2} = {5^2}$

${\omega ^2}{L^2} = 9$

${L^2} = {9 \over {{\omega ^2}}}$

$L = {3 \over \omega }$

$= {3 \over {2050}}$ $\therefore \omega = 2 \pi f = 2 \pi 50 = 100 \pi$

$= 9.5\,mH$

An ac source of angular frequency $\omega$ is fed across a resistor $R$ and a capacitor $C$ in series. The current registered is $I$ . If now the frequency of source is changed to $\omega/3$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of the reactance to resistance at the original frequency $\omega$ is given as $\sqrt{\displaystyle\frac{x}{5}}$ . Find $x$

$at\quad \omega ,\\ { X }_{ C }=\dfrac { 1 }{ C\omega } \\ I=\dfrac { V }{ \sqrt { { R }^{ 2 }+{ X }_{ C }^{ 2 } } } \\ at\quad \omega /3,\\ { X }_{ C }^{ " }=\dfrac { 3 }{ C\omega } =3{ X }_{ C }\\ { I }^{ " }=\dfrac { V }{ \sqrt { { R }^{ 2 }+9{ X }_{ C }^{ 2 } } } =\dfrac { I }{ 2 } =\dfrac { V }{ 2\sqrt { { R }^{ 2 }+{ X }_{ C }^{ 2 } } } \\ \Rightarrow { R }^{ 2 }+9{ X }_{ C }^{ 2 }=4{ R }^{ 2 }+4{ X }_{ C }^{ 2 }\\ \Rightarrow 5{ X }_{ C }^{ 2 }=3{ R }^{ 2 }\Rightarrow \dfrac { { X }_{ C } }{ R } =\sqrt { \dfrac { 3 }{ 5 } }$
$x$ =3

Calculate the current (in $A$ ) in the circuit.

$z=\sqrt { { X_L }^{ 2 }+{ R }^{ 2 } }$

$= \sqrt { { (\dfrac { 9 }{ 100\pi } \times2\pi \times50 })^{ 2 }+{ 12 }^{ 2 } }$

$=\sqrt { { 9 }^{ 2 }+{ 12 }^{ 2 } }$

$=\sqrt { 81+144 } =\sqrt { 225 } =15\Omega$

$I=\dfrac{V}{Z}$

$=\dfrac{225}{15}$

$=15 A$

If the inductance of the choke coil is $x\times10^{-4}H$ . Find $x$

$V=160\sqrt { 2 } sin(2\pi 50t)\quad V\\ \omega =2\pi 50\quad rad/sec\\ in\quad phasor,\quad \\ { I }_{ rms }=\frac { 160 }{ \sqrt { { R }^{ 2 }+{ L }^{ 2 }{ \omega }^{ 2 } } } A\\ therefore,\quad \\ { I }_{ rms }^{ 2 }={ 10 }^{ 2 }\\ \Rightarrow \frac { { 160 }^{ 2 } }{ { R }^{ 2 }+{ L }^{ 2 }{ \omega }^{ 2 } } =100\\ \Rightarrow \frac { { 160 }^{ 2 } }{ 100 } ={ R }^{ 2 }+{ L }^{ 2 }{ \omega }^{ 2 }\\ \Rightarrow 256={ 5 }^{ 2 }+{ L }^{ 2 }{ (2\pi 50) }^{ 2 }\\ \Rightarrow L=\frac { \sqrt { 256-25 } }{ 2\pi 50 } =484 \times { 10 }^{ -4 }\ H$

The maximum current in the coil is given as $\dfrac{x}{100} A$ . Find $x$ .

$V=240\sqrt { 2 } sin(2\pi 50t)\quad V\\ \omega =2\pi 50\quad rad/sec\\ in\quad phasor,\quad \\ { I }_{ max }=\frac { 240\sqrt { 2 } }{ \sqrt { { R }^{ 2 }+{ L }^{ 2 }{ \omega }^{ 2 } } } =1.823A$

$x$ = 182

The maximum current in the circuit is $x\times 10^{-2} A$ . Find $x$ .

$V=110\sqrt { 2 } sin(2\pi 60t)\quad V\\ \omega =2\pi 60\quad rad/sec\\ in\quad phasor,\quad \\ { I }_{ max }=\frac { 110\sqrt { 2 } }{ \sqrt { { R }^{ 2 }+\frac{1}{{ C }^{ 2 }{ \omega }^{ 2 } }} } =3.24A$

$x$ =324

Leading is $\dfrac{x}{1000}H$

$\displaystyle P=V_{rms}i_{rms}\cos\phi$
or

$200=230\times 8\times \cos\phi$

$\therefore \cos\phi=0.108$
or

$\phi=83.8^o$
Further,

$P=i^2_{rms}R$

$\therefore R=\frac{P}{i^2_{rms}}=\frac{200}{(8)^2}=3.125\Omega$

(a)

$\tan\phi=\displaystyle\frac{X_C-X_L}{R}$

$\therefore \displaystyle\frac{1}{2\pi fC}-(2\pi fL)=R\tan\phi$

$\therefore \displaystyle L=\frac{1}{(2\pi f)^2C}-\frac{R \tan \phi}{2\pi f}$

$\displaystyle =\frac{1}{(2\pi\times 50)^2\times 20\times 10^{-6}}-\frac{3.125\tan 83.8^o}{2\pi \times 50}$

$=0.416H$

Match the following column

For (A) Inductance depends on shape because shape will be the deciding factor in th increasing and decreasing rate of current through the inductor.And

$L=\frac { e }{ di/dt }$
Also, the rate of increasing and decreasing of current is also depends on the medium inserted through the inductor.therefore, 2&3
For,(B)It definitely depends on shape. Because

$C=\epsilon A/d$ and also medium inserted because of presence of

$\epsilon$ in the expression of C. therefore, 2&3
For (C)Since impedance depends on resistance, capacitacnce and inductance therefore, it depends on inductor and capacitor and they both depends on shape and medium inserted.
And impedance is also depends on frequency of the applied source since,

$(X_L-X_C)=\omega L-1/\omega C$
therefore, 1,2,3&4
For (D)

$X_C=1/\omega C$ therefore, 2,3&4

In the circuit shown in figure, match the following two columns. In column $II$ quantities are given in $SI$ units.

(a)

$R=\displaystyle \frac{V_R}{I}=\frac{40}{2}=20\Omega$
(b)

$V_C=IX_C=2\times 30=60V$
(c)

$V_L=IX_L=2\times 15$

$=30V$

$(d)$

$V=\displaystyle \sqrt{V^2_R+(V_C-V_L)^2}$

$=50V$ $

A $300\Omega$ resistor is connected in series with a $0.800H$ inductor. The voltage across the resistor as a function of time is $V_R=(2.50V)\cos[(950 rad/s)t]$ . Determine the inductive reactance of the inductor.

$X_L=\omega L=950\times 0.8=760\Omega$

Power loss (in W) in case of dc is

In AC mode

$P = V _{rms} I_{rms} = 160*10 = 1600 W$

Power loss in lamp =

$i^{2}R = 10^{2}*5 = 500 W$

Power loss in dc mode = 1600 - 500 = 1100 W

An $L-C-R$ series circuit with $L=0.120H, R=240\Omega$ , and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$ . What is the impedance of the circuit?

$R=240\Omega, C=7.3\mu F, L=0.12H,\omega=2\pi\times400=2512\ Hz$

RMS current is

$i_r=0.45A$

Impedance is given as

$Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$ .

So,

$Z=344\Omega$

The source voltage $V$ ,

$\displaystyle V_R=IR=80\ V$

$cos(\theta) = 0.8 = \cfrac{V_R}{V}$

Solving,

$\displaystyle V=100\ V$

What is meant by rms (effective) value of alternating current?

RMS value of alternating current is defined as that value of the steady current, which when passed through a resistor for a given time, will generate the same amount of heat as generated by an alternating current when passed through the same resistor for the same time.

(a)The pointer in the galvanometer deflects, when a bar magnet pushed or away from the coil connected to a galvanometer. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depends. State the laws describing this phenomenon.
(b) Sketch the change in flux, emf and force when a conducting rod PQ of resistance R and length l moves freely to and fro between A and C with speed v on a rectangular conductor placed in uniform magnetic field as shown in the figure.

a) When the magnet shown below is moved “towards” the coil, the pointer or needle of the Galvanometer, which is basically a very sensitive center, will deflect away from its center position in one direction only. When the magnet stops moving and is held stationary with regards to the coil the needle of the galvanometer returns back to zero as there is no physical movement of the magnetic field. This is as per Michael Faraday’s law of electromagnetic induction which states: “that a voltage is induced in a circuit whenever relative motion exists between a conductor and a magnetic field and that the magnitude of this voltage is proportional to the rate of change of the flux”.

Factors on which amount and direction of deflection depends are as under-

1) Increasing the number of turns of wire in the coil. – By increasing the amount of individual conductors cutting through the magnetic field, the amount of induced emf produced will be the sum of all the individual loops of the coil.

2) Increasing the speed of the relative motion between the coil and the magnet. – If the same coil of wire passed through the same magnetic field but its speed or velocity is increased, the wire will cut the lines of flux at a faster rate so induced emf would be produced.

3) Increasing the strength of the magnetic field. – If the same coil of wire is moved at the same speed through a stronger magnetic field, there will be more emf produced because there are more lines of force to cut.

In a series LCR circuit connected to an ac source of variable frequency and voltage $v=v_{m}sin\omega t$ , draw a plot showing the variation of current (I) with angular frequency $(\omega)$ for two different values of resistance $R_{1}\;and \;R_{2}(R_{1}>R_{2})$ , Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves a sharper resonance is produced? Define Q-factor of the circuit and give its significance

Figure shows the variation of

$i_{m}$ with

$\omega$ in a LCR series circuits for two values of Resistance

$RF_{1}\;and \;R_{2}(R_{1}>R_{2})$ ,
The condition for resonance in the LCR circuit is,

$\displaystyle \omega_{0}=\frac{1}{\sqrt{LC}}$
We can observe that the current amplitude is maximum at the resonant frequency

$\omega$ . Since

$i_{m}=V_{m}lR$ at resonance, the current amplitude fro case

$R_{2}$ is sharper to that for case

$R_{1}$ Quality factor or simply the Q-factor of a resonance LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor ) to that of applied voltage
It is given by

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$
The Q factor determines the sharpness of the resonance curve and if the resonance is less sharp, the maximum current decrease and also the circuit is close to the resonance for a larger range

$\Delta \omega$ of frequencies and the regulation of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit

An alternative voltage given by $V=140( \sin{314} t)$ is connected across a pure resistor of $50 \Omega$ . Find
(i) The frequency of the source.
(ii) The rms current through the resistor.

(i)
If

$V=140\sin314t$
Comparing it with

$V=V_{0}\sin\omega t$

$\omega=314$ or

$2\pi f=314$
or

$f=314/2\pi=314/(2\times3.14)=50Hz$
(ii)

$I_{0}=V_{0}/R=140/50=2.8A$
therefore

$I_{rms}=I_{0}/\sqrt2=2.8/\sqrt2=1.98A$

Adjoining figure shows a series $RCL$ circuit connected to an ac source which generates an alternating emf of frequency $50\ Hz$ . The reading of the voltmeters $V_{1}$ and $V_{2}$ are $80\ V$ and $60\ V$ respectively.
Find:
The current in the circuit

$f = 50\ Hz$
Let the current in the circuit be

$I$

$V = IR$

$80 = I\times 100$

$I = \dfrac {80}{100} = 0.8\ amp$ .

At resonance, what is the relation between impedance of a series $LCR$ circuit and its resistance $R$ ?

At resonance,

$X_L=X_C$ . So, total reactance

$=X_T=X_L-X_C=0$

So, the relation between impedance of a series

$LCR$ circuit and its resistance

$R$ at resonance is

$Z=R$ .

Adjoining figure shows a series $RCL$ circuit connected to an ac source which generates an alternating emf of frequency $50\ Hz$ . The reading of the voltmeters $V_{1}$ and $V_{2}$ are $80\ V$ and $60\ V$ respectively.
Find:
The capacitance $C$ of the capacitor

Let the voltage across the capacitor be

$V_C$

$V_{C} = I\times X_{C}$

$60 = 0.8\times X_{C}$

$X_{C} = \dfrac {60\times 10}{80} = 75\Omega$

$X_{C} = \dfrac {1}{2\pi fC}$

$C = \dfrac {1}{2\pi f \times X_{C}}$

$= \dfrac {1}{2\pi \times 50\times 75}$

$= 4.25\times 10^{-5}F$

Impedance of the circuit

$Z = \sqrt{R^2 + X_L^2} = \sqrt{80^2 + 60^2}$

$= 100 \Omega$

Inductive reactance

$R = 80 \Omega$

$E = 110 V$

$\theta = tan^{-1}(3/4)$

$tan \theta = 3/4$

$tan \theta = \dfrac{X_L}{80}$

$X_L = \dfrac{80 \times 3}{4} = 60 \Omega$

In an alternating circuit applied voltage is $220V$ . If $R=8\Omega$ , ${X}_{L}={X}_{C}=6\Omega$ then write the values of the following:
(a) Root mean square value of voltage.'
(b) Impedance of circuit.

(a)

$V_{rms}=\dfrac{V_P}{\sqrt{2}}$

$=\dfrac{220V}{\sqrt{2}}V$

$=155.56V$

(b) Impedance of an RLC circuit is given as

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$=\sqrt{8^2+(6-6)^2}\Omega$

$=8\Omega$

Arrive at the expression for the impedance of a series $LCR$ circuit using phasor diagram method and hence write the expression for the current through the circuit.

Consider a series LCR circuit connected to an AC source

AC voltage applied is

$v={ v }_{ m }\sin { \omega t }$
Let the current through the series be

$i={ i }_{ m }\sin { \left( \omega t+\phi \right) }$
where

$\phi$ is phase difference between voltage and current. (Ref. image 1)
Let

$\vec {I}$ be the phasor representing current and

${ \vec { V } }_{ R },{ \vec { V } }_{ L },{ \vec { V } }_{ C }$ and

${ \vec { V } }$ be the phasors representing voltage across

$R,L,C$ and source respectively and we assume that

${ \vec { V } }_{ C }> { \vec { V } }_{ L }$ . (Ref. image 2)
The length of the phasors

${ \vec { V } }_{ R },{ \vec { V } }_{ L },{ \vec { V } }_{ C }$ are

${ V }_{ Rm }={ i }_{ m }R\quad$

${ V }_{ cm }={ i }_{ m }{ X }_{ C }$ and

${ V }_{ Lm }={ i }_{ m }{ X }_{ L }$
From Kirchhoff's loop rule

$V_L+V_R+V_C=V$
and the phasor relation is represented as,

${ \vec { V } }_{ L }+{ \vec { V } }_{ R }+{ \vec { V } }_{ C }\quad ={ \vec { V } }$
Expression for Impedence:
Applying Pythagorean Theorem to the right angle triangle,

${ V }_{ m }^{ 2 }={ V }_{ Rm }^{ 2 }+{ \left( { v }_{ cm }-{ v }_{ Lm } \right) }^{ 2 }$

${ V }_{ m }^{ 2 }={ \left( { i }_{ m }R \right) }^{ 2 }+{ \left( { i }_{ m }{ X }_{ c }-{ i }_{ m }{ X }_{ L } \right) }^{ 2 }\quad$

${ V }_{ m }^{ 2 }={ i }_{ m }^{ 2 }\left[ { R }^{ 2 }+{ \left( { X }_{ c }-{ X }_{ L } \right) }^{ 2 } \right] \quad$

${ V }_{ m }={ i }_{ m }\sqrt { { R }^{ 2 }+{ \left( { X }_{ c }-{ X }_{ L } \right) }^{ 2 } } \quad \quad$

${ i }_{ m }=\cfrac { { V }_{ m } }{ \sqrt { { R }^{ 2 }+{ \left( { X }_{ c }-{ X }_{ L } \right) }^{ 2 } } }$

${ i }_{ m }=\cfrac { { V }_{ m } }{ Z }$
Where

$Z$ is called impedance of the coil

$Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ c }-{ X }_{ L } \right) }^{ 2 } }$

A $20$ volt $5$ watt lamp is used on a.c mains of $200$ volts $50$ c.p.s. Calculate the value of
(a) capacitance (b) inductance to be put in series to run the lamp
(c) how much pure resistance should be included in place of the above device so that lamp can run on its voltage?

For lamp

$I=\cfrac { 5 }{ 20 } 0.25A;R=\cfrac { 20 }{ 0.25 } =80\Omega \quad$
Current through the lamp should be

$0.25A$
(a) when condenser

$C$ is placed in series

$\quad i=\cfrac { 200 }{ \sqrt { { R }^{ 2 }+{ \left( \cfrac { 1 }{ \omega C } \right) }^{ 2 } } } =0.25$
putting the value of

$\omega =2\pi \times 50$

$\therefore C=4.0\mu F$
(b) when inductor is used

$I=\cfrac { 200 }{ \sqrt { { R }^{ 2 }+{ \left( \omega L \right) }^{ 2 } } } =0.25\Rightarrow L=2.53H$
(c) when resistance is used

$I=\cfrac { 200 }{ R+r } =0.25\Rightarrow r=720\Omega \quad$

$200V$ A.C is applied at the ends of an LCR circuit. The circuit consists of an inductive reactance ${X}_{L}=50\Omega$ , capacitive reactance ${X}_{C}=50\Omega$ and ohmic resistance $R=10\Omega$ . Calculate the impedance of the circuit and also potential difference across $L$ and $R$ . What will be the potential difference across L-C?

The impedance of L-C-R circuit is

$Z=\sqrt { { \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 }+{ R }^{ 2 } } =\sqrt { { \left( 50-50 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 } } =10\Omega$
Now the rms current flowing in the circuit is
ie

$\cfrac { E }{ Z } =\cfrac { 200 }{ 10 } =20amp\quad$
The potential difference across the inductance is

${ V }_{ L }=I{ X }_{ L }=20\times 50\times =1000V,{ V }_{ R }=iR=20\times 10=200V\quad$
The potential difference across L-C is

$\left( 50-50 \right) \times 20=0$

A $220\ volt$ input is supplied to a transformer. The output circuit of $2.0\ ampere$ at $440\ volts$ . If the efficiency of the transformer is $80 \%$ , the current drawn by the primary windings of the transformer is

Efficiency,

$\eta=\dfrac{output power}{input power}$

$\eta=\dfrac{V_s I_s}{V_p I_p}$

$I_p=\dfrac{440\times 2}{\dfrac{80}{100}\times 220}$

$I_p=5A$

A capacitor of $20 \mu F$ is connected in series with a $25 \Omega$ resistance to a peak e.m.f $240v$ , $50Hz$ A.c. Calculate
(i)the capacitive reactance of the circuit.
(ii)an impedance of the circuit

(i)

Capacitive reactance of the circuit

$= \dfrac{1}{2\pi f C}$

$= \dfrac{1}{2 \times 3.14 \times 50 \times 20 \times 10^{-6}} = 159 \; \Omega$

(ii)

Impedance of the circuit

$= 25+159 \approx 184\; \Omega$

In the circuit shown, initially the switch is in position 1 for a long . Then the switch is shifted to position 2 for a long time. Find the total heat produced in $R_2$ .

When switch

$S$ is at position

$1$ for a long time then inductor store energy.

$U=\cfrac{1}{2}Li^2$

$i=\cfrac{E}{R_1}$

$U=\cfrac{1}{2}L\left(\cfrac{E^2}{R_1^2}\right)$

When switch

$S$ is shifted to position

$2$ then the stored energy in inductor is dissipate in resistor

$R_2$

Total heat produced in

$R_2=U=\cfrac{1}{2}\cfrac{LE^2}{R_1^2}$

970451_1020107_ans_2b2d35126d1c4a8281ef1b16b53da3f2.png

The peak voltage of an ac supply is $300V$ . What is the rms voltage?

Denote peak voltage as

$V_0$

then

$rms$ voltage is

$V_0$

$divided$

$by$

$\sqrt [2]{2}$

$V_{rms} = \dfrac{300}{1.414} = 212.16 V$

$\sqrt[2]{2} =1.414$

The frequency of applied ac emf applied to an ac circuit is quadrupled What is the effect on the (i) resistance (ii) inductive resistance and (iii) capacitive resistance.

Resistance will not change
Inductive resistance $X_L = 2\pi f L$ so it will become $4$ times when $f$ is quadrupled.

capacitive resistance

$X_c = \dfrac{1}{2\pi f c}$ so it will become

$\dfrac{1}{4}$ times of its old value.

An $LCR$ circuit has $L=10\ mH, R=150\ \Omega$ and $C=1\ \mu F$ connected in series to a source of $150\sqrt{2} \cos \omega t$ volt. At a frequency that is $50\%$ of the resonant frequency, calculate the average power (in $watt$ ) dissipated per cycle.

${ \omega }_{ 0 }=$ resonance frequency.

${ \omega }_{ 0 }=\cfrac { 1 }{ \sqrt { LC } } =\cfrac { 1 }{ \sqrt { 10\times { 10 }^{ -3 }\times 1\times { 10 }^{ -6 } } } =\cfrac { 1 }{ { 10 }^{ -4 } } ={ 10 }^{ 4 }Hz$

For,

$\omega=\cfrac{\omega_0}{2}=5\times10^3Hz$

$Z=\sqrt { { R }^{ 2 }+({ \omega L-\cfrac { 1 }{ C\omega } ) }^{ 2 } } =\sqrt { { (150) }^{ 2 }+(50-{ 200) }^{ 2 } }$

$=150\sqrt2$

$P$ decipated

$=\cfrac { Vrm{ s }^{ 2 } }{ R } cos\phi =\cfrac { { 150 }^{ 2 } }{ 150\sqrt { 2 } } \times cos\cfrac { \pi }{ 4 }$

$=\cfrac { 150\times 150 }{ 150\sqrt { 2 } } \times \cfrac { 1 }{ \sqrt { 2 } } =\cfrac { 150 }{ 2 } =75W$

In the adjoining circuit, calculate the inductive reactance and the capacitive reactance. What should be the source frequency so that the two reactance become equal in magnitude? What would then be the impedance(in ohms) of the circuit?

Inductive reactance =

$w\times L$

$0.05\times 1000 \Omega$

$50 \Omega$

Capicitive reactance =

$\dfrac{1}{w\times C}$

$\dfrac{1}{1000\times2\times10^-6}$

$500 \Omega$

Source frequency so that the two reactance become equal in magnitude

$\Rightarrow w L = \dfrac{1}{wC}$

$\Rightarrow w^2 = \dfrac{1}{LC}$

$\Rightarrow w =\dfrac{1}{\sqrt {LC}}$

$\dfrac{1}{\sqrt{ 2\times10^-6\times0.05}}$

$1000\sqrt{10}$ Hz

Then the impedance of circuit becomes equal to resistance

$\Rightarrow$ Impedance of circuit = 100

$\Omega$

When a coil is connected to a $100\ V\ dc$ supply. The current is $2\ A$ . When the same coil is connected to a.c. source $E=100\sqrt { 2 } \sin { \omega t }$ , the current is $1\ A$ . Find the inductive reactance used :

Given,

$Voltage=100v, I=2A$

We know that in Dc circuit,

$I=\dfrac{v}{R}\Rightarrow R=\dfrac{V}{I}=\dfrac{100}{2}=50\Omega$

Or in the Ac circuit

$I_{rms}=\dfrac{V_{rms}}{z}\Rightarrow 1=\dfrac{100}{z}\Rightarrow z=100\Omega$

Now,

$z=\sqrt{R^2+x_L^2}=\sqrt{50^2+x_L^2}=100\Rightarrow x_L=50\sqrt3$

Draw the graphs showing variation of Inductive reactance and Capacitive reactance with frequency of applied A.C source.

$X_L-$ Capacitive Reactance

$X_C-$ Unductive reactance

1421081_1073682_ans_ab110499151b40559e434519d194fe30.png

A bulb and a capacitor are connected in series to a source of alternating current. If its frequency increases keeping the voltage constant, the bulb will

The capacitive reactance of a circuit is given by:

${{\chi }_{c}}=\dfrac{1}{f}$

And the impedance of the circuit is given by:

$Z=\sqrt{R^2+\chi_c^2}$

When a bulb and a capacitor are connected in series to a source of alternating current. If the frequency of the source increases, then capacitive reactance decreases and thus the impedance of the circuit will also decrease.

If the impedance of the circuit decreases, the current in the circuit will increase and thus the bulb will glow brighter.

When a coil is connected to a $100$ $V$ DC supply, the current is $2$ $A$ . When the same coil is connected to AC source $E=100\sqrt{2}\sin\omega t$ , the current is $1$ $A$ . Find the inductive reactance used in the circuit.

$R=\cfrac{V}{I}$

$\Rightarrow R=\cfrac{100}{2}$

$\Rightarrow R=50$

$\Omega$

$\cfrac{100}{1A}=Z=\cfrac{V}{I}$

$\sqrt{R^2+X^2}=100$

$\Rightarrow X^2=100^2-50^2$

$\Rightarrow X=\sqrt{100^2-50^2}=\sqrt{7500}=50\sqrt{3}$

$\Rightarrow X=86.60$

$\Omega$

977170_1076410_ans_8932b9091d934814a0c727e4caf8eb32.png

The r.m.s current in an ac circuit is $1\ A$ . If the watt less current be $\sqrt {3}A$ , Then what is the power factor

Given that,

rms current ${{i}_{rms}}=1\,A$

Watt less current ${{i}_{w}}=\sqrt{3}\,A$

Now, the watt full current

${{i}_{f}}=\sqrt{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}$

${{i}_{f}}=\sqrt{2}$

Now, the power factor is

$P=\dfrac{{{i}_{f}}}{i_{rms}}$

$P=\dfrac{\sqrt{2}}{1}$

$P=\sqrt{2}$

Hence, the power factor is $\sqrt{2}$

1011897_1078454_ans_670023e66f6f4667834f8b56d0bead0f.png

Plot a graph showing variation of capacitive reactance with the change in the frequency of the $AC$ source.

Capacitive reactance

${{X}_{c}}=\dfrac{1}{2\pi fC}$

$f=\dfrac{1}{2\pi {{X}_{c}}C}$

${{X}_{c}}=Capacitive\,Reactance$

$f=frequency$

Graph between capacitive reactance and change in frequency of AC source is shown in above figure.

1032754_1079815_ans_832c1aa5cdc340b0926ca72845f28615.jpg

In the given circuit find out :
(i) maximum value of current in the circuit,
(ii) root mean square value of current in the circuit,

Given,

Inductive reactance $L=0.1\ H$

Resistance $R=40\,\Omega$

Input voltage $V=400\,\sin 300t$

Impedance $Z=\sqrt{{{R}^{2}}+{{\left( \omega L \right)}^{2}}}=\sqrt{{{40}^{2}}+{{\left( 300\times 0.1 \right)}^{2}}}=50\,\Omega$

Maximum value of current, ${{I}_{\max }}=\dfrac{{{V}_{\max }}}{Z}=\dfrac{400}{50}=8\,A$

RMS Current, ${{I}_{rms}}=\dfrac{{{I}_{\max }}}{\sqrt{2}}=\dfrac{8}{\sqrt{2}}=4\sqrt{2}\,A$

Consider the $LCR$ circuit shown in Fig. Find the net current $i$ and the phase of $i$ . Show that $i = \dfrac {V}{Z}$ . Find the impedance $Z$ for this circuit.

Total current

$i$ from the source

$V_{m}\sin \omega t$ is divided at

$B$ in two part,

$i_{1}$ through capacitor and inductor and part

$i_{2}$ through resistance.
Potential across

$R =$ Potential of source

$P.D.$ across

$R = V_{m}\sin \omega t$

$i_{2}R = V_{m}\sin \omega t$

$i_{2} = \dfrac {V_{m}\sin \omega t}{R} .... (I)$

$q_{1}$ is charge on the capacitor at any time

$t$ , then for series combination of

$C, L$ applying Kirchhoff's voltage law in loop

$ABEFA$ .

$V_{C} + V_{L} = V_{M}\sin \omega t$

$\dfrac {q_{1}}{C} + L \dfrac {di_{1}}{dt} = V_{m}\sin \omega t$

$\dfrac {q_{1}}{C} + L \dfrac {d^{2}q_{1}}{dt^{2}} = V_{m}\sin \omega t .... II$
Let

$q_{1} = q_{m} \sin (\omega t + \phi) .... III$

$i_{1} = \dfrac {dq_{1}}{dt} = q_{m}\omega \cos (\omega t + \phi) ...IV$

$\dfrac {d^{2}q_{1}}{dt^{2}} = -q_{m}\omega^{2} \sin (\omega t + \phi) ...V$
Substitute the values of equations (III) and (V) in equation (II)

$\dfrac {q_{m}\sin (\omega t + \phi)}{C} - Lq_{m}\omega^{2} \sin (\omega t + \phi) = V_{m}\sin \omega t$

$q_{m}\sin (\omega t + \phi) \left [\dfrac {1}{C} - L\omega^{2}\right ] = V_{m}\sin \omega t$
at

$\phi = 0$ ,

$q_{m}\sin (\omega t + \phi) \left [\dfrac {1}{C} - L\omega^{2}\right ] = V_{m}\sin \omega t$

$q_{m}\left [\dfrac {1}{C} - L\omega^{2}\right ] \sin \omega t = V_{m}\sin \omega t$

$q_{m} \left [\dfrac {1}{C} - L\omega^{2}\right ] = V_{m}$

$q_{m} = \dfrac {V_{m}}{\omega \left [\dfrac {1}{C\omega - L\omega}\right ]} .... (VI)$
Applying Kirchhoff's junction rule as junction

$B, i = i_{1} + i_{1}$ using relation

$I, IV$

$i = \dfrac {V_{m}\sin \omega t}{R} + q_{m} \omega \cos (\omega t + \phi)$
Now using relation VI for

$q_{m}$ and at

$\phi = 0$

$i = \left [\dfrac {V_{m}\sin \omega t}{R} + \dfrac {V_{m}\omega \cos \omega t}{\omega \left [\dfrac {1}{\omega C} - \omega L\right ]}\right ]$

$i = \dfrac {V_{m}}{R}\sin \omega t + \dfrac {V_{m}}{\left (\dfrac {1}{\omega C} - \omega L\right )} \cos \omega t$
Let

$A = \dfrac {V_{m}}{r} = C\cos \phi .... (VII)$

$B = \dfrac {V_{m}}{\dfrac {1}{\omega C} - \omega L} = C \cos \phi .... (VIII)$

$i = C\cos \phi \sin \omega t + C \sin \phi . \cos \omega t$

$= C [\cos \phi \sin \omega t + \sin \phi \cos \omega t]$

$i = C \sin (\omega t + \phi)$
Squaring and adding (VII), (VIII)

$A^{2} + B^{2} = C^{2}\cos^{2}\phi + C^{2}\sin^{2} \phi$

$= C^{2}[\cos^{2} \phi + \sin^{2}\phi)$

$A^{2} + B^{2} = C^{2}$
or

$C = \sqrt {A^{2} + B^{2}}$

$\phi = \tan^{-1} \dfrac {B}{A} = \tan^{-1} \dfrac {\dfrac {V_{m}}{\dfrac {1}{\omega C} - \omega L}}{\dfrac {V_{m}}{R}}$

$\therefore \tan \phi = \dfrac {R}{\left (\dfrac {1}{\omega C} - \omega L\right )}$

$\because C^{2} = A^{2} + B^{2} = \dfrac {V_{m}^{2}}{R^{2}} + \dfrac {V_{m}^{2}}{\left (\dfrac {1}{\omega C} - \omega L\right )^{2}}$

$C = \left [\dfrac {V_{m}^{2}}{R^{2}} + \dfrac {V_{m}^{2}}{\left (\dfrac {1}{\omega C} - \omega L\right )^{2}}\right ]^{\dfrac {1}{2}}$

$\therefore i = \left [\dfrac {V_{m}^{2}}{R^{2}} + \dfrac {V_{m}^{2}}{\left (\dfrac {1}{\omega C} - \omega L\right )^{2}}\right ]_{2} \sin (\omega t + \phi)$

$i = V_{m} \left [\dfrac {1}{R^{2}} + \dfrac {1}{\left (\dfrac {1}{\omega C} - \omega L\right )^{2}} \right ]^{\dfrac {1}{2}} \sin (\omega t + \phi) .... (IX)$
And

$\phi = \tan^{-1} \dfrac {R}{\left (\dfrac {1}{\omega C} - \omega L\right )}$

$\because I = \dfrac {V}{R}$ or

$i = \dfrac {V}{Z}$
For ac

$i = \dfrac {V}{Z} \sin (\omega t + \phi) .... (X)$
Comparing (IX) and (X)
So,

$\dfrac {1}{Z} = \left [\dfrac {1}{R^{2}} + \dfrac {1}{\left (\dfrac {1}{\omega C} - \omega L\right )^{2}}\right ]^{\dfrac {1}{2}}$
This is the impedance

$Z$ for the circuit.
1681874_1223913_ans_d35d4500235c4c92b4a7e8b2afb2bc8c.png

Find the time dependence of the current flowing through the inductance $L$ of the circuit shown in Fig. after the switch $Sw$ is shorted at the moment $t = 0$ .

Clearly,

$L \dfrac {di}{dt} = R(I - i) = \xi - RI$
So,

$2L \dfrac {di}{dt} = \xi - Ri$
This equation has the solution

$i = \dfrac {\xi}{R} (1 - e^{-tR/ 2L})$ .
1797830_1209863_ans_9e033a2d9095435898f899d222108dae.jpg

What is impedance?

Impedance, denoted Z, is an expression of the opposition that an electronic component, circuit, or system offers to alternating and/or direct electric current.Impedance is a vector (two-dimensional)quantity consisting of two independent scalar (one-dimensional) phenomena: resistance and reactance.

A circuit is set up by connecting inductance L=100mH resistor R=100 $\Omega$ and a capacitor of reactance 200 $\Omega$ in series An alternating emf of 150 $\sqrt{2}V,500m^{-1}$ Hz is applied across this series combination. Calculate the power dissipated.

Given the r.m.s value of emf in the circuit

$E_v = 150\sqrt2 V$

The impedance of the LCR circuit is given by

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

$Z = \sqrt{R^2 + (2\pi f L - 200)^2}$

$Z = \sqrt{100^2 + (2\pi \times \dfrac{500}{\pi} \times L - 200)^2}$

$Z = 100\sqrt2 \Omega$

The r.m.s value of current

$I_v$ in the circuit is given by

$I_V = \dfrac{E_V}{Z} = \dfrac{150\sqrt2}{100\sqrt2} = 1.5 A$

Power dissipated in the resistor

$= E_VI_V = 318.2 W$

Define impedance. Arrive at the expression for impedance of a series LCR circuit using a phasor diagram.

Impedance denoted by Z is an expression for the opposition that an electronic component (or) circuit offers to alternating current.

Impedance is a vector quantity.

$\in$ (ac source) has voltage

$v=v_m \sin wt$

q= charge on the the capacitor

i=current , t=time

Apply Kirchhoff's loop rule, we get.

$L \dfrac{di}{dt}+iR+\dfrac{q}{C}=V.$

$L \dfrac{di}{dt}+iR+\dfrac{q}{C}=V_m \sin wt.$ .........(1)

Let,

$q=q_m\sin (wt+\theta)$

$\therefore \dfrac{dq}{dt}=q_mw \cos (wt+\theta)$ ...........(2)

$\dfrac{d'q}{dt^2}=^-q_mw^2\sin (wt+\theta)$ .............(3)

Putting the values (2) & (3) in (1).

$L(-q_mw^2\sin (wt+\theta))+R(qn^w\cos (wt+\theta))+\dfrac{q_m\sin (wt+\theta)}{C}V_m\sin wt$

$\because X_c=\dfrac{1}{wC};X_L=wL$ ...........(4)

2133323_1571385_ans_cc2743cc6d95400fa3dc07333b05ae8d.png

An ac circuit consist of a series combination of circuit elements $X$ and $Y$ . The Current is ahead of the voltage in phanse by $\dfrac{\pi}{4}$ . If element $X$ is a pure resistor of $100\Omega$ .
(a) name the circuit element $Y$ .
(b) calculate the rms value of current, if rms value of voltage is $141V$ .
(c) what will happen if the ac source is replaced by a dc source?

As the current is ahead then the circuit is capacitive.

hence Y has to be a capacitor.

As impedence =

$\sqrt{R^2+X_c^{2}}$

thus,

R=100

$\Omega$

$\dfrac{X_c}{R}=tan \dfrac{\pi }{4}$

$X_C=R=100$

$100\sqrt2$

Therefore,

$I_{rms }=\dfrac{V_{rms }}{Z}=\dfrac{141}{100\sqrt2}=09987$

Nearly

$1\Omega$

1219307_1575289_ans_1a87826c5ba54b459a1b0c93e15393e4.png

What is wattless current?

The current in an AC circuit is said to be Wattless Current when the average power consumed in such circuit corresponds to Zero.Such current is also called as Idle Current.

What are resonating frequency for two circuits.

For ist circuit For IInd circuit

$W_r=\dfrac{1}{\sqrt{L_1C_1}}$

$W_r=\dfrac{1}{\sqrt{L_2C_2}}$

$f_r=\dfrac{1}{2\pi\sqrt{L_1C_1}}$

$f_r=\dfrac{1}{2\pi\sqrt{L_2C_2}}$

Explain the term 'sharpness of resonance' in ac circuit.

Sharpness of resonance in ac circuit :

The sharpness of resonance is defined using the 'Q' factor which explain how fast energy decay in an oscillating system.

The figure shows the graphical variation of the reactance of a capacitor with frequency of ac source.
An ideal inductor has the same reactance at $100 Hz$ frequency as the capacitor has at the same frequency. Find the value of inductance of the inductor.

Let the capacitance of the capacitor is

$C$

we know

Reactant

$X_c = \dfrac{1}{\omega C}$

$\omega = 2\pi f$

$X_c = \dfrac{1}{2\pi fC}$

From the graph

$X_c = 2 \Omega, f = 300Hz$

$2 = \dfrac{1}{2\pi \times 300 \times C}$

$C = \dfrac{1}{2\pi \times 2 \times 300} = \dfrac{1}{1200\pi} = 0.0002652 = 265.2\mu F$

Given that reactance of the inductor at

$100H_z = 6\Omega$ ,

We know

$f = 100H_z$

Reactance

$X_L = w_L$ (L = inductoance of coil)

$\omega = 2\pi f$

$6 = 2\pi fL$

$6 = 2\pi \times 100L$

$\boxed{L = \dfrac{3}{100\pi} H}$

1497214_1700259_ans_22e761dd3a024218a63efa890e79995c.png

A $0.50 kg$ body oscillates in SHM on a spring that, when extended $2.0 mm$ from its equilibrium position, has an $8.0 N$ restoring force. What are (a) the angular frequency of oscillation, (b) the period of oscillation and (c) the capacitance of an $LC$ circuit with the same period if $L$ is $5.0 H$ ?

(a) The angular frequency is.

$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{F/x}{m}}=\sqrt{\dfrac{8.0N}{\left ( 2.0\times10^{-13}m \right )\left ( 0.50kg \right )}}=89rad/s$ .

(b) The period is

$1/f$ and

$f=\omega /2\pi$ . Therefore,

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{89rad/s}=7.0\times10^{-2}s$ .

(c) From

$\omega=\left ( LC \right )^{-1/2}$ , we obtain

$C=\dfrac{1}{\omega ^{2}L}=\dfrac{1}{\left ( 89rad/s \right )^{2}\left ( 5.0Hz \right )}=2.5\times10^{-5}F$ .

To construct an oscillating $LC$ system, you can choose from a $10 mH$ inductor, a $5.0\mu F$ capacitor, and a $2.0\mu F$ capacitor. What are the (a) smallest, (b) second smallest, (c) second largest and (d) largest oscillation frequency that can be set up by these elements in various combinations?

The capacitors

$C_{1}$ and

$C_{2}$ can be used in four different ways: (1)

$C_{1}$ only; (2)

$C_{2}$ only; (3)

$C_{1}$ and

$C_{2}$ in parallel; and (4)

$C_{1}$ and

$C_{2}$ in series.

(a) The smallest oscillation frequency is,

$f_{3}=\dfrac{1}{2\pi \sqrt{L\left ( C_{1}+C_{2} \right )}}=\dfrac{1}{2\pi \sqrt{\left ( 1.0*10^{-2}H \right )\left ( 2.0*10^{-6}F+5.0*10^{-6}F \right )}}=6.0*10^{2}Hz$ .

(b) The second smallest oscillation frequency is,

$f_{1}=\dfrac{1}{2\pi\sqrt{LC_{1}}}=\dfrac{1}{2\pi \sqrt{\left ( 1.0*10^{-2}H \right )\left ( 5.0*10^{-6}F \right )}}=7.1*10^{2}Hz$ .

(c) The second largest oscillation frequency is,

$f_{2}=\dfrac{1}{2\pi \sqrt{LC_{2}}}=\dfrac{1}{2\pi \sqrt{\left ( 1.0*10^{-2}H \right )\left ( 2.0*10^{-6}F \right )}}=1.1*10^{3}Hz$ .

(d) The largest oscillation frequency is,

$f_{4}=\dfrac{1}{2\pi \sqrt{LC_{1}C_{2}/\left ( C_{1}+C_{2} \right )}}=\dfrac{1}{2\pi}\sqrt{\dfrac{2.0*10^{-6}F+5.0*10^{-6}F}{\left ( 1.0*10^{-2}H \right )\left ( 2.0*10^{-6}F \right )\left ( 5.0*10^{-6}F \right )}}=1.3*10^{3}Hz$ .

The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which can change from the rms value to zero.

$E_{rms}=220V$
Frequency = 50 Hz
(a)

$E_{rms}=220V$ =

$\displaystyle\frac{E_{0}}{\sqrt{2}}$

$\Rightarrow E_{0}=E_{rms}\sqrt{2}=\sqrt{2}\times 220=1.414\times 220=311.08V=311V$
(b) Time taken for the current to reach the r.m.s value = Time taken to reach the 0 value from r.m.s.

$\displaystyle I=\frac{I_{0}}{\sqrt{2}}\Rightarrow \frac{I_{0}}{\sqrt{2}}=I_{0}\sin wt$

$\displaystyle\Rightarrow wt=\frac{\pi }{4}$

$\displaystyle\Rightarrow t=\frac{\pi }{4}=\frac{\pi }{2\pi f}=\frac{\pi }{8\pi 50}=\frac{1}{400}=2.5ms$

Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.

f=50Hz

$I=I_{0}\sin Wt$
peak value

$I=\displaystyle\frac{I_{0}}{\sqrt{2}}$

$\displaystyle\frac{I_{0}}{\sqrt{2}}$ =

$I=I_{0}\sin Wt$

$\displaystyle\Rightarrow \frac{1}{\sqrt{2}}=\sin Wt=\sin \frac{\pi }{4}$

$\displaystyle\Rightarrow \frac{\pi }{4}=Wt$
or,

$\displaystyle t=\frac{\pi }{400}=\frac{\pi }{4\times 2\pi f}=\frac{1}{8f}=\frac{1}{8\times 50}=0.0025=2.5ms$

In a certain oscillating $LC$ circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor in $1.50\mu s$ . What are (a) the period of oscillation and (b) the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again?

(a) The period is

$T=4\left ( 1.50\mu s \right )=6.00\mu s$ .

(b) The frequency is the reciprocal of the period:

$f=\dfrac{1}{T}=\dfrac{1}{6.00\mu s}=1.67\times10^{5}Hz$ .

(c) The magnetic energy does not depend on the direction of the current (since

$U_{B}\propto i^{2}$ ),

so this will occur after one-half of a period, or

$3.00\mu s$ .

A variable capacitor with a range from $10$ to $365 pF$ is used with a coil to form a variable-frequency $LC$ circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from $0.54MHz$ to $1.60 MHz$ , the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what capacitance should be added and (c) what inductance should the coil have?

(a) Since the frequency of oscillation

$f$ is related to the inductance

$L$ and capacitance

$C$ by

$f=1/2\pi \sqrt{LC}$ , the smaller value of

$C$ gives the larger value of

$f$ . Consequently,

$f_{max}=1/2\pi \sqrt{LC_{min}}, f_{min}=1/2\pi \sqrt{LC_{max}}$ , and

$\dfrac{f_{max}}{f_{min}}=\dfrac{\sqrt{C_{max}}}{\sqrt{C_{min}}}=\dfrac{\sqrt{365pF}}{\sqrt{10pF}}=6.0$ .

(b) An additional capacitance

$C$ is chosen so the ratio of the frequencies is

$r=\dfrac{1.60MHz}{0.54MHz}=2.96$ .

Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that of the tuning capacitor.

$C$ is in picofarads (

$pF$ ), then,

$\dfrac{\sqrt{C+365pF}}{\sqrt{C+10pF}}=2.96$ .

The solution for

$C$ is,

$C=\dfrac{\left ( 365pF \right )-\left ( 2.96 \right )^{2}\left ( 10pF \right )}{\left ( 2.96 \right )^{2}-1}=36pF$ .
(c) We solve

$f=1/2\pi \sqrt{LC}$ for

$L$ .

For the minimum frequency,

$C = 365 pF + 36 pF =401 pF$ and

$f = 0.54 MHz$ . Thus

$L=\dfrac{1}{\left ( 2\pi\right )^{2}Cf^{2}}$

$L=\dfrac{1}{\left ( 2\pi\right )^{2}\left ( 401*10^{-12}F\right )\left ( 0.54*10^{-6}Hz \right )^{2}}=2.2\times10^{-4}H$ .

In the circuit shown in figure, the battery has negligible internal resistance. Show that the current in the circuit through the battery rises instantly to its steady state value $E/R$ when the switch is closed, provided that resistance R is $\sqrt{L/C}$

Let the currents through inductive branch and capacitive branch be

$I_L$ and

$I_C$ , respectively.Then the current through the battery, from KCL , is

$I=I_L+I_C$ .

Since the battery is connected in parallel to the RL and RC branches of the circuit, the current in the RL branch is unaffected by the presence of the RC branch, so

$I_L=\dfrac{E}{R}[1-e^{-(R/L)t}]$ and

$I_C=\dfrac{E}{R}e^{-(t/RC)}$

Hence,the current through battery is

$I=\dfrac{E}{R}[1-e^{-(R/L)t+e^{-(t/RC)}}]$

If the current has to reach its final value E/R instantaneously, the exponential terms must cancel out, i.e

$e^{-t/ \tau_L}$ ,which is possible if

$\tau_L=\tau_C$

$L/R=RC,R=\sqrt{(L/C)}$

For all t > 0 this is the desired result.

An inductor is connected across a capacitor whose capacitance can be varied by turning a knob. We wish to make the frequency of oscillation of this $LC$ circuit vary linearly with the angle of rotation of the knob, going from $210^{5}$ to $410^{5}Hz$ as the knob turns through $180^{0}$ . If $L=1.0 mH$ , plot the required capacitance $C$ as a function of the angle of rotation of the knob.

The linear relationship between

$\theta$ (the knob angle in degrees) and frequency

$f$ is,

$f=f_{0}\left ( 1+\dfrac{\theta }{180^{0}} \right )\Rightarrow \theta =180^{0}\left ( \dfrac{f}{f_{0}}-1 \right )$ .

where

$f_{0}=2*10^{5}Hz$ . Since

$f=\omega /2\pi =1/2\pi \sqrt{LC}$ , we are able to solve for C in terms of

$\theta$ :

$C=\dfrac{1}{4\pi ^{2}Lf_{0}^{2}\left ( 1+\theta /180^{0} \right )^{2}}=\dfrac{81}{400000\pi ^{2}\left ( 180^{0}+\theta \right )^{2}}$ .
with SI units understood. After multiplying by

$10^{12}$ (to convert to picofarads), this is plotted below:
1650576_1775167_ans_636e146ebe0b47e1ba49caaf189e46a6.png

In an oscillating $LC$ circuit in which $C=4.00 mF$ , the maximum potential difference across the capacitor during the oscillations is $1.50V$ and the maximum current through the inductor is $50.0mA$ . What are (a) the inductance $L$ and (b) the frequency of the oscillations? (c) How much time is required for the charge on the capacitor to rise from zero to its maximum value?

(a) We use

$U=\dfrac{1}{2}LI^{2}=\dfrac{1}{2}Q^{2}/C$ to solve for

$L$ :

$L=\dfrac{1}{C}\left ( \dfrac{Q}{I} \right )^{2}=\dfrac{1}{C}\left ( \dfrac{CV_{max}}{I} \right )^{2}=C\left ( \dfrac{V_{max}}{I} \right )^{2}=\left ( 4.00*10^{-6}F \right )\left ( \dfrac{1.50V}{50.0*10^{-3}A} \right )^{2}=3.60\times10^{-3}H$ .

(b) Since

$f=\omega/2\pi$ , the frequency is,

$f=\dfrac{1}{2\pi \sqrt{LC}}=\dfrac{1}{2\pi \sqrt{\left ( 3.60*10^{-3}H \right )\left ( 4.00*10^{-6}F \right )}}=1.33\times10^{3}Hz$ .

(c) Referring to figure below, we see that the required time is one-fourth of a period (where the period is the reciprocal of the frequency). Consequently,

$t=\dfrac{1}{4}T=\dfrac{1}{4f}=\dfrac{1}{4\left ( 1.33*10^{3}Hz \right )}=1.88\times10^{-4}s$ .
1650593_1775220_ans_dd24c86c611544b7b25dd8b5a8ed1b13.png

An oscillating $LC$ circuit has an inductance of $3.00 mH$ and a capacitance of $10.0\mu F$ . Calculate the (a) angular frequency and (b) period of the oscillation. (c) At time $t=0$ , the capacitor is charged to $200\mu C$ and the current is zero. Describe the graph of charge on the capacitor as a function of time.

(a) Equation

$\omega=\frac{1}{\sqrt{LC}}$ directly gives,

$1/\sqrt{LC}\approx 5.77\times10^{3}rad/s$ .

(b) Period of oscillation,

$T=2\pi /\omega =1.09ms$ .

(c)It is a cosine curve with amplitude

$200 μC$ and period given in part (b).

When under load and operating at an rms voltage of $220V$ , a certain electric motor draws an rms current of $3.00 A$ . It has a resistance of $24.0\Omega$ and no capacitive reactance. What is its inductive reactance?

We have,

$\left ( 220V/3.00A \right )^{2}=R^{2}+X_{L}^{2}\Rightarrow X_{L}=69.3\Omega$ .

(a) At what frequency would a $6.0 mH$ inductor and a $10\mu F$ capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same $L$ and $C$ .

(a) The inductive reactance for angular frequency

$\omega _{d}$ is given by

$X_{L}=\omega _{d}L$ , and the capacitive reactance is given by

$X_{C}=\dfrac{1}{\omega _{d}C}$ .

The two reactances are equal if

$\omega _{d}L=\dfrac{1}{\omega _{d}C}$ , or

$\omega _{d}=\dfrac{1}{\sqrt{LC}}$ .

The frequency is,

$f_{d}=\dfrac{\omega _{d}}{2\pi}=\dfrac{1}{2\pi \sqrt{LC}}=\dfrac{1}{2\pi \sqrt{\left ( 6.0 \times 10^{-3}H \right )\left ( 10 \times 10^{-6}F \right )}}=6.5 \times 10^{2}Hz$ .

(b) The inductive reactance is,

$X_{L}=\omega _{d}L=2\pi f_{d}L=2\pi \left ( 650Hz \right )\left ( 6.0 \times 10^{-3}H \right )=24\Omega$ .
The capacitive reactance has the same value at this frequency.

(c) The natural frequency for free

$LC$ oscillations is

$f=\dfrac{\omega} {2\pi} =\dfrac{1}{2\pi \sqrt{LC}}$ , the same as we found in part (a).

An inductor $200 mH$ , a capacitor $100 \mu F$ and a resistor $10 \Omega$ are connected in series to an a.c. source of $100 V$ , having variable frequency. At what frequency of the applied voltage will the power factor of the circuit be $1$ ?

Since the power factor

$\varphi =1$ it means the phase difference between voltage and the current is Zero.

(

$\varphi = cos\phi = 1 \Rightarrow \phi = 0$ )

This is possible when

$\omega L = \dfrac{1}{\omega C}$

$\omega^{2} = \dfrac{1}{LC}$

$\implies \nu = \dfrac{1}{2\pi \sqrt{LC}}$

$=35.58 Hz$

A series circuit containing inductance $L_{1}$ and capacitance $C_{1}$ oscillates at angular frequency $\omega$ . A second series circuit, containing inductance $L_{2}$ and capacitance $C_{2}$ , oscillates at the same angular frequency. In terms of $\omega$ , what is the angular frequency of oscillation of a series circuit containing all four of these elements? Neglect resistance. (Hint: Use the formulas for equivalent capacitance and equivalent inductance).

For the first circuit

$\omega=\left ( L_{1}C_{1} \right )^{-1/2}$ , and for the second one

$\omega=\left ( L_{2}C_{2} \right )^{-1/2}$ . When the two circuits are connected in series, the new frequency is,

$\omega '=\dfrac{1}{\sqrt{L_{eq}C_{eq}}}=\dfrac{1}{\sqrt{\left ( L_{1}+L_{2} \right )C_{1}C_{2}/\left ( C_{1}+C_{2} \right )}}=\dfrac{1}{\sqrt{\left ( L_{1}C_{1}C_{2}+L_{2}C_{2}C_{1} \right )/\left ( C_{1}+C_{2} \right )}}$ ,

$=\dfrac{1}{\sqrt{L_{1}C_{1}}}\dfrac{1}{\sqrt{\left ( C_{1}+C_{2} \right )/\left ( C_{1}+C_{2} \right )}}=\omega$ .

where we use

$\omega ^{-1}=\sqrt{L_{1}C_{1}}=\sqrt{L_{2}C_{2}}$ .

A $1.50\mu F$ capacitor has a capacitive reactance of $12.0\Omega$ . (a) What must be its operating frequency? (b) What will be the capacitive reactance if the frequency is doubled?

(a) Equation

$X_{C}=\frac{1}{\omega _{d}C}$ gives,

$f=\omega /2\pi =\left ( 2\pi CX_{C} \right )^{-1}=8.84kHz$ .

(b) Because of its inverse relationship with frequency, the reactance will go down by a factor of

$2$ when

$f$ increases by a factor of

$2$ . The answer is

$X_{C}=6.00\Omega$ .

A generator with an adjustable frequency of oscillation is wired in series to an inductor of $L=2.50 mH$ and a capacitor of $C=3.00\mu F$ . At what frequency does the generator produce the largest possible current amplitude in the circuit?

We get

$f=1/2\pi \sqrt{LC}=1.84kHz$ .

A coil of inductance $88 mH$ and unknown resistance and a $0.94\mu F$ capacitor are connected in series with an alternating emf of frequency $930 Hz$ . If the phase constant between the applied voltage and the current is $75$ , what is the resistance of the coil?

The resistance of the coil is related to the reactances and the phase constant by equation

$\tan \phi =\dfrac{X_{L}-X_{C}}{R}$ . Thus,

$\dfrac{X_{L}-X_{C}}{R}=\dfrac{\omega _{d}L-1/\omega _{d}C}{R}=\tan \phi$ .

which we solve for

$R$ :

$R=\dfrac{1}{\tan \phi }\left ( \omega _{d}L-\dfrac{1}{\omega _{d}C} \right )=\dfrac{1}{\tan 75^{o}}\left [ \left ( 2\pi \right )\left ( 930Hz \right )\left ( 0.94 \times 10^{-6}F \right ) \right ]=89\Omega$ .

In above figure, $R=14.0\Omega$ , $C=6.20\mu F$ , and $L=54.0 mH$ , and the ideal battery has emf $\xi=34.0V$ . The switch is kept at a for a long time and then thrown to position $b$ . What are the (a) frequency and (b) current amplitude of the resulting oscillations?

(a) After the switch is thrown to position

$b$ the circuit is an

$LC$ circuit. The angular frequency of oscillation is

$\omega =1/\sqrt{LC}$ . Consequently,

$f=\dfrac{\omega}{2\pi}=\dfrac{1}{2\pi \sqrt{LC}}=\dfrac{1}{2\pi \sqrt{\left ( 54.0*10^{-3}H \right )\left ( 6.20*10^{-6}F \right )}}=275Hz$ .

(b) When the switch is thrown, the capacitor is charged to

$V = 34.0 V$ and the current is zero.

Thus, the maximum charge on the capacitor is

$Q=VC=\left ( 34.0V \right )\left ( 6.20*10^{-6}F \right )=2.11*10^{-4}C$ .

The current amplitude is,

$I\omega Q=2\pi fQ=2\pi \left ( 275Hz \right )\left ( 2.11*10^{-4}C \right )=0.365A$ .

Sketch a graph showing variation of reactance of a capacitor with frequency in an AC circuit.

The graph of capacitive reactance

$X_C$ and frequency

$v$ is shown in figure.

1676977_1783189_ans_ae97ac3e2e844d259d099b967b532f6a.png

A resistor of $100 \Omega$ and a capacitor of $100/\pi\ \mu F$ are connected in series to a $220 V$ , $50 Hz$ a.c. supply. Calculate the current in the circuit. Calculate the (rms) voltage across the resistor and the capacitor. Do you find the algebraic sum of these voltages more than the source voltage? If yes, how do you resolve the paradox?

(a) Capacitive resistance

$X_c = \dfrac{1}{wC} = \dfrac{1}{2 \pi vC} = \dfrac{1}{2 \pi \times 50 \times \dfrac{100}{ \pi} \times 10^{-6} } = 100\Omega$

Impedance of the circuit ,

$Z = \sqrt{R^2 + X^2 _C} = \sqrt{100^2 + 100^2} = 100\sqrt2$

Current in the circuit

$I_{rms} = \dfrac{E_{rms}}{Z} = \dfrac{220}{100\sqrt2} = 1.56 A$

(b) Voltage across resistor

$V_R = I_{rms} R = 1.56 \times 100 = 156 V$

Voltage across capacitor,

$V_C = I_{rms} \times C = 1.56 \times 100 = 156 V$

The algebric sum of voltages across the combination is

$V_{rms} = V_R + V_c = 156 + 156 = 312 V$

While

$V_{rms}$ of the source is

$220V$ .

Yes the voltage across the combination is more than the voltage of the source. The voltage across the resistor and the capacitor are not in phase. This paradox can be resolved as when the current passes through the capacitor , it lead the voltage

$V$ , phase

$\dfrac{\pi}{2}$ . So, voltage of the source can be given as

$V_{rms}$ =

$\sqrt{{V_R}^2 + {V_C}^2}$ =

$\sqrt{(156)^2 + (156)^2} = 156\sqrt2 = 220V$

The figure shows a series $LCR$ circuit with $L=5.0H$ , $C=80 \mu F$ , $R=40 \Omega$ connected to a variable frequency $240V$ source. Calculate the angular frequency of the source which drives the circuit at resonance.

We know that

$\omega_{r} =$ Angular frequency at resonance

$=\dfrac{1}{\sqrt{LC}}$

$=\dfrac{1}{\sqrt{5 \times 80 \times 10^{-6}}}$

$=50 rad/s$

Prove that the power dissipated in an ideal resistor connected to an ac source is $\dfrac{{V_{eff}}^{2}}{R}$ .

Power in an ac circuit,

$P = V_{rms}I_{rms}cos \varphi$

As rms values of current and voltage are also called effectivevalues i.e.

$P = V_{eff}I_{eff} cos \varphi$

But

$cos \varphi = \text{power factor} = \dfrac{R}{Z}$

In a purely resistive circuit

$Z=R,\ cos \varphi = 1$

and

$i_{eff} = \dfrac{V_{eff}}{Z} = \dfrac{V_{eff}}{R}$

Substituting these values in (i), we get

$P = V_{eff}\cdot \dfrac{V_{eff}}{R} \times 1 = \dfrac{{V_{eff}}^{2}}{R}$ .

Show that the current leads the voltage in phase by $\pi/2$ in an ac circuit containing an ideal capacitor.

The instantaneous voltage,

$V = V_{0}sin (\omega t)$ ...(i)

Let

$q$ be the charge on capacitor and

$I$ , the current in the circuit at any instant, then instantaneous potential difference,

$V=\dfrac{q}{c}$ ...(ii)

From (i) and (ii)

$\dfrac{q}{C} = V_{0}sin(\omega t) \implies q = CV_{0} sin(\omega t)$

The instantaneous current,

$I = \dfrac {dq}{dt} = (CV_{0}sin(\omega t))=CV_{0}\dfrac{d}{dt}(sin(\omega t))=CV_{0}\omega cos(\omega t)$

$I=\dfrac{V_{0}}{1/\omega C} cos(\omega t)$

$I=I_{0}sin(\omega t + \dfrac{\pi}{2})$

Hence, the current leads the applied voltage in phase by

$\dfrac{\pi}{2}$

1676412_1782992_ans_cc748e1025a440f08dad782717cc0085.png

A capacitor of unknown capacitance, a resistor of $100 \Omega$ and an inductor of self-inductance $L= (\dfrac{4}{\pi^{2}})$ henry are connected in series to an ac source of $200 V$ and $50 Hz$ . Calculate the value of the capacitance and impedance of the circuit when the current is in phase with the voltage.

Given:

Resistance =

$100\Omega$

L =

$\dfrac{4}{\pi^2} Henry$

Capacitance,

$C =\dfrac{1}{L\omega^{2}}$

$=\dfrac{1}{\dfrac{4}{\pi^{2}}(2\pi \times 50)^{2}} F$

$=\dfrac{1}{40000}F$

$=2.5 \times 10^{-5}F$

Since,

$V$ and

$I$ are in the same phase

$Impedance = Resistance = 100 \Omega$

In a series $LCR$ circuit, obtain the conditions under which the impedance of the circuit is minimum,

Impedance of series

$LCR$ circuit is given by

$Z= \sqrt{R_{2} + (X_{L}-X_{C})^{2}}$

For the impedance,

$Z$ to be minimum

$X_{L} = X_{C}$

A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is $360 \mu C$ . when potential across the capacitor is reduced by $120 V$ , the charge stored in it becomes $120 \mu C$ .
The potential V and the unknown capacitance C.

Given - initial voltage

$V_1 = V,$ Initial charge

$q_1 = 360 \mu C$

Therefore , capacitance

$C = \dfrac{q_1}{V_1} =\dfrac{360}{V}$ -- (i)

When , voltage

$V_2 = V - 120,$ charge

$q_2 = 120 \mu C$

Therefore , capacitance

$C = \dfrac{q_2}{V_2} = \dfrac{120}{V-120}$ --(ii)

Capacitance of the capacitor remains constant as there is no change in area of the plates , medium and distance between plates ,

(i) Dividing (i) by (ii) ,

$\dfrac{360}{V}=\dfrac{120}{(V−120)}$

$or, 3(V−120)=V$

$or, V=360/2=180volt$

(ii)Capacitance

$C=360/180=2μF$

When a circuit element $X$ is connected across an a.c. source, a current of $\sqrt{2} A$ flows through it and this current is in phase with the applied voltage. When another element $Y$ is connected across the same a.c. source, the same current flows in the circuit but it leads the voltage by $\dfrac{\pi}{2}$ radians. Find the net impedance.

As current through X is in phase with the applied voltage, therefore, element X must be resistance T only.

As current through Y leads the applied voltage by a phase angle ϕ=π/2, therefore, element Y must be capacitor (C ) only

$I_V = \dfrac{E_V }{R} = \sqrt2$ ,

$R = E_V/ \sqrt2$

similarly,

$X_C = E_V/ \sqrt2$

When

$X$ and

$Y$ are connected in series, net impedance of circuit,

$Z = \sqrt{R^2 X^2 _C} = \sqrt{(\dfrac{E_V}{\sqrt2})^2 + (\dfrac{E_V}{\sqrt2})^2} = E_V$

$\therefore$

$I_V = \dfrac{E_V}{Z} = \dfrac{E_V}{E_V} = 1A$

$Z_{net} = \sqrt{R^2 + X^2_C} = \sqrt{R^2 + \dfrac{1}{(wC)^2}}$

Calculate the wattless current of the given ac circuit.

The phase lead (

$\varphi$ ) of current over applied voltage is

$\varphi = \dfrac{X_{C}}{R}$

$\text{Wattless Current, } I_{wattless} = Isin \varphi = I(\dfrac{X_{C}}{Z}) = 2 \times \dfrac{20}{25}A = 1.6A$

1678331_1783477_ans_7411f5d6931e4debbf7d97299923efad.png

The instantaneous voltage from an ac source is given by $E=300 sin 314t$ ; what is the rms voltage of the source?

Given equation is

$E=300 sin 314t$

Comparing with standard equation

$E =E_{0} sin\omega t$ , we have

$E_{0} = 300 V$

So,

$E_{rms} = \dfrac{E_{0}}{\sqrt{2}}$

$= \dfrac{300}{\sqrt{2}}$

$= 150\sqrt{2}V$

$= 150 \times 1.414$

$= 212 V$

What is the power dissipated in an ac circuit in which voltage and current are given by $V = 230sin (\omega t +\dfrac{\pi}{2}) \text{ and } I = 10sin\omega t$ ?

Power dissipated

$P =\dfrac{1}{2}V_{0}I_{0}cos\ \varphi$

Here

$V_{0} =230V, I_{0}=10A, \varphi = \dfrac{\pi}{2}$

$\therefore P=\dfrac{1}{2} \times 230 \times 10cos\dfrac{\pi}{2} = 0$ .

The instantaneous current in an ac circuit is $i=0.5sin314t$ , what is the frequency of the current?

Given:

$I = 0.5sin 314 t$ ...(i)

Standard equation of current is

$I=I_{0}sin \omega t$ ...(ii)

Comparing (i) and (ii), we get

$I_{0} = 0.5A$ ,

$\omega = 314$

$\therefore$ Frequency

$\nu = \dfrac{\omega}{2\pi} = \dfrac{314}{2 \times 3.14} = 50 Hz$

Derive an expression for impedance of an a.c. circuit consisting of an inductor and a resistor.

Let a circuit contain a resistor of resistance

$R$ and an inductor of inductance

$L$ connected in series.

The applied voltage is

$V = V_{0} sin \omega t$ .

Suppose the voltage across resistor

$V_{R}$ and that across inductor is

$V_{L}$ .

The voltage

$V_{R}$ and current

$I$ are in the same phase, while the voltage

$V_{L}$ leads the current by an angle

$\dfrac{\pi}{2}$ . Thus,

$V_{R}$ and

$V_{I}$ are mutually perpendicular.

The resultant of

$V_{R}$ and

$V_{L}$ is the applied voltage i.e.,

$V = \sqrt{V_{R}^{2}+V_{L}^{2}}$

But

$V_{R} = Ri$ ,

$V_{L} = X_{L}i = \omega Li$

$\therefore \text{where} X_{L} =\omega L \text{ is inductive reactance}$

$\therefore V = \sqrt{(Ri)^{2}+(X_{L}i)^{2}}$

$\therefore \text{Impedance, } Z=\dfrac{V}{i} = \sqrt{R^{2}+X_{L}^{2}} \implies Z = \sqrt{R^{2}+(\omega L)^{2}}$

1678719_1783992_ans_78df01531dca426887998882e5e1fee5.png

An ac voltage $V = V_{0} sin \omega t$ is applied across a pure inductor of inductance $L$ . Draw graphs of V and i versus $\omega t$ for the circuit.

Graph for

$V$ and

$I$ versus

$\omega t$ for the circuit is as shown in figure
1678679_1783495_ans_a37203aa013b4d7482cc0a295420c787.png

A $2 \mu F$ capacitor, $100 W$ resistor and $8 H$ inductor are connected in series with an AC source.
What should be the frequency of the source such that current drawn in the circuit is maximum? What is this frequency called?

For maximum frequency,

$\omega L = \dfrac{1}{\omega C}$

$\implies 2\pi \nu 8 = \dfrac{1}{2\pi \nu \times 10^{-6} \times 2} \implies (2\pi \nu)^{2} = \dfrac{1}{16 \times 10^{-6}}$

$\implies 2\pi \nu = \dfrac{1}{4 \times 10^{-3}} \implies 2\pi \nu = \dfrac{10^{3}}{4}$

$\implies \nu = \dfrac{250}{2\pi} = 39.77 s^{-1}$

This frequency is called resonance frequency.

Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.

A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e., if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency; it is given by

$\dfrac {1}{\omega C}$ .

Explain the term inductive reactance.

Inductive Reactance: The opposition offered by an inductor to the flow of alternating current through it is called the inductive reactance. It is denoted by

$X_{L}$ . Its value is

$X_{L}. = \omega L = 2\pi fL$ where.

$L$ is inductance and

$f$ is the frequency of the applied voltage.

Define the term Sharpness of Resonance. Under what condition, does a circuit become more selective?

$\dfrac{\omega _{0}}{2 \Delta \omega}$ is measure of sharpness of resonance, where

$\omega _{0}$ is the resonant frequency and

$2\Delta \omega$ is the bandwidth.

Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

An inductor opposes flow of current through it by developing an induced emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency, being given by

$\omega L$ .

State the condition for resonance to occur in series $LCR$ a.c. circuit and derive an expression for resonant frequency.

Condition for resonance to occur in series LCR ac circuit:

For resonance the current produced in the circuit and emf applied must always be in the same phase.

Phase difference (

$\varphi$ ) in series

$LCR$ circuit is given by

$tan \varphi =\dfrac{X_{C}-X_{L}}{R}$

For resonance,

$\varphi = 0 \implies X_{C}-X_{L} =0\ or\ X_{C}=X_{L}$

$\omega_{r}$ is resonant frequency, then

$X_{C} = \dfrac{1}{\omega _{r} C}$

and

$X_{C} = \omega _{r} L$

$\dfrac{1}{\omega _{r} C} = \omega _{r} L \implies \omega _{r} = \dfrac{1}{\sqrt{LC}}$

Linear resonant frequency,

$f_{r} = \dfrac{\omega}{2 \pi} = \dfrac{1}{2\pi \sqrt{LC}}$

Define the term capacitive reactance.

Capacitive Reactance: The opposition offered by a capacitor alone to the flow of alternating current through it is called the capacitive reactance.

It is denoted by

$X_{C}$ . Its value is

$X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{2 \pi f C}$ .

A device $X$ is connected to an ac source $V =V_{0}sin \omega t$ . The variation of voltage, current and power in one cycle is shown in the graph (see figure). Identify the device $X$ .

Phase difference between V and I is $\dfrac{\pi}{2}$ . So the device can be a capacitor.

A coil of $0.01$ henry inductance and $1\ ohm$ resistance is connected to $200\ volt, 50\ Hz$ ac supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current.

$R = 1\omega, L = 0.01\ H, V = 200\ V, v = 50\ Hz$ .

Impedance of the circuit

$Z = \sqrt {R^{2} + X_{L}^{2}} = \sqrt {R^{2} + (2\pi vL)^{2}}$

$= \sqrt {(1)^{2} + (2\times 3.14 \times 50 \times 0.01)^{2}}$

$= \sqrt {1 + 9.86} = \sqrt {10.86}$

$= 3.3\Omega$ .

$\therefore$ For phase angle,

$\tan \phi = Z/R$

$\tan \phi = \dfrac {X_{L}}{R} = \dfrac {2\pi vL}{R} = \dfrac {2\times 3.14 \times 50 \times 0.01}{1}$

$\tan \phi = 3.14$

$\phi = \tan^{-1} 3.14 = 72^{\circ}$

Phase difference

$\phi = \dfrac {72\times \pi}{180^{\circ}} rad$

$\phi = 1.20\ radian$

Time lag between alternating voltage and current

$\phi = \omega t$

$t = \dfrac {\phi}{\omega} = \dfrac {72\times \pi}{180\times 2\times \pi \times 50} = \dfrac {1}{150} \sec$ .

A series $LCR$ circuit is connected to an ac source having voltage $V=V_{0} sin \omega t$ . Derive the expression for resonant frequency.

Resonant frequency: For resonance

$\varphi=0$ , so

$X_{C}-X_{L}=0$

$$\dfrac{1}{\omega C} = \omega L

\implies \omega^{2} = \dfrac{1}{LC}$$

$\therefore \text{Resonant frequency} \omega _{r} = \dfrac{1}{\sqrt{LC}}$

Phase difference (

$\varphi$ ) in series

$LCR$ circuit is given by

$tan \varphi =\dfrac{V_{C}-V_{L}}{V_{R}} = \dfrac{i_{m}(X_{C}-X_{L})}{i_{m}R} = \dfrac{(X_{C}-X_{L})}{R}$

When current and voltage are in phase,

$\varphi =0 \implies X_{C}-X_{L} =0 \implies X_{C}=X_{L}$

This condition is called resonance and the circuit is called resonant circuit.

A series $LCR$ circuit is connected to an ac source having voltage $V=V_{0} sin \omega t$ . Derive the expression for the impedance.

Expression for Impedance in

$LCR$ series circuit: Suppose resistance

$R$ , inductance

$L$ and capacitance

$C$ are connected in series and an alternating source of voltage

$V = V_{0} sin \omega t$ is applied across it. (fig. a) On account of being in series, the current (

$i$ ) flowing through all of them is the same.

Suppose the voltage across resistance

$R$ is

$V_{r}$ voltage across inductance

$L$ is

$V_{L}$ and voltage across capacitance

$C$ is

$V_{C}$ . The voltage

$V_{R}$ and current

$i$ are in the same phase, the voltage

$V$ will lead the current by angle

$90^{\circ}$ while the voltage

$V_{C}$ will lag behind the current by angle

$90^{\circ}$ (fig. b). Clearly

$V_{C}$ and

$V_{L}$ are in opposite directions, therefore their resultant potential difference =

$V_{C} — V_{L}$ (if

$V_{C}$ >

$V_{L}$ )

Thus

$V_{R}$ and (

$V_{C} — V_{L}$ ) are mutually perpendicular and the phase difference between them is

$90^{\circ}$ . As applied voltage across the circuit is

$V$ , the resultant of

$V_{R}$ and (

$V_{C}$ -

$V_{L}$ ) will also be

$V$ .

From figure.

$V^{2} = V_{R}^{2}+(V_{C}-V_{L})^{2} \implies V = \sqrt{V_{R}^{2}+(V_{C}-V_{L})^{2}}$ ...(i)

But,

$V_{R} = Ri$ ,

$V_{C} = X_{C}i$ ,

$V_{L}=X_{L}i$ ...(ii)

where,

$X_{C} = \dfrac{1}{\omega C} = \text{Capacitance reactance}$ and

$X_{L} = \omega L =\text{inductive reactance}$

$V =\sqrt{(Ri)^{2}+(X_{C}i-X_{L}i)^{2}}$

Impedance of circuit,

$Z = \dfrac{V}{i} = \sqrt{R^{2}+(X_{C}-X_L)^{2}}$

i.e.,

$Z = \dfrac{V}{i} = \sqrt{R^{2}+(X_{C}-X_L)^{2}}=\sqrt{R^{2}+(\dfrac{1}{\omega C}-\omega L)^{2}}$

1678869_1784343_ans_fcdb12d8acff4ba2a04e694b4c7a9010.png

When all the three elements are connected in series across the same source, determine the impedance of the circuit.

Let

$I_{eff}$ be the current that flows through each element

$X$ ,

$Y$ , and

$Z$ . The voltage

$V_{x}$ ,

$V_{y}$ and

$V_{z}$ drop across the elements. However,

$V_{x}+V_{y}+V_{z} > E_{eff}$

It is not possible. So, on plotting phasor diagram, we have

$|E_{eff}|^{2} = V_{R}^{2} + (V_{L}-V_{C})^{2}$ ... (1)

$E_{eff} = I_{eff}\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$

Since,

$E_{eff} = I_{eff}Z$ ... (2)

The impedance of the given circuit can be given by,

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$

1678892_1794492_ans_b63b684b7dc94cc6884d3fa3dfe23ecc.png

The inductance of a coil is $0.1H$ . It is connected to an alternating current of $50Hz$ frequency. Find out the reactance.

1883421_1836445_ans_7d84258091434b52a70f299c652811ec.jpg

Answer in brief:
For a very high frequency AC supply, a capacitor behaves like a pure conductor.Why?

Resistance by the capacitor is inversely proportional to the frequency of the electric signal.

Whenever any component offers zero resistance to an electric signal it behaves like a conductor therefore at a very high-frequency AC supply, the capacitor behaves like a pure conductor.

At which time, sinusoidal alternating current will be
(i) $\cfrac{1}{2}$ time
(ii) $\cfrac { \sqrt { 3 } }{ 2 }$ time of its peak value

The instantaneous value of current is

$I={ I }_{ m }\sin { \omega t } ....(1)\quad$
(I) When

$I=\cfrac { { I }_{ m } }{ 2 }$
From equation (1)

$\cfrac { { I }_{ m } }{ 2 } ={ I }_{ m }\sin { \omega t } \Rightarrow \cfrac { 1 }{ 2 } =\sin { \omega t } \Rightarrow \omega t=\cfrac { \pi }{ 6 } \quad \quad \cfrac { 2\pi }{ T } t=\cfrac { \pi }{ 6 } \Rightarrow t=\cfrac { T }{ 12 }$
(ii) When

$\quad I=\cfrac { \sqrt { 3 } }{ 2 } { I }_{ m }$
From equation (1)

$\cfrac { \sqrt { 3 } }{ 2 } { I }_{ m }={ I }_{ m }\sin { \omega t } \Rightarrow \omega t=\cfrac { \pi }{ 3 } \quad \quad$

$\cfrac { 2\pi t }{ T } =\cfrac { \pi }{ 3 } \Rightarrow t=\cfrac { T }{ 6 }$

The inductance of the coil is $1H$
(i) At which frequency will its reactance be $3140\Omega$
(ii) What will be the capacity of the capacitor so that its reactance remain same at the same frequency?

Inductance

$(L)=1H$
Inductive reactance

$\left( { X }_{ L }=2\pi fL \right)$

$3140=2\times 3.14\times f\times 1\Rightarrow f=\cfrac { 3140 }{ 2\times 314 } =500Hz$
(ii) Capacitive reactance

${ X }_{ C }=3140\Omega$

${ X }_{ C }=\cfrac { 1 }{ \omega C } =\cfrac { 1 }{ 2\pi fC } \Rightarrow C=\cfrac { 1 }{ 2\pi f{ X }_{ C } } =\cfrac { 1 }{ 2\times 3.14\times 500\times 3140 } =0.1014\times { 10 }^{ -6 }F=0.1014\mu F$

An alternating source of ${V}_{rms}=120V,f=60Hz$ is connected to an series circuit containing $L=200mH$ , $C=40\mu F$ , $R=20\Omega$ . Determine
(i) total reactance
(ii) impedance
(iii) power factor
(iv) average power

Voltage

$({V}_{rms})=120V$
and frequency

$(f)=60Hz$
Inductance

$(L)=200mH=200\times { 10 }^{ -3 }H$
Capacitance

$(C)=40\mu F=40\times { 10 }^{ -6 }F$
Resistance

$(R)=20\Omega$
Inductive reactance

$\left( { X }_{ L } \right) =\omega L=2\pi fL=2\times 3.14\times 60\times 200\times { 10 }^{ -3 }=75.36\Omega$
Capacitive reactance

$\left( { X }_{ C } \right) =\cfrac { 1 }{ \omega C } =\cfrac { 1 }{ 2\pi fC } =\cfrac { 1 }{ 2\times 3.14\times 60\times 40\times { 10 }^{ -6 } } =\cfrac { { 10 }^{ 6 } }{ 2\times 3.14\times 2400 } =66.34\Omega \quad$
(i) Resultant reactance

$=\left( { X }_{ L }-{ X }_{ C } \right) =75.36-66.34=9.02=9\Omega$
(ii) The impedance of circuit

$(Z)=\sqrt { { \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 }+{ R }^{ 2 } } =\sqrt { { 9 }^{ 2 }+{ (20) }^{ 2 } } =\sqrt { 81+400 } =21.93\Omega$
(iii) Power factor

$\cos { \phi } =\cfrac { R }{ Z } =\cfrac { 20 }{ 21.93 } =0.9119=0.912$
(iv) Power dissipated in circuit

$(P)={ I }_{ rms }{ V }_{ rms }\cos { \phi } =\cfrac { 120 }{ 21.93 } \times 120\times 0.912=598.850W$

A series LCR circuit connected through an alternating voltage $230V$ . If $L=5H,C=80\mu F, R=40\Omega$ , then find out the:
(i) resonant frequency
(ii) impedance of the circuit and peak value of current at resonant frequency
(iii) square mean root value of voltages at three components.

Alternating voltage

$(V)=230V$
Inductance

$(L)=5H$
Capacitance

$C)=80\mu F=80\times { 10 }^{ -6 }F$ s
Resistance

$(R)=40\Omega$
(i) Resonance frequency

$(f)=\cfrac { 1 }{ 2\pi \sqrt { LC } } =\cfrac { 1 }{ 2\times 3.14\sqrt { 5\times 80\times { 10 }^{ -6 } } } =\cfrac { 1000 }{ 2\times 3.14\times 20 }$

$2\pi f=\omega =\cfrac { 1000 }{ 20 } =50rad/s$
Inductive reactance

$\left( { X }_{ L } \right) =\omega L=50\times 5=250\Omega \quad$
Capacitive reactance

$\left( { X }_{ C } \right) =\cfrac { 1 }{ \omega C } =\cfrac { 1 }{ 50\times 80\times { 10 }^{ -6 } } =250\Omega$
(ii) Impedance of circuit

$(Z)=\sqrt { { \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 }+{ R }^{ 2 } } =\sqrt { { (250-250) }^{ 2 }+{ (40) }^{ 2 } } =40\Omega \quad$
Current flow in the circuit

${ I }_{ rms }=\cfrac { { V }_{ rms } }{ Z } =\cfrac { 230 }{ 40 } =5.75A\quad$
Peak value of current

${ I }_{ m }=\sqrt { 2 } { I }_{ rms }=1.414\times 5.75=8.13A\quad$
(iii) Potential drop across inductor

${ V }_{ L }={ I }_{ rms }{ X }_{ L }=5.75\times 250=1437.5V$
Potential drop across capacitor

${ V }_{ C }={ I }_{ rms }\times { X }_{ C }=5.75\times 250=1437.50V$

A consider of $120\mu F$ is connected through an ac source of frequency $50Hz$ . Calculate its capacitive reactance. If frequency changes to $5MHz$ , then what will be the effect on reactance?

Capacitive reactance

$\left( { X }_{ C } \right) =\cfrac { 1 }{ \omega C }$

$\quad \therefore { X }_{ C }=\cfrac { 1 }{ 2\pi fC }$
Given:

$f=50Hz;C=120\mu F=120\times { 10 }^{ -6 }F$

${ X }_{ C }=\cfrac { 1 }{ 2\times 3.14\times 50\times 120\times { 10 }^{ -6 } } =\cfrac { { 10 }^{ 6 } }{ 2\times 3.14\times 50\times 120 } ={ 10 }^{ 6 }\times 2.6539\times { 10 }^{ -5 }=26.54\Omega \quad \quad$
Capacitive reactance

${ X }_{ C }\propto \cfrac { 1 }{ f }$

$\therefore \cfrac { { X }_{ { C }_{ 1 } } }{ { X }_{ { C }_{ 2 } } } =\cfrac { { f }_{ 2 } }{ { f }_{ 1 } }$

$\cfrac { { X }_{ { C }_{ 1 } } }{ { X }_{ { C }_{ 2 } } } =\cfrac { 26.54 }{ { X }_{ { C }_{ 2 } } } =\cfrac { 5MHz }{ 50Hz } =\cfrac { 5\times { 10 }^{ 6 } }{ 50 } \Rightarrow \cfrac { 26.54 }{ { X }_{ { C }_{ 2 } } } ={ 10 }^{ 5 }\Rightarrow { X }_{ { C }_{ 2 } }=26.54\times { 10 }^{ -5 }=2.654\times { 10 }^{ -4 }\Omega$

The ohmic resistance of a coil is $6\Omega$ . If impedance of the coil is $10\Omega$ , then find inductive reactance $\left( { X }_{ L } \right)$ of the coil.

Ohmic resistance

$(R)=6\Omega$
Impedence of coil

$(Z)=10Q$
For inductor and resistance circuit

$L=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } \Rightarrow 10=\sqrt { { (6) }^{ 2 }+{ \left( { X }_{ L } \right) }^{ 2 } } \Rightarrow 10=36+{ X }_{ L }^{ 2 }\Rightarrow { X }_{ L }^{ 2 }=100-36=64\Rightarrow { X }_{ L }=8$

Write value of wattless current in an ac circuit.

1883441_1836460_ans_638fccf0ded2410fae3b36f3343e5c14.jpg

Anwser in brief.
What is wattles current?

A Wattless Current can be defined in the following way:
The current in an AC circuit is said to be Wattless Current when the average power consumed in such a circuit corresponds to Zero. Such current is also called as Idle Current.

An inductance, a capacitance and a resistance are in series. If $L=0.1H,C=20\mu F, R=10\Omega$ , then at which frequency, the circuit will be in resonance?

Inductance of circuit

$(L)=0.1H$
Capacitance

$(C)=20\mu F=20\times { 10 }^{ -6 }F$
Resistance

$(R)=10\Omega$
At the condition of resonance

${ X }_{ L }={ X }_{ C }\quad$

$\omega L=\cfrac { 1 }{ \omega C }$

$\therefore \omega =\cfrac { 1 }{ \sqrt { LC } }$

$2\pi f=\cfrac { 1 }{ \sqrt { LC } }$

$\therefore f=\cfrac { 1 }{ 2\pi \sqrt { LC } } =\cfrac { 1 }{ 2\times 3.14\times \sqrt { 0.1\times 20\times { 10 }^{ -6 } } } =\cfrac { 1000 }{ 2\times 3.14\times \sqrt { 2 } } =112.6Hz$

A circuit consist of a capacitor with capacitance $C$ and a coil with active resistance $R$ and inductance $L$ connected in parallel. Find the impedance of the circuit at frequency $\omega$ of alternating voltage.

From the previous problem

$z=\dfrac{R^2+\omega ^2L^2}{\sqrt{R^2+{\omega C(R^2+\omega ^2L^2)-\omega L}^2}}$

$=\dfrac{R^2+\omega ^2L^2}{\sqrt{(R^2+\omega ^2L^2)(1-2\omega ^2LC)+\omega ^2C^2(R^2+\omega ^2L^2)^2}}$

$\dfrac{\sqrt{R^2+\omega ^2L^2}}{\sqrt{(1-2\omega ^2LC)+\omega ^2C^2(R^2+\omega ^2L^2)}}=\dfrac{\sqrt{R^2+\omega ^2L^2}}{\sqrt{(1-\omega ^2LC)^2+(\omega RC)^2}}$

A cell of emf $50 V$ , with negligible resistance is connected in parallel through a switch $S$ to two branches: one is R-L branch and other is R-C branch as shown in the figure. Now in Column I, certain numerical values are given as A, B, C and D, which may correspond to one or more physical quantities, given in Column II.
$[R_{1}=25\Omega ,R_{2}=5\Omega ,L=10H,C=20\mu F]$ . Use $e^{x}\simeq 1+x,x< < 1$ .

$i_1=\dfrac{50}{25}(1-e^{-t\dfrac{25}{10}})$

$i_2=\dfrac{50}{5}(1-e^{\dfrac{-t}{10^{-4}}})$

1) at

$t\ll0.4s, i_1=2(1-e^{-2.5t})$

i.e.

$2.5t\ll1\Rightarrow i_1=2(1-1+2.5t)$

$\approx 0$

2) at

$t\gg 0.1s,$ current will stop flowing

$\therefore$ whole voltage will drop across capacitor.

$\therefore V_C=50V$

3) at

$t\ll0.15, i_2=10(1-e^{-10\dfrac{3}{t}})\approx 10A$

$V_L=L\dfrac{di}{dt}=(2\times10\times 2.5\times e^{-2.5t})=50e^{-2.5t}$

$t\ll 0.4s\Rightarrow V_L\approx 50V$

$i_1=i_2\Rightarrow 2(1-e^{-2.5t})=10(1-e^{-10^{4}+})$

$1-e^{-2.5t}=5-5e^{-10^4t}$

$4e^{-10^4t}-e^{-2.5t}=4$

A series LRC circuit is activated by a sinusoidal ac volage e = E $_{o}$ sin $\omega$ t, with varying frequency $\omega$ . In Column I, different frequency values are given and in Column II, different responses of the circuit are shown. One or more of items in Column II may match with any of the frequencies given in Column I. $[ \omega _{o}=\frac{1}{\sqrt{LC}}]$

$e=E_oSinwt$

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$W_o=\frac{1}{\sqrt{LC}}\Rightarrow X_L=X_C$
A) if

$w>w_o,$ the circuit will be inductive

$\therefore$ voltage will lead current
B)

$W=\sqrt{\frac{1}{LC}+\frac{R}{4L^2}}-\frac{R}{2L}>w_o$

$\therefore w>w_o\Rightarrow$ voltage leads current because circuit is inductive.
C)

$w<w_o \Rightarrow$ circuit is capacitive

$\therefore$ current leads the voltage
D)

$w=w_o\Rightarrow$ P are resistive impedance also,

$cos\phi =1$

$\therefore Power=Pmax$

In the circuit shown aside the switch $S$ can be moved from $a$ to $b,$ connecting either inductance $L$ or capacitance $C$ to a resistor $R$ .when the key is pressed the switch $S$ will be at either $a$ or $b$ .In column $I,$ the variation of physical quantities charge,current,voltage etc.in time are given in terms of constants $C's$ and $K's$ the values of which depends on emf $V,R,L$ and $C$ .In column $II,$ the physical quantities under different switch positions are given. the variation in column $I$ may correspond to one or more quantities of Column $II$ .

Ans

$: A\rightarrow 5, B\rightarrow 3, B \rightarrow 4, C \rightarrow 1, D \rightarrow 2$

Initially current

$=0\ A$

After switch at

$a$

$E=R i +L (di/dt)$

so, Forced response is

$i= \dfrac{E}{R}$

and Natural Response

$i =Ae^{ -\left(\dfrac{R}{L}\right)t}$

$\Rightarrow i= \dfrac{E}{R} + A e^ {\left(-\dfrac{R}{L}\right)t}$

$V_L = L \dfrac{di}{dt}= - \left(\dfrac{R}{L} \right)A e^{\left(-\dfrac{R}{L}\right)t}$

$\Rightarrow A\rightarrow 5$

Energy stored in inductor

$= \dfrac{1}{2} LI^2$

$\Rightarrow D \rightarrow 2$

Voltage across the resistor

$= i R= \dfrac{E}{R} + A e^{\left(-\dfrac{R}{L}\right)t}$

$\Rightarrow B\rightarrow 3$

charge across the inductor

$=$ integration of i w.r.t

$t.$

$\Rightarrow C \rightarrow 1$

In the shown circuit, $R_{1}=10\Omega ,L=\dfrac{\sqrt{3}}{10}H,R_{2}=20\Omega,C=\dfrac{\sqrt{3}}{2}\ milli-farad$ and $t$ is time in seconds. Then at the instant current through $R_1 is 10\sqrt2\ A$ ;
find the current through resistor $R_2$ in amperes.

Given ,

$V=200\sqrt{2}\sin(100t)\ volt$ .

thus,

$V_0=200\sqrt 2, \omega =100$

For

$R_1-L$ branch :

$X_L=\omega L=100\left(\dfrac{\sqrt 3}{10}\right)=10\sqrt 3 \Omega$ ,

$R_1=10\ \Omega$

$\therefore \tan\phi_1=\dfrac{X_L}{R_1}=\dfrac{10\sqrt 3}{10}=\sqrt 3 \Rightarrow \phi_1=60^o$

Hence the current

$I_1$ lags voltage by

$60^o$ .

For

$R_2-C$ branch

$:\ X_C=\dfrac{1}{\omega C}=\dfrac{1}{100\left(\dfrac{\sqrt 3}{2}\times 10^{-3}\right)}=\dfrac{20}{\sqrt 3}\ \omega$ and

$R_2=20\ \Omega$

$\therefore \tan\phi_2=\dfrac{X_C}{R_2}=\dfrac{1}{\sqrt 3} \Rightarrow \phi_2=30^o$

The phase difference between

$I_1$ and

$I_2$ is

$\phi=\phi_1+\phi_2=60+30=90^o$

The maximum current from source is

$I=\dfrac{V_0}{\sqrt{R_1^2+\omega^2L^2}}=\dfrac{200\sqrt 2}{10^2+(10\sqrt 3)^2}=10\sqrt{2}\ A$ .

Hence

$I=I_1+I_2$

$10\sqrt 2=10\sqrt 2+I_2 \Rightarrow I_2=0\ A$

If the the current amplitude is $\dfrac {x}{1000}A$ , find the value of $x$ .

$\displaystyle X_L=\omega L=100\Omega$

$\displaystyle X_C=\frac{1}{\omega C}=312.5\Omega$

$\displaystyle Z=\sqrt{R^2+(X_C-X_L)^2}$

$=\displaystyle \sqrt{(300)^2+(312.5-100)^2}=368\Omega$

$\displaystyle I_0=\frac{V_0}{Z}=\frac{120}{368}=0.326A$

What are the values of the elements?

$\displaystyle Z=\frac{170}{805}=20\Omega$ ...(i)

$\displaystyle \cos\phi=\frac{\sqrt 3}{2}=\frac{R}{Z}=\frac{R}{20}$

$\therefore \displaystyle r=10\sqrt 3\Omega=17.32\Omega$

$\displaystyle X_C=\sqrt{Z^2-R^2}$

$=10\Omega$

$C=\displaystyle\frac{1}{\omega X_C}$

$=\displaystyle\frac{1}{6280\times 10}F$

$=15.92\times 10^{-6}F$

Derive an expression for the circuit current.

$\displaystyle I_0=\frac{(V_0)_R}{R}=\frac{2.5}{300}=8.33\times 10^{-3}A$

$=8.33mA$
Current function and

$V_R$ function are in phase. Hence,

$I=(8.33mA)\cos[(950 rad/s)t]$

A coil of inductance 0.50 H and resistance $100 \Omega$ is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?

(a)

Peak velocity

$V_0=\sqrt 2 V=339.41$

$\omega=2\pi f=100\pi\ rad/s$

$I_o=\dfrac{V_o}{\sqrt{R^2+\omega^2 L^2}}=1.82\ A$

(b)

The voltage is given as $V=V_ocos\omega t$

The current is given as $I=I_ocos(\omega t-\phi)$

$\phi$ is the phase angle between the voltage and current.

The time lag is given as $\phi/\omega$ .

Now phase angle is given as $tan\phi=\omega L/R=2\pi\times50\times 0.5/100=1.57$ .

So $\phi=tan^{-1}1.57=57.5^o=57.5\pi/180\ radians$

time lag is $t=\phi/\omega=57.5\pi/180\times2\pi\times 50=3.19\times10^{-3}\ s$

In a series LCR circuit connected to an a.c. source of voltage $v = v_m sin \omega t$ , use phasor diagram to derive an expression for the current in the circuit. Hence, obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum.

Phasor diagram is shown for a series LCR circuit.

Using definition of reactance,

Voltage across capacitor is

$V_C = IX_C = \dfrac{I}{\omega C}$

Voltage across inductor is

$V_L = IX_L = I\omega L$

Voltage across resistor is

$V_R = IR$

From definition of impedance,

$V_S = IZ$

From the phasor diagram,

$V_S^2 = V_R^2+(V_L-V_C)^2$

$\left( \dfrac{V_m}{\sqrt 2} \right)^2 = I^2\left(R^2+(\omega L - \dfrac{1}{\omega C})^2\right)$

$I = \dfrac{V_m}{\sqrt {2(R^2+(\omega L -\frac{1}{\omega C})^2)}}$

Power dissipated in the circuit is:

$P = I^2 R$

$P = \dfrac{V_m^2 R}{2(R^2 + (\omega L - \frac{1}{\omega C})^2)}$

At resonance,

$\omega L = \dfrac{1}{\omega C}$

Denominator of power is minimum and hence, power is maximum.

Maximum power is given by:

$P_{max} = \dfrac{V_m^2}{2R}$

If a direct current of value $A$ ampere is superimposed on an alternating current $I = b \sin \omega t$ flowing through a wire, what is the effective (rms) value of the resulting current in the circuit?

Total current,

$i = a + b \sin (\omega t)$

By definition of rms value,

$\displaystyle i_{rms} = \sqrt{ \cfrac{ \int_0^T i^2\ dt }{ T } }$

$\displaystyle i_{rms} = \sqrt{ \cfrac{ \int_0^T [a^2 + b^2\sin^2(\omega t) + 2ab\sin (\omega t) ] \ dt }{ T } }$

$\displaystyle i_{rms} = \sqrt{ \cfrac{ \int_0^T [a^2 + b^2/2 (1-\sin(2\omega t)) + 2ab\sin(\omega t)]\ dt }{ T } }$

Average value of

$\sin(\omega t)$ and

$\sin(2\omega t)$ is

$0$ over a time period.

$\implies i_{rms} = \sqrt {a^2 + \cfrac{b^2}{2}}\ A$

A choke coil is needed to operate an arc lamp at $160V$ (rms) and $50Hz$ . The arc lamp has an effective resistance of $5\Omega$ when running at $10A$ (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on $160V$ DC, what additional resistance would be required. Compare the power losses in both the cases.

We know that

${ I }_{ rms }=\cfrac { { V }_{ rms } }{ \sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } } }$

$\therefore 10=\cfrac { 160 }{ \sqrt { 25+{ \left( 2\pi \times 50\times L \right) }^{ 2 } } } \quad or\quad 25+{ \left( 100\pi L \right) }^{ 2 }={ 16 }^{ 2 }$

$\Rightarrow 25+{ 10 }^{ 5 }{ L }^{ 2 }=256\Rightarrow L=\sqrt { \cfrac { 231 }{ { 10 }^{ 5 } } } =0.05H\quad \quad$
Let

${R}_{A}$ be the additioal resistance required for operating the arc lamp with

$160V$ DC source, then

$10=\sqrt { \cfrac { 160 }{ { R }_{ A }+5 } } \Rightarrow { R }_{ A }+5=16\quad \therefore { R }_{ A }=16-5=11\Omega \quad$
AC power consumed by the arc lamp

$={ V }_{ rms }{ I }_{ rms }\cos { \phi } ={ P }_{ ac }$
where

$\cos { \phi } =\cfrac { R }{ Z } =\cfrac { R }{ \sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } } } =\cfrac { 5 }{ 16 }$

${ P }_{ ac }=160\times 10\times \cfrac { 5 }{ 16 } =500W\quad \quad$
Now DC power consumed by arc lamp

${ P }_{ dc }=160\times 10=1600W$
Power loss

$=1600=500=110W=68.75$ %

For resistance R and capacitance C in series, the impedance is twice that of a parallel combination of the same elements. What is the frequency of applied emf?

${ Z }_{ s }=\sqrt { { R }^{ }+{ X }_{ C }^{ 2 } } ={ \left[ { R }^{ 2 }+{ \left( \cfrac { 1 }{ \omega C } \right) }^{ 2 } \right] }^{ 1/2 }$
In case of parallel combination

${ I }_{ R }=\left( V/R \right) \sin { \omega t } ;{ I }_{ C }=\left( V/{ X }_{ C } \right) \sin { \left( \omega t+\cfrac { \pi }{ 2 } \right) }$
or

$I={ I }_{ R }+{ I }_{ C }={ I }_{ 0 }\sin { \left( \omega t+\phi \right) }$
With

${ I }_{ 0 }\cos { \phi } =V/R;{ I }_{ 0 }\sin { \phi } =V/{ X }_{ C }$
So,

${ I }_{ 0 }={ \left[ { \left( \cfrac { V }{ R } \right) }^{ 2 }+{ \left( \cfrac { V }{ { X }_{ C } } \right) }^{ 2 } \right] }^{ 1/2 }=\cfrac { V }{ { Z }_{ p } }$
ie.,

$\cfrac { 1 }{ { Z }_{ p } } ={ \left[ { \left( \cfrac { 1 }{ { R }^{ 2 } } \right) }^{ }+{ \left( \cfrac { 1 }{ { X }_{ C } } \right) }^{ } \right] }^{ 1/2 }\Rightarrow \cfrac { 1 }{ { Z }_{ p } } =\cfrac { R }{ \sqrt { 1+{ \omega }^{ 2 }{ C }^{ 2 }{ R }^{ 2 } } }$
and as according to given problem

${ Z }_{ S }=2{ Z }_{ p }\Rightarrow { Z }_{ S }^{ 2 }=4{ Z }_{ P }^{ 2 }\quad R$

$\cfrac { { \omega }^{ 2 }{ C }^{ 2 }{ R }^{ 2 }+1 }{ { \omega }^{ 2 }{ C }^{ 2 } } =4\cfrac { { R }^{ 2 } }{ 1+{ \omega }^{ 2 }{ C }^{ 2 }{ R }^{ 2 } } \Rightarrow \omega =\cfrac { 1 }{ RC }$
or

$f=\cfrac { 1 }{ 2\pi RC }$

For the circuit shown in the figure. Find the expressions for the impedance of the circuit and phase of current:

Let ac source

$V={ V }_{ 0 }\sin { \omega t }$
ie.,

${ I }_{ R }=\cfrac { { V }_{ 0 } }{ R } \sin { \omega t } ,{ I }_{ L }=\cfrac { { V }_{ 0 } }{ { X }_{ L } } \sin { \left( \omega t-\cfrac { \pi }{ 2 } \right) } ,{ I }_{ C }=\cfrac { { V }_{ 0 } }{ { X }_{ C } } \sin { \left( \omega t+\cfrac { \pi }{ 2 } \right) } \quad$
And as all these are in parallel

$I={ I }_{ R }+{ I }_{ L }+{ I }_{ C }=\left[ \cfrac { { V }_{ 0 } }{ R } \sin { \omega t } +{ V }_{ 0 }\left( \omega C-\cfrac { 1 }{ \omega L } \right) \cos { \omega t } \right]$

$\cfrac { { V }_{ 0 } }{ R } ={ I }_{ 0 }\sin { \phi } { V }_{ 0 }\left( \omega C-\cfrac { 1 }{ \omega L } \right) ={ I }_{ 0 }\cos { \phi } \quad I={ I }_{ 0 }\sin { \left( \omega t-\phi \right) }$

$\tan { \phi } =\left( \omega C-\cfrac { 1 }{ \omega L } \right) /\cfrac { 1 }{ R } =R\left( \omega C-\cfrac { 1 }{ \omega L } \right)$

$\quad { I }_{ 0 }=\cfrac { { V }_{ 0 } }{ Z } ={ V }_{ 0 }{ \left[ \cfrac { 1 }{ { R }^{ 2 } } +{ \left( \omega C-\cfrac { 1 }{ \omega L } \right) }^{ 2 } \right] }^{ 1/2 }\quad \quad$

$Z=V_{ 0 }{ \left[ \cfrac { 1 }{ { R }^{ 2 } } +{ \left( \cfrac { 1 }{ { X }_{ C } } -\cfrac { 1 }{ { X }_{ L } } \right) }^{ 2 } \right] }^{ 1/2 }$

Obtain the relation $I = I_0 \sin ( \omega t + \pi / 2 )$ and $X_C = 1 / \omega C$ for a pure capacitor across which an altering $emf= V = V_0 \sin \omega t$ is applied. Draw a phasor diagram showing $emf V$ , current $I$ and their phase difference $\phi$

The phase difference between the current and voltage is $90^\circ$ .

It can be shown in the above phasor diagram,

997751_1033990_ans_d6007817a8444c86aec176b30c55c43a.PNG

In the given circuit, the value of resistance effect of the coil L is exactly equal to the resistance R. Bulbs B1 and B2 are exactly identical. Answer the following question.
a) Which one of the two bulbs lights up earlier, when key k is closed and why?
b) What will be the comparative brightness of the two bulbs after sometime if the key K is kept closed and why?

a) As given in the question a coil of resistance

$R$ and inductance

$L$ connected in parallel to a resistor

$R$ and in turn the combination connected to a battery. So the bulb which is connected with the arm of resistor

$R$ lights up earlier because the induced emf generated by the coil opposes the current which produced it. it takes time for bulb in the

$r$ of inductor to light up.

b) Both bulbs will h up with same brightness because an inductor does not offer resistance in the pat of

$D. C$ current. Inductor only produces induced emf at the time of make or break circuit.

A resistance of $20 \Omega$ is connected to a source of alternating current rated $110 V , 50 Hz$ Find the $rms$ current.

The root mean square value of the current is given as,

${I_{rms}} = \dfrac{{{V_{rms}}}}{R}$

$= \dfrac{{110}}{{20}}$

$= 5.5\;{\rm{A}}$

Thus, the rms value of the current is $5.5\;{\rm{A}}$ .

Calculate the impedance of the circuit.

Since $f = 50\;{\rm{Hz}}$

$\omega = 2\pi f$

$\omega = 2\pi \times 50 = 100\pi$

Now

${X_L} = \omega L = 100\pi \times \dfrac{2}{\pi } = 200\;\Omega$

${X_C} = \dfrac{1}{{\omega C}} = \dfrac{\pi }{{100\pi \times 100 \times {{10}^{ - 6}}}} = 100\;\Omega$

Impedance of the circuit is,

$Z = \sqrt {{R^2} +{{\left( {{X_L} - {X_C}} \right)}^2}}$

$= \sqrt {{{100}^2} + {{\left( {200 - 100} \right)}^2}}$

$= 100\sqrt 2$

$= 141.4\;\Omega$

An inductance of $2.0$ H,a capacitance of $18$ and a resistance of $10$ are connected to an AC source of $20$ with adjustable frequency (a) What frequency should be chosen to maximum the current (RMS) in the circuit? (b) What is the value of this maximum current (RMS)?

For current to be maximum in a circuit

${ X }_{ l }={ X }_{ C }$ (Resonant Condition)

${ W }_{ l }={ \frac { 1 }{ { W }_{ c } } }$

${ W }^{ 2 }=\frac { 1 }{ { L }_{ c } } =\frac { 1 }{ 2\ast 18\ast { 10 }^{ -6 } } =\frac { { 10 }^{ 6 } }{ 36 }$

$W=\frac { { 10 }^{ 3 } }{ 6 } =2\pi f$

$f=\frac { 1000 }{ 6\ast 2\pi } =27Hz$

$Max.current=\frac { 20 }{ 10\ast { 10 }^{ 3 } } =2mA.$

Obtain an expression for impedance of resistor, pure inductor and capacitor connected in series across alternating $e.m.f$ . State formula for phase difference.

Impedance

$(Z) = \dfrac{V_{rms}}{I_{rms}}$

Let

$I = I_p \sin (\omega t)$ ; when

$I_p =$ peak current and

$\omega = 2 \pi f$

So,

$V_R = RI_p \sin (wt)$

$\dfrac{dQ}{dt} = I = I_p \sin (wt)$

$\Rightarrow Q = - \dfrac{I_p}{w} \cos (wt)$ ; assuming no charge at beginning.

$\Rightarrow V_c = \dfrac{Q}{C} = -\dfrac{I_p}{w_c} \cos (wt)$

Now,

$\dfrac{dI}{dt} = \dfrac{d(I_p \sin wt)}{dt} = I_p w \cos w t$

$V_L = L \dfrac{dI}{dt} = L I_p w \cos wt$

Let source voltage be

$V(t)$ ,

then ,

$V(t) = V_R(t) + (V_L (t) + V_C (t))$

$= I_P \left(R \sin (wt) +\left(Lw - \dfrac{1}{wc} \right) \cos wt \right)$

$\Rightarrow I_P. Z \sin (wt = \phi ) = I_p . \sqrt{R^2 + \left(wL - \dfrac{1}{wc}\right)^2} \sin (wt = \phi);$

$\phi = \tan^{-1} \left(\dfrac{wL - \dfrac{1}{wc}}{R} \right)$

$\Rightarrow Z = \sqrt{R^2 + \left(wL - \dfrac{1}{wc}\right)^2}$ and phase difference between current and voltage of the source is

$\phi \equiv \tan^{-1} \left(\dfrac{wL - \dfrac{1}{wc}}{R} \right)$

1525125_1125053_ans_deaed6479eee4493a4e2eb7b5fade75f.png

In a Zener regulated power supply a Zener diode with $V_z=6.0V$ is used for regulation. The load current is to be $4.0mA$ and the unregulated input is $10.0V$ . What should be the value of series resistor $R_s$

The value of the $R_s$ will depend on the power rating of the Zener diode to be used.

If $\dfrac{6v}{0.5w}$ is used , the maximum safe current thru Zener will be $\dfrac{[0.5] }{6}$ = 83 mA

Then minimum value of Rs will be $\dfrac{10-6}{0.083}$ = =48 ohms .

You may use a standard resistor value 0f 51 ohm or 56 ohm.

(If you want to reduce the power dissipation in Zener diode, 120 ohms may be used which will limit the current to 33mA. (29mA thru Zener & 4mA thru Load Resistance (1.5K).

The effective resistance in between the points A and B in the circuit given is?

When the current is flowing from A to B, it passes through 14 ohm resistor then divides into to ways first through the 9 ohm resistor and the other through the 10 and 8 ohm resistors. Thus 10 and 8 ohm resistors are in series while the setup of 10 and 8 ohm resistors is parallel with the resistor 9 ohm . Therefore when we calculate the effective resistance here is 6 ohm which in turn is in series connection with the 14 ohm resistor.

Next the current divides into two ways first through the 48 ohm resistor and other through the 6 ohm resistor.

The resistor of 20 ohm is connected in parralel to the setup of 15 and 5 ohm resistors. As we calculate we get the effective resistance of this part to be 10 ohm which in turn is connected in series with 6 ohm resistor. The 48 ohm resistor is now parallel to the 16 ohm calculated effective resistance and their resulting effective resistance is equal to 12 ohm.

Lastly the 14 ohm resistor, the calculated resistance of 6ohm and the calculated resistance of 12 ohm resistance are in series. Hence the total effective resistance is equal to 32 ohm

When does LCR series circuit has minimum impedance.

2043119_1313832_ans_98672ace5270499a92bfdc57e17c7245.jpg

In a series LCR circuit with an AC source, $R=300\ \Omega,\ C=20\ \mu F,\ L=10\ henry\ \epsilon _{max}=50\ V$ and $v=50/\pi\ Hz$ . Find
(a) the $rms$ current in the circuit.
(b) the $rms$ potential differences across the capacitor, the resistor and the inductor, Note that the sum of the $rms$ potential difference across the three elements is greater than the $rms$ voltage of the source.

Given:

$R=300\Omega$ ,

$C=20\mu\ F=20\times {10}^{-6}$ F and

$L=1$ Henry,

$E=50$ V,

$V=\dfrac{50}{\pi}$ Hz.

$(a) {I}_{0}=\dfrac{{E_0}}{Z}\$

where

$Z=\sqrt{{R}^{2}+{\left({X}_{c}-{X}_{L}\right)}^{2}}$

$=\sqrt{{300}^{2}+{\left(\dfrac{1}{2\pi fC}-2\pi fL\right)}^{2}}$

$=\sqrt{{300}^{2}+{\left(\dfrac{1}{2\pi\times\dfrac{50}{\pi}\times 20\times{10}^{-6}}-2\pi\times\dfrac{50}{\pi}\times 1\right)}^{2}}$

$=\sqrt{{300}^{2}+{\left(\dfrac{{10}^{4}}{20}-100\right)}^{2}}=500$

${I}_{0}=\dfrac{{E}_{0}}{Z}=\dfrac{50}{500}=0.1$ A

$(b)$ Potential across the capacitor

$={i}_{0}\times{X}_{c}=0.1\times 500=50$ V

Potential difference across the resistor

${i}_{0}\times R=0.1\times 300=30V$

Potential difference across the inductor

${i}_{0}\times{X}_{L}=0.1\times 100=10V$

Rms.Potential

$=50$ V

Net sum of all potential drops

$=50$ V

$+30$ V

$+10$ V

$=90$ V

Sum of potential drops

$>$ R.M.S potential applied.

Assuming expression for impedance in a parallel resonant circuit, state the condition for parallel resonance. Define resonant frequency and obtain and expression for it.

Let R,C,L be the re-stance capacitance and inductance connected parallel to the source of voltage V.
Admittare

$\gamma=\gamma_R+\gamma_c+\gamma_2$

$=\dfrac{1}{R}+jcw+\dfrac{1}{jLw}$
Resonance occurs when the current and voltage are in phase. To achieve this ,the imaginary past of admittance (or importance) must be zero

$\therefore jcw+\dfrac{1}{jLw}=0,w=\dfrac{1}{\sqrt{Lc}}=2\pi f_R$ .
Resonant frequency the frequency at which the voltage and current are in phase in the circuit
We know

$w=\dfrac{1}{\sqrt{Lc}}=2\pi f_{R1}$

$\therefore =\dfrac{1}{2 \pi \sqrt{Lc}}$
1212339_1555844_ans_927cf4568c004b969ffc251406c9ac91.png

Draw the effective equivalent circuit of the circuit shown in Fig., at very high frequencies and find the effective impedance.

$\because$ We know that reactance due to inductance

$X_{L} = 2\pi vL$ . As frequency is high so

$X_{L}$ will be high or

$L$ can be considered as open circuit for high frequency of

$ac$ circuit.
In similar way,

$X_{C} = \dfrac {1}{2\pi vC}$
At high frequency,

$X_{C}$ will be low.
So reactance of capacitance can be considered negligible and capacitor can be considered as short circuited.
Here the above figure is equivalent circuit to given circuit.
Total impedance,

$Z = R_{1} + R_{3}$ .
1682699_1796591_ans_9ce7ea5410444d84ba661e7732aa80e1.png

In above figure, a generator with an adjustable frequency of oscillation is connected to resistance $R=100\Omega$ , inductances $L_{1}=1.70mH$ and $L_{2}=2.30mH$ , and capacitances $C_{1}=4.00\mu F$ , $C_{2}=2.50\mu F$ , and $C_{3}=3.50\mu F$ . (a) What is the resonant frequency of the circuit? What happens to the resonant frequency if (b) $R$ is increased, (c) $L_{1}$ is increased and (d) $C_{3}$ is removed from the circuit.

(a) Since

$L_{eq}=L_{1}+L_{2}$ and

$C_{eq}=C_{1}+C_{2}+C_{3}$ for the circuit, the resonant frequency is,

$\omega=\frac{1}{2\pi \sqrt{L_{eq}C_{eq}}}=\frac{1}{2\pi \sqrt{\left ( L_{1}+L_{2} \right )\left ( C_{1}+C_{2}+C_{3} \right )}}$ ,

$=\frac{1}{2\pi \sqrt{\left ( 1.70\times10^{-3}H+2.30\times10^{-3}H \right )\left ( 4.00\times10^{-6}F+2.50\times10^{-6}F+3.50\times10^{-6}F \right )}}=796Hz$ .

(b) The resonant frequency does not depend on

$R$ so it will not change as

$R$ increases.

(c) Since

$\omega \propto \left ( L_{1}+L_{2} \right )^{-1/2}$ , it will decrease as

$L_{1}$ increases.

(d) Since

$\omega \propto C_{eq}^{-1/2}$ and

$C_{eq}$ decreases as

$C_{3}$ is removed,

$ω$ will increase.

For a certain driven series $RLC$ circuit, the maximum generator emf is $125 V$ and the maximum current is $3.20 A$ . If the current leads the generator emf by $0.982 rad$ , what are the (a) impedance and (b) resistance of the circuit? (c) Is the circuit predominantly capacitive or inductive?

(a) The impedance is

$Z=\frac{\varepsilon _{m}}{I}=\frac{125V}{3.20A}=39.1\Omega$ .

(b) From

$V_{R}=IR=\varepsilon _{m}\cos \phi$ , we get

$R=\frac{\varepsilon _{m}\cos \phi}{I}=\frac{\left ( 125V \right )\cos \left ( 0.982rad \right )}{3.20A}=21.7\Omega$ .

(c) Since

$X_{L}-X_{C}\propto \sin \phi =\sin \left ( -0.982rad \right )$ , we conclude that

$X_{L}<X_{C}$ .

The circuit is predominantly capacitive.

A series circuit with resistorinductorcapacitor combination $R_{1},L_{1},C_{1}$ has the same resonant frequency as a second circuit with a different combination $R_{2},L_{2},C_{2}$ . You now connect the two combinations in series. Show that this new circuit has the same resonant frequency as the separate circuits.

Resonance occurs when the inductive reactance equals the capacitive reactance. Reactances of a certain type add (in series). Thus, since the resonance

$ω$ values are the same for both circuits, we have for each circuit:

$\omega L_{1}=\frac{1}{\omega C_{1}}$ ,

$\omega L_{2}=\frac{1}{\omega C_{2}}$ .

and adding these equations we find,

$\omega \left ( L_{1}+L_{2} \right )=\frac{1}{\omega}\left ( \frac{1}{C_{1}}+\frac{1}{C_{2}} \right )$

Since

$L_{eq}=L_{1}+L_{2}$ and

$C_{eq}^{-1}=\left ( C_{1}^{-1}+C_{2}^{-1} \right )$ ,

$\omega L_{eq}=\frac{1}{\omega C_{eq}}$

$\Rightarrow$ resonance in the combined circuit.

In an RC series circuit ( see Fig. 7.29 ) , the rms voltage of source is $200 \,V$ and its frequency is $50 \,Hz$ . If $R = 100 \Omega$ and $C$ = $\dfrac{100}{\pi} \mu F$ , find
(i) Impedance of the circuit ,
(ii) Power factor angle ,
(iii) Power factor ,
(iv) Current ,
(v) Maximum current ,
(vi) Voltage across R ,
(vii) Voltage across C ,
(viii) Maximum voltage across R ,
(ix) Maximum voltage across C ,
(x) $$ ,
(xi) $< P_R >$ ,
(xii) $< P_C >$
(xiii) $i(t) , V_R (t)$ and $V_C(t)$

$X_C = \dfrac{10^6}{\dfrac{100}{\pi}(2\pi 50)} = 100 \, \Omega$

i.

$Z = \sqrt{R^2 + X_{C}^{2}} = \sqrt{100^2 + (100)^2} = 100\sqrt{2} \, \Omega$

ii.

$tan \, \phi = \dfrac{X_C}{R} = 1$

$\therefore \, \phi = 45^{\circ}$

iii. Power factor =

$cos \phi = \dfrac{1}{\sqrt{2}}$

iv. Current

$I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{200}{100\sqrt{2}} = \sqrt{2} \, A$

v. Maximum current =

$I_{rms} \,\sqrt{2} = 2\,A$

vi. Voltage across R =

$V_{R , rms} = I_{rms} R = \sqrt{2} \times 100 \,V$

vii. Voltage across C =

$V_{C,rms} = I_{rms}X_C = \sqrt{2} \times 100 \,V$

viii. Max voltage across

$R = \sqrt{2}\,V_{R , rms} = 200\,V$

ix. Max voltage across

$C = \sqrt{2} \,V_{C , rms} = 200\,V$

x.

$ = V_{rms} I_{rms} \,cos \phi = 200 \times \sqrt{2} \times \dfrac{1}{\sqrt{2}} = 200\,W$

xi.

$<P_R> = I^2_{rms}R = 200\,W$

xii.

$<P_C> = 0$

xiii. Applied voltage can be written as :

$V_s(t) = V_0 \,sin \, \omega t$

$(\omega = 2 \pi f = 2 \pi 50 = 100 \pi \, rad/s )$

$= 200\sqrt{2} \, sin(100 \pi t)$

$i (t) = I_0 \, sin(100 \pi t + \phi)$

$= \dfrac{V_0}{Z} \, sin \left ( 100 \pi t + \dfrac{\pi}{4} \right ) = \dfrac{200 \sqrt{2}}{100\sqrt{2}} \, sin \left ( 100 \pi t + \dfrac{\pi}{4} \right )$

$= 2 sin \,\left ( 100 \pi t + \dfrac{\pi}{4} \right )$

$V_R(t) = V_{R_0} sin \,\left ( 100 \pi t + \dfrac{\pi}{4} \right ) = I_0R sin\,\left ( 100 \pi t + \dfrac{\pi}{4} \right )$

$= 2 \times 100 sin\,\left ( 100 \pi t + \dfrac{\pi}{4} \right ) = 200 \, sin \, \left ( 100 \pi t + \dfrac{\pi}{4} \right )$

$V_C (t) = V_{C0} sin \left ( 100 \pi t + \dfrac{\pi}{4} - \dfrac{\pi}{2} \right )$

$= I_0 X_C \, sin \,\left ( 100 \pi t - \dfrac{\pi}{4} \right ) = 200 sin \,\left ( 100 \pi t - \dfrac{\pi}{4} \right )$

A $45.0mH$ inductor has a reactance of $1.30k\Omega$ . (a) What is its operating frequency? (b) What is the capacitance of a capacitor with the same reactance at that frequency? If the frequency is doubled, what is the new reactance of (c) the inductor and (d) the capacitor?

(a) We find

$L$ from

$X_{L}=\omega L=2\pi fL$ :

$f=\frac{X_{L}}{2\pi L}=\frac{1.30\times10^{3}\Omega}{2\pi \left ( 45.0\times10^{-3}H \right )}=4.60*10^{3}Hz$ .

(b) The capacitance is found from

$X_{C}=\left ( \omega C \right )^{-1}=\left ( 2\pi fC \right )^{-1}$ :

$C=\frac{1}{2\pi fX_{C}}=\frac{1}{2\pi \left ( 4.60\times10^{3}Hz \right )\left ( 1.30\times10^{3}\Omega \right )}=2.66\times10^{-8}F$ .

(c) Noting that

$X_{L}\propto f$ and

$X_{C}\propto f^{-1}$ , we conclude that when

$f$ is doubled,

$X_{L}$ doubles and

$X_{C}$ reduces by half. Thus,

$X_{L}=2\left ( 1.30\times10^{3}\Omega \right )=2.60\times10^{3}\Omega$ .

(d)

$X_{C}=1.30\times10^{3}\Omega /2=6.50\times10^{2}\Omega$ .

The resonance frequency.

At resonance

$\omega_0L=(\omega_0C)^{-1}$ or

$\omega_0=\dfrac{1}{\sqrt{LC}}$
and

$(I_m)_{res}=\dfrac{V_m}{R}$
Now

$\dfrac{V_m}{nR}=\dfrac{V_m}{R^2+\left(\omega_1L-\dfrac{1}{\omega_1C}\right)^2}=\dfrac{V_m}{R^2+\left(\omega_2L-\dfrac{1}{\omega_2C}\right)^2}$
Then

$\omega_1L-\dfrac{1}{\omega_1C}=\sqrt{n^2-1}R$

$\omega_2L-\dfrac{1}{\omega_2C}=+\sqrt{n^2-1}R$ (assuming

$\omega_2<\omega_1$ )
or

$\omega_1-\dfrac{\omega_0^2}{\omega_1}=-\omega_1+\dfrac{\omega_0^2}{\omega_1}=-\sqrt{n^2-1}\dfrac{R}{L}$
or

$\omega_1+\omega_2=\dfrac{\omega_0^2}{\omega_1\omega_2}(\omega_1+\omega_2)\Rightarrow \omega_0=\sqrt{\omega_1\omega_2}$
and

$\omega_1-\omega_2=\sqrt{n^2-1}\dfrac{R}{L}$

$\beta =\dfrac{R}{2L}=\dfrac{\omega_2-\omega_1}{2\sqrt{n^2-1}}$
and

$Q=\sqrt{\dfrac{\omega_0^2}{4\beta}-\dfrac{1}{4}}=\sqrt{\dfrac{(n^2-1)\omega_1\omega_2}{(\omega_2-\omega_1}}$

In a series oscillating $RLC$ circuit, $R=16.0\Omega$ , $C=31.2\mu F$ , $L=9.20 mH$ and $\xi _{m}=\xi _{m}\sin \omega _{d}t$ with $\xi _{m}=45.0V$ and $\omega _{d}=3000rad/s$ . For time $t=0.442 ms$ find (a) the rate $P_{g}$ at which energy is being supplied by the generator, (b) the rate $P_{c}$ at which the energy in the capacitor is changing, (c) the rate $P_{L}$ at which the energy in the inductor is changing and (d) the rate $P_{R}$ at which energy is being dissipated in the resistor. (e) Is the sum of $P_{C}$ , $P_{L}$ and $P_{R}$ greater than, less than or equal to $P_{g}$ ?

The current in the circuit satisfies

$i\left ( t \right )=I\sin \left ( \omega _{d}t-\phi \right )$ , where,

$I=\frac{\varepsilon _{m}}{Z}=\frac{\varepsilon _{m}}{\sqrt{R^{2}+\left ( \omega _{d}L-1/\omega _{d}C \right )^{2}}}$

$=\frac{45.0V}{\sqrt{\left ( 16.0\Omega \right )^{2}+\left \{ \left ( 3000rad/s \right )\left ( 9.20mH \right )-1/\left [ \left ( 3000rad/s \right )\left ( 31.2\mu F \right ) \right ] \right \}^{2}}}=1.93A$ .
and,

$\phi=\tan ^{-1}\left ( \frac{X_{L}-X_{C}}{R} \right )=tan^{-1}\left ( \frac{\omega _{d}L-1/\omega _{d}C}{R} \right )$

$=\tan ^{-1}\left [ \frac{\left ( 3000rad/s \right )\left ( 9.20mH \right )}{16.0\Omega}-\frac{1}{\left ( 3000rad/s \right )\left ( 16.0\Omega \right )\left ( 31.2\mu F \right )} \right ]=46.5^{0}$ .

(a) The power supplied by the generator is

$P_{g}=i\left ( t \right )\varepsilon \left ( t \right )=I\sin \left ( \omega _{d}t-\phi \right )\varepsilon _{m}\sin \omega _{d}t$

$=\left ( 1.93A \right )\left ( 45.0V \right )\sin \left [ \left ( 3000rad/s \right )\left ( 0.442ms \right ) \right ]\sin \left [ \left ( 3000rad/s \right )\left ( 0.442ms \right )-46.5^{0} \right ]=41.4W$ .

(b) With,

$v_{c}\left ( t \right )=V_{c}\sin \left ( \omega _{d}t-\phi -\pi /2 \right )=-V_{c}\cos \left ( \omega _{d}t-\phi \right )$
where

$V_{c}=I/\omega _{d}C$ the rate at which the energy in the capacitor changes is

$P_{c}=\frac{d}{dt}\left ( \frac{q^{2}}{2C} \right )=i\frac{q}{C}=iv_{c}$

$=-I\sin \left ( \omega _{d}t-\phi \right )\left ( \frac{I}{\omega _{d}C} \right )\cos \left ( \omega _{d}t-\phi \right )=-\frac{I^{2}}{2\omega _{d}C}\sin \left [ 2\left ( \omega _{d}t-\phi \right ) \right ]$

$=\frac{\left ( 1.93A \right )^{2}}{2\left ( 3000rad/s \right )\left ( 31.2\times10^{-6}F \right )}\sin \left [ 2\left ( 3000rad/s \right )\left ( 0.442ms \right )-2\left ( 46.5^{0} \right ) \right ]=-17.0W$ .

(c) The rate at which the energy in the inductor changes is,

$P_{L}=\frac{d}{dt}\left ( \frac{1}{2}Li^{2} \right )=Li\frac{di}{dt}=LI\sin \left ( \omega _{d}t-\phi \right )\frac{d}{dt}\left [ I\sin \left ( \omega _{d}t-\phi \right ) \right ]=\frac{1}{2}\omega _{d}LI^{2}\sin \left [ 2\left ( \omega _{d}t-\phi \right ) \right ]$

$=\frac{1}{2}\left ( 3000rad/s \right )\left ( 1.93A \right )^{2}\left ( 9.20mH \right )\sin \left [ 2\left ( 3000rad/s \right )\left ( 0.442ms \right )-2\left ( 46.5^{0} \right ) \right ]=44.1W$ .

(d) The rate at which energy is being dissipated by the resistor is,

$P_{R}=i^{2}R=I^{2}R\sin ^{2}\left ( \omega _{d}t-\phi \right )=\left ( 1.93A \right )^{2}\left ( 16.0\Omega \right )\sin ^{2}\left [ \left ( 3000rad/s \right )\left ( 0.442ms \right )-46.5^{0} \right ]=14.4W$ .

(e) Equal.

$P_{L}+P_{R}+P_{C}=44.1W-17.0W+14.4W=41.5W=P_{g}$ .

An air conditioner connected to a $120 V$ $(rms)$ $AC$ line is equivalent to a $12.0\Omega$ resistance and a $1.30\Omega$ inductive reactance in series. Calculate (a) the impedance of the air conditioner and (b) the average rate at which energy is supplied to the appliance.

(a) Using equation

$Z=\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}$ , the impedance is,

$Z=\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}=\sqrt{\left ( 12.0\Omega \right )^{2}+\left ( 1.30\Omega -0 \right )^{2}}=12.1\Omega$ .

(b) The average rate at which energy has been supplied is,

$P_{avg}=\frac{\varepsilon _{rms}^{2}R}{Z^{2}}=\frac{\left ( 120V \right )^{2}\left ( 12.0\Omega \right )}{\left ( 12.07\Omega \right )^{2}}=1.186\times10^{3}W\approx 1.19\times10^{3}W$ .

What will be value of impedence, frequency and power factor for a resonant LCR circuit? Write the expression

1883447_1836488_ans_1a0b964c87704f72ae6049bdf84d885f.jpg

What will be the effect on inductive reactance and capacitive reactance on the increasing frequency of the alternating current?

1883418_1836444_ans_e2495940da7a49189165471bd0752866.jpg

The natural oscillation frequency.

The equations of the

$L-C$ circuit are

$L\dfrac{d}{dt}(I_1+I)=\dfrac{C_1V-\displaystyle \int {I_1dt}}{C_1}=\dfrac{C_2V-\displaystyle \int {I_2} dt}{C_2}$
Differentiating again

$L(I_1+I_2)=-\dfrac{1}{C_1}I=\dfrac{1}{C_2}I_2$
Then

$I_1=\dfrac{C_1}{C_1-C_2}I,I_2\dfrac{C_2}{C_1+C_2}I$
I=I_1+I_2$$
so

$L(C_1+C_2)I+I=0$
or I=I_0\sin(\omega_0t+\alpha)$$
where

$\omega_0^2=\dfrac{1}{L(C_1+C_2)}$
(Hence

$T=\dfrac{2\pi}{\omega_0}=0.7\ ms$ )
At

$t=0,I=0$ so

$\alpha =0$

$I=I_0\sin \omega_0t$
The peak value of the current is

$I_0$ and its related to the voltage

$V$ by the first equation

$LI=V-\displaystyle \int Idt/(C_1+C_2)$
or

$+L\omega_0I_0\cos \omega_0t=V=\dfrac{1}{C_1+C_2}\displaystyle \int _0^1I_0\sin \omega_0t\ dt$
(The

$P.D$ across the inductance is

$V$ at

$t=0$ )

$=V+\dfrac{1}{C_1+C_2}.\dfrac{I_0}{\omega_0}(\cos \omega_0t-1)$
Hence

$I_0=(C_1+C_2)\omega_0V=V\sqrt{\dfrac{C_1+C_2}{L}}=8.05$

What inductance must be connected to a $17 pF$ capacitor in an oscillator capable of generating $550 nm$ (i.e.,visible) electromagnetic waves? Comment on your answer.

$f$ is the frequency and

$\lambda$ is the wavelength of an electromagnetic wave, then

$f\lambda =c$ . The frequency is the same as the frequency of oscillation of the current in the

$LC$ circuit of the generator. That is,

$f=1/2\pi \sqrt{LC}$ , where

$C$ is the capacitance and

$L$ is the inductance. Thus,

$\frac{\lambda}{2\pi \sqrt{LC}}=c$ .

The solution for

$L$ is,

$L=\dfrac{\lambda ^{2}}{4\pi ^{2}Cc^{2}}=\dfrac{\left ( 550*10^{-9}m \right )^{2}}{4\pi ^{2}\left ( 17*10^{-2}F \right )\left ( 2.998*10^{8}m/s \right )^{2}}=5.00*10^{-21}H$ .

This is exceedingly small.

An inductor $L$ of reactance $X_{L}$ is connected in series with a bulb $B$ to an ac source as shown in figure. Explain briefly how does the brightness of the bulb change when a capacitor of reactance $X_{C} = X_{L}$ is included in the circuit.

Brightness of the bulb depends on square of the

$i_{rms}$ ( i.e

${i_{rms}}^{2}$ )

$\text{Impedance of the circuit}, Z = \sqrt{R^{2}+(\omega L)^{2}}$

and

$\text{Current in the circuit,} I=\dfrac{V}{Z}$

When capacitor of reactance

$X_{C} = X_{L}$ is introduced, the net reactance of circuit becomes zero, so impedance of circuit decreases; it becomes

$Z = R$ ; so current in circuit increases; hence brightness of bulb increases. Thus brightness of bulb in both cases Increases.

What will be impedance in a series LCR circuit?

1883427_1836451_ans_07386ff93e1746b29bf34552876c7f48.png

A source of sinusoidal emf with constant voltage is connected in series with an oscillating circuit with quality factor $Q=100$ . At a certain frequency of the external voltage the heat power generated in the circuit reaches the maximum value. How much (in per cent) should this frequency be shifted to decrease the power generated $n=2.0$ times?

$p=\dfrac{V^2R}{R^2+(X_L-X_C)^2}$
At resonance

$X_L-X_C \Rightarrow \omega_0=\dfrac{1}{\sqrt{LC}}$
Power generated will decrease

$n$ times when

$(X_L-X_C)^2=\left(\omega L-\dfrac{1}{\omega C}\right)^2=(n-1)R^2$
or

$\omega -\dfrac{\omega _0^2}{\omega }=\pm \sqrt{n-1}\dfrac{R}{L}=\pm \sqrt{n-1}2\beta$
Thus

$\omega ^2\bar + 2\sqrt {n-1}\beta \omega -\omega _0^2=0$

$(\omega \bar +\sqrt{n-1}\beta)^2=\omega _0^2+(n-1)\beta^2$
or

$\dfrac{\omega }{\omega _0}=\sqrt{1+(n-1)\beta^2.\omega _0^2}\pm \sqrt{n-1}\beta /\omega _0$
(taking only the positive sign in the first term to ensure positive value for

$\dfrac{\omega }{\omega _0})$
Now

$Q=\dfrac{\omega }{2\beta }=\dfrac{1}{2}\sqrt{\left(\dfrac{\omega _0}{\beta}\right)^2-1}$

$\dfrac{\omega _0}{\beta}=\sqrt{1+4Q^2}$
Thus

$\dfrac{\omega }{\omega _0}=\sqrt{1+\dfrac{n-1}{(1+4Q^2)}}\pm \sqrt{n-1}/\sqrt{1+4Q^2}$
For large

$Q$

$\left|\dfrac{\omega -\omega _0}{\omega _0}\right|=\dfrac{\sqrt{en-1}}{2Q}=\dfrac{\sqrt{}en-1}{2Q}\times 100\%=0.5\%$

Conceptual Questions(7)
The open switch in Figure is thrown closed at $t =0$ . Before the switch is closed, the capacitor is uncharged and all currents are zero. Determine the currents in $L, \,C$ , and $R$ , the emf across $L$ , and the potential differences across $C$ and $R$
(a) at the instant after the switch is closed and
(b) long after it is closed.

1822460_1875652_ans_5b569f7d230c4835829d58715e8f06f6.jpg

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance $R$ is equal to the inductive reactance. If the plate separation of the parallel-plate capacitor is reduced to one-half its original value, the current in the circuit doubles. Find the initial capacitive reactance in terms of $R$ .

1908505_1879033_ans_8834b02d06964c27997ac0902a27343c.jpeg

What mean power should be fed to an oscillating circuit with active resistance $R=0.44\Omega$ to maintain undamped harmonic oscillations with current amplitude $I_m=30mA$ ?

Energy is lost across the resistance and the mean power lass is

$=R<I^2>=\dfrac{1}{2}RI_m^2=0.2\ mW$ .
This power should be fed to the circuit to maintain undamped oscillations.

Oscillations in an LC Circuit(53)
The switch in Figure is connected to position a for a long time interval. At $t = 0$ , the switch is thrown to position "b". After this time, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at $t = 3.00 \,s$ ?

1979965_1877180_ans_16efce624c864687ad5a3e365480abef.jpg

The RLC Circuit(58)
Consider an LC circuit in which $L = 500 \,mH$ and $C = 0.100 \mu F$ . (a) What is the resonance frequency $\omega_0$ ?(b) If a resistance of $1.00 k\Omega$ is introduced into this circuit, what is the frequency of the damped oscillations? (c) By what percentage does the frequency of the damped oscillations differ from the resonance frequency?

1981204_1877187_ans_01490263b3c442658b283273276b6bec.jpg

An oscillating circuit consist of capacitance $C=10\mu F$ , inductance $L=25mH$ , and active resistance $R=1.0\Omega$ . How many oscillation frequency period does it take for the current amplitude to decrease $\theta -$ fold?

Current decreases

$e$ fold in time

$t=\dfrac{1}{\beta }=\dfrac{2L}{R}sec=\dfrac{2L}{RT}$ oscillations

$\dfrac{2L}{R}\dfrac{\omega}{2\pi}$

$=\dfrac{L}{\pi R}\sqrt{\dfrac{1}{LC}-\dfrac{R^2}{4L^2}}=\dfrac{1}{2\pi}\sqrt{\dfrac{4L}{R^2C}-1}=15.9$ oscillations

A loop Fig is formed by two parallel conductors connected by a solenoid with inductance $L$ and a conducting rod of mass $m$ which can freely (without friction) slide over the conductors. The conductors are located in a horizontal plane in a uniform vertical magnetic field with induction $B$ . The distance between the conductors is equal to $l$ . At the moment $t=0$ the rod imparted an initial velocity $\nu_0$ directed to the right. Find the law of its motion $x(t)$ if the electric resistance of the loop is negligible.

We have in the circuit at a certain instant of time

$(t)$ , from Faraday's law of electromagnetic induction:

$L \dfrac{di}{dt}= Bl \dfrac{dx}{dt}$ or

$L di=B l dx$
As at

$t=0,\ xc=0,$ so

$L i=B l x$ or

$i=\dfrac{B l}{L}x (1)$
For the rod from the second law of motion

$F_x=m w_x$

$- i l B= m \ddot x$
Using Eqn.

$(1)$ , we get :

$\ddot x=- \left( \dfrac{l^2 B62}{mL}\right)x=- -\omega^2_0 x ...(2)$
Where
The solution of the above differential equation is of the from

$x= \sin (\omega_0 t+ \alpha)$
From the initial cosition art

$t=0, x=0$ so

$\alpha=0$
Hence,

$x=a \sin \omega_0 t \ \ (3)$
Differentiating w.r.t. time ,

$\dot x= a \omega_0 \cos\omega_0 t$
But from the initial condition of the problem at

$t=0, \ \dot x=v_0$
Thus

$v_0-a \omega_0$ or

$a=v_0/ \omega_0\ \ \ (4)$
Putting the value of

$a$ from Eqsn

$(4)$ into Eqn

$(3)$ , we obtained

$x=\dfrac{v_0}{\omega_0} \sin \omega_) t \left( where\ \omega_0=\dfrac{lB}{\sqrt{mL}}\right)$

An oscillating circuit consist of a capacitor of capacitance $C$ and a solenoid with inductance $L_1$ . The solenoid is inductively connected with a short-circuited coil having an inductance $L_2$ and a negligible active resistance. Their mutual inductance coefficient is equal to $L_{12}$ . find the natural frequency of the given oscillating circuit.

$L_1\dfrac{dI_1}{dt}+\dfrac{\displaystyle \int I_1dt}{C}=-L_{12}\dfrac{dI_2}{dt}$

$L_2\dfrac{dI_2}{dt}=-L_{12}\dfrac{dI_2}{dt}$
from the second equation

$L_2I_2=-L_{12}I_1$
Then

$\left(L_1-\dfrac{L_{12}^2}{L_2}\right)\ddot {I}+\dfrac{I_1}{C}=0$
Thus the current oscillates with frequency

$\omega =\dfrac{1}{\sqrt{\left(L_1-\dfrac{L_{12}^2}{L_2}\right)}}$

1787618_1853341_ans_2882d9bffe4142edb32b0de24540a704.png

Additional Problems(64)
Show that the rms value for the sawtooth voltage shown in Figure is $\Delta V_{max}/\sqrt{3}$ .

1914338_1879956_ans_145c9c1262664b6bb92e257883a02268.jpeg

Additional Problems(68)
A series RLC circuit has resonance angular frequency $2.00 \times 10^3 \,rad/s$ . When it is operating at some input frequency, $X_L = 12.0 \Omega$ and $X_C = 8.00 \Omega$ . (a) Is this input frequency higher than, lower than, or the same as the resonance frequency? Explain how you can tell.(b) Explain whether it is possible to determine the values of both L and C. (c) If it is possible, find L and C.If it is not possible, give a compact expression for the condition that L and C must satisfy.

1914342_1880009_ans_d5885e31b28147c1a2c9296e0113b157.jpeg

Additional Problems(63)
A $400-\Omega$ resistor, an inductor, and a capacitor are in series with an AC source. The reactance of the inductor is $700 \,\Omega$ , and the circuit impedance is $760 \,\Omega$ . (a) What are the possible values of the reactance of the capacitor? (b) If you find that the power delivered to the circuit decreases as you raise the frequency, what is the capacitive reactance in the original circuit? (c) Repeat part (a) assuming the resistance is $200 \,\Omega$ instead of $400 \,\Omega$ and the circuit impedance continues to be $760 \,\Omega$ .

1914334_1879952_ans_d1957c3dfe5a468c8af046ac2fc104a5.jpeg

Resonance in a Series RLC Circuit(42)
A series RLC circuit has components with the following values: $L = 20.0 \,mH, \,C = 100 \,nF, \,R = 20.0 \,\Omega$ , and $\Delta V_{max} = 100 \,V$ , with $\Delta v = \Delta V_{max} \sin \omega t$ . Find (a) the resonant frequency of the circuit, (b) the amplitude of the current at the resonant frequency, (c) the Q of the circuit, and (d) the amplitude of the voltage across the inductor at resonance.

1909895_1879362_ans_85cfade6f16e4348b0311abb2680e13b.jpeg

An AC voltage of the form $\Delta v = 90.0 \sin \,350 t$ , where $\Delta v$ is in volts and $t$ is in seconds, is applied to a series RLC circuit. If $R = 50.0 \,\Omega, \,C = 25.0 \mu F$ , and $L = 0.200 \,H$ , find (a) the impedance of the circuit, (b) the rms current in the circuit, and (c) the average power delivered to the circuit.

1908512_1879357_ans_8625b259c96b4d86aaa9d8f4ff9dff7c.jpeg

Power in an AC Circuit(41)
A diode is a device that allows current to be carried in only one direction (the direction indicated by the arrowhead in its circuit symbol). Find the average power delivered to the diode circuit of Figure in terms of $\Delta V_{rms}$ and $R$ .

1909884_1879361_ans_435865387bec4faba4ada0deb2b96922.jpeg

Alternating Current - Class 12 Medical Physics - Extra Questions

In a series LCR circuit, obtain an expression for the resonant frequency.

When a series combination of inductance and resistance are connected with a 10V.50Hz10V.50Hz a.c source, a current of 1A1A flows in the circuit.The voltage leads the current by phase angle of π3\cfrac { \pi }{ 3 } radian. Calculate the values of resistance and inductive reactance.

Calculate the impedance of a condenser in order to run a bulb rated at 10 watt and 60 volt when connected in series to an A.C.Source of 100 volt.

The current in the shown circuit is found to be 4 \sin\left(314t-\dfrac{\pi}{4}\right)A4 \sin\left(314t-\dfrac{\pi}{4}\right)A. Find the value of inductance.

An alternating voltage E=200 \,sin \,(300t) E=200 \,sin \,(300t) volt is applied across a series combination of R=10 \OmegaR=10 \Omega and inductance of 800\ mH800\ mH. Calculate the impedance of the circuit.

A coil of reactance 100 \Omega100 \Omega and resistance 100 \Omega100 \Omega is connected to a 240 V, 50 Hz a.c. supply. Find the impedance of the circuit.

Draw the graph of I_{rms} \rightarrow \omegaI_{rms} \rightarrow \omega for an A.C., L-C-R series circuit and hence explain Q-factor.

The current in the shown circuit is found to be 4\sin { \left( 314t-\cfrac { \pi }{ 4 } \right) A } 4\sin { \left( 314t-\cfrac { \pi }{ 4 } \right) A } . Find the value of inductance.

A 15.0\ \mu\ F 15.0\ \mu\ F capacitor is connected to a 220\ V, 50\ Hz220\ V, 50\ Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled , what happens to the capacitive reactance and the current ?

A capacitor of capacitance 0.5\ \mu F0.5\ \mu F is connected to a source of alternating electromagnetic force of frequency 100\ Hz100\ Hz. What is the capacitive reactance? (\pi = 3.142\pi = 3.142)

An AC rms voltage of 2V2V having a frequency of 50\ KHz50\ KHz is applied to a condenser of the capacity of 10\mu F10\mu F. The maximum value of the magnetic field between the plates of the condenser if the radius of a plate is 10 cm10 cm is _____.

Calculate Impedance of the circuit at the given frequency.

Calculate inductive reactance of coil in the following figure.

Using the expressions for charge and current for L-C oscillator, explain L-C oscillations.

In the circuit diagram shown, X_C=100 \Omega,X_L=200 \omegaX_C=100 \Omega,X_L=200 \omega and R=100 \omegaR=100 \omega. Find the effective current through the source.

What do you mean by resonance in LCRLCR series circuit? Write the formula for resonant frequency.

An alternating voltage E = 200\sin 300tE = 200\sin 300t is applied across a series combination of R = 10\ ohmR = 10\ ohm and an inductor of 800\ mH800\ mH calculate.(i) impedance of the circuit.(ii) Peak value of current in the circuit.

What is wattles current?

A series LCRLCR circuit has an inductance of 100\ mH100\ mH, a capacitor 0.1\mu F0.1\mu F and a resistance of 200\Omega200\Omega. Find the impedance at resonance and the resonant frequency.

An inductor 2/\pi 2/\pi F and a resistor 100/\pi 100/\pi are connected in series across a source of emf v=1010sin 100100 [\pi\pi t]. Here t is in second. (a) find the impedance of the circuit find the energy dissipated in the circuit in 2020 minutes.

A voltage V = 200 sin 377 t volt is applied across an inductance L having a resistance of 1.0 \Omega 1.0 \Omega The maximum current is found to be 10 A. Find the value of L.

Calculate the total reactance if two inductor of 10mH and 50 mH are connected in series with 10 kHz AC.

A sinusodial voltage of peak value 283 V and frequency 50 Hz i applied across a LCR circuit in which R = 3 \Omega R = 3 \Omega L = 25.48\space mHL = 25.48\space mH C = 796 \space \mu F C = 796 \space \mu F. Find the impedance of the circuit.

Explain the production of the electrical oscillation in LC circuit. Under what conditions are the oscillation produce in the circuit.

A 20\Omega20\Omega resistor, 1.51.5H inductor and 35\mu F35\mu F capacitors are connected in series with 200200V, 5050Hz ac supply. Calculate the impedance of the circuit and also find the current through the circuit?

If an LC circuit has C= 1\mu F C= 1\mu F & L = 0.25 L = 0.25 henry. neglecting any resistance in the circuit, what is the frequency of oscillation ?

The figure shows the graphical variation of the reactance of a capacitor with frequency of ac source. Find the capacitance of the capacitor.

An ac source of emf V = V_0 \, \sin \omega tV = V_0 \, \sin \omega t is connected to a capacitor of capacitance C. Deduce the expression for the current (I) flowing in it. Plot the graph of (i) V vs \omega t\omega t, and (ii) I vs. \Omega t\Omega t.

Show that in an ACAC circuit containing a pure inductor, the voltage is ahead of current by \pi /2\pi /2 in phase

Answer in brief:What is the natural frequency of the LC circuit? What is the reactance of this circuit at this frequency?

Alternating emfemf of e = 220\sin 100 \pi te = 220\sin 100 \pi t is applied to a circuit containing an inductance of (1/\pi)henry(1/\pi)henry. Write an equation for instantaneous current through the circuit. What will be the reading of the ACAC galvanometer connected in the circuit?

How soon does the current amplitude in an oscillating circuit with quality factor Q=5000Q=5000 decrease \eta =2.0\eta =2.0 times if the oscillation frequency is \nu=2.2MHz\nu=2.2MHz?

Conceptual Questions(a) Why does a capacitor act as a short circuit at high frequencies?(b) Why does a capacitor act as an open circuit at low frequencies?

Box may have any series combination of L, C and R. Column-I represents source current and column-II represents possible statements.

If a 0.03H0.03H inductor, a 10\Omega10\Omega resistor and a 2\mu F2\mu F capacitor are connected in series. it is observed that they resonate at frequency of 325 \times\times s Hz. Find the value of s.

An electric lamp which runs at 100V100V dc and consumes 10A10A current is connected to ac mains at 150V, 50 Hz150V, 50 Hz cycles with a choke coil in series. The inductance of voltage across the choke is given as x\times 10^{-3} Hx\times 10^{-3} H.Find xx Neglect the resistance of choke.

What is the reactance in ohms of a 2.00H2.00H inductor at a frequency of 50.0Hz50.0Hz?

Match the following two columns:

A series LCR circuit with R = 20 \Omega\Omega , L = 1.5 H and C = 35 \mu\muF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

What will be the effect on Inductive reactance X_{L}X_{L} and capacitive reactance X_{C}X_{C} if frequency of ac source is increased?

What is meant by wattless component of the current?

In the given circuit, the potential difference across the inductor LL and resistor RR are 200 V200 V and 150 V150 V respectively and the rms value of current is 5 A5 A. Calculate the impedance of the circuit.

Calculate the resonant frequency and Q-factor (Quality factor) of a series L-C-R circuit containing a pure inductor of inductance 4 H, capacitor of capacitance 27\mu F27\mu F and resistor of resistance 8\cdot 4\Omega8\cdot 4\Omega

Calculate resonant frequency and Q-factor of a series L-C-R circuit containing a pure inductor of inductance 3 H, Capacitor of capacitance 27\mu F27\mu F and resistor of resistance 7.4\Omega7.4\Omega.

Calculate Impedance of the given ac circuit.

What is the power dissipated by an ideal inductor in ac circuit? Explain.

Distinguish between the terms effective value and peak value of alternating current.

Answer the following:What is the impedance of a capacitor of capacitance C in an AC circuit using source of frequency n Hz?

An alternating current of frequency \omega =314s^{-1}\omega =314s^{-1} is fed to a circuit consisting of a capacitor of capacitance C=73\mu FC=73\mu F and an active resistance R=100\OmegaR=100\Omega connected in parallel. Find the impedance of the circuit.

If a circuit made up of a resistance 1 \Omega1 \Omega and inductance 0.01H, and alternating emf 200 V at 50 H is connected, then the phase difference between the current and the emf in the circuit is :

The current is given as \dfrac{x}{10} A\dfrac{x}{10} A. Find xx

An L-C-RL-C-R series circuit with L=0.120H, R=240\OmegaL=0.120H, R=240\Omega, and C=7.30\mu FC=7.30\mu F carries an rms current of 0.450A0.450A with a frequency of 400Hz400Hz. What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor?

Power losses in case of ac (in W) is

A charged 30 \mu\muF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

The reactance of a 2.00\mu F2.00\mu F capacitor at a frequency of 50.0Hz50.0Hz is \dfrac{x}{10}k\Omega\dfrac{x}{10}k\Omega. Find xx

The capacitance of a capacitor whose reactance is 2.00\Omega2.00\Omega at 50.0Hz50.0Hz is \dfrac{x}{10}mF\dfrac{x}{10}mF Find xx

The inductance of an inductor whose reactance is 2.00\Omega2.00\Omega at 50.0Hz50.0Hz is \dfrac{x}{100} mH\dfrac{x}{100} mH. Find xx

Define capacitor reactance. Write its S.I. units

What is the phase difference between AC e.m.f and current in the following?Pure resistor and pure inductor.

Alternating current flowing through a certain electrical device leads over the potential difference across it by 90^{\circ}90^{\circ}. State whether this device is a resistor, capacitor or an inductor

In modern days incoming frequency of radio receiver is superposed with a locally produced frequency produce intermediate frequency which is always constant. This makes tuning of the receiver very simple. This is used in superheterodyne oscillators.

A light bulb is rated at 100W100W for a 220V220V ac supply. Calculate the resistance of the bulb.

When 10V, DC is applies across a coil current through it is 2.5 A, if 10V, 50Hz A. C. is applied current reduces to 2 A. Calculate reluctance of the coil.

A 9/100\pi9/100\pi inductor and a 12\omega12\omega resistance are connected in series to a 225\space V, \space 50\space Hz225\space V, \space 50\space Hz ac source. Calculate the current (in AA) in the circuit.

If the inductance of the choke coil is x\times10^{-4}Hx\times10^{-4}H. Find xx

A coil of inductance 0.50\space H0.50\space H and resistance 100\Omega100\Omega is connected to a 240\space V,\space 50\space Hz240\space V,\space 50\space Hz ac supply. The maximum current in the coil is given as \dfrac{x}{100} A\dfrac{x}{100} A. Find xx.

A 100\space \mu F100\space \mu F capacitor in series with a 40\space \Omega40\space \Omega resistor is connected to a 110\space V, \space 60\space Hz110\space V, \space 60\space Hz supply. The maximum current in the circuit is x\times 10^{-2} Ax\times 10^{-2} A. Find xx.

Leading is \dfrac{x}{1000}H\dfrac{x}{1000}H

Match the following column

In the circuit shown in figure, match the following two columns. In column IIII quantities are given in SISI units.

A 300\Omega300\Omega resistor is connected in series with a 0.800H0.800H inductor. The voltage across the resistor as a function of time is V_R=(2.50V)\cos[(950 rad/s)t]V_R=(2.50V)\cos[(950 rad/s)t]. Determine the inductive reactance of the inductor.

Power loss (in W) in case of dc is

An L-C-RL-C-R series circuit with L=0.120H, R=240\OmegaL=0.120H, R=240\Omega, and C=7.30\mu FC=7.30\mu F carries an rms current of 0.450A0.450A with a frequency of 400Hz400Hz. What is the impedance of the circuit?

A circuit element shown in the figure as a box is having either a capacitor or an inductor. The power factor of the circuit is 0.80.8, which current lags behind the voltage. The source voltage VV,

What is meant by rms (effective) value of alternating current?

An alternative voltage given by V=140( \sin{314} t)V=140( \sin{314} t) is connected across a pure resistor of 50 \Omega50 \Omega. Find(i) The frequency of the source.(ii) The rms current through the resistor.

Adjoining figure shows a series RCLRCL circuit connected to an ac source which generates an alternating emf of frequency 50\ Hz50\ Hz. The reading of the voltmeters V_{1}V_{1} and V_{2}V_{2} are 80\ V80\ V and 60\ V60\ V respectively.Find:The current in the circuit

At resonance, what is the relation between impedance of a series LCRLCR circuit and its resistance RR?

When a series combination of inductance and resistance are connected with a $10V.50Hz$ a.c source, a current of $1A$ flows in the circuit.The voltage leads the current by phase angle of $\cfrac { \pi }{ 3 }$ radian. Calculate the values of resistance and inductive reactance.

The current in the shown circuit is found to be $4 \sin\left(314t-\dfrac{\pi}{4}\right)A$ . Find the value of inductance.

An alternating voltage $E=200 \,sin \,(300t)$ volt is applied across a series combination of $R=10 \Omega$ and inductance of $800\ mH$ . Calculate the impedance of the circuit.

A coil of reactance $100 \Omega$ and resistance $100 \Omega$ is connected to a 240 V, 50 Hz a.c. supply. Find the impedance of the circuit.

Draw the graph of $I_{rms} \rightarrow \omega$ for an A.C., L-C-R series circuit and hence explain Q-factor.

The current in the shown circuit is found to be $4\sin { \left( 314t-\cfrac { \pi }{ 4 } \right) A }$ . Find the value of inductance.

A $15.0\ \mu\ F$ capacitor is connected to a $220\ V, 50\ Hz$ source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled , what happens to the capacitive reactance and the current ?

A capacitor of capacitance $0.5\ \mu F$ is connected to a source of alternating electromagnetic force of frequency $100\ Hz$ . What is the capacitive reactance? ( $\pi = 3.142$ )

An AC rms voltage of $2V$ having a frequency of $50\ KHz$ is applied to a condenser of the capacity of $10\mu F$ . The maximum value of the magnetic field between the plates of the condenser if the radius of a plate is $10 cm$ is _____.

In the circuit diagram shown, $X_C=100 \Omega,X_L=200 \omega$ and $R=100 \omega$ . Find the effective current through the source.

What do you mean by resonance in $LCR$ series circuit? Write the formula for resonant frequency.

An alternating voltage $E = 200\sin 300t$ is applied across a series combination of $R = 10\ ohm$ and an inductor of $800\ mH$ calculate.
(i) impedance of the circuit.
(ii) Peak value of current in the circuit.

A series $LCR$ circuit has an inductance of $100\ mH$ , a capacitor $0.1\mu F$ and a resistance of $200\Omega$ . Find the impedance at resonance and the resonant frequency.

An inductor $2/\pi$ F and a resistor $100/\pi$ are connected in series across a source of emf v= $10$ sin $100$ [ $\pi$ t]. Here t is in second. (a) find the impedance of the circuit find the energy dissipated in the circuit in $20$ minutes.

A voltage V = 200 sin 377 t volt is applied across an inductance L having a resistance of $1.0 \Omega$ The maximum current is found to be 10 A. Find the value of L.

A sinusodial voltage of peak value 283 V and frequency 50 Hz i applied across a LCR circuit in which $R = 3 \Omega$ $L = 25.48\space mH$ $C = 796 \space \mu F$ . Find the impedance of the circuit.

A $20\Omega$ resistor, $1.5$ H inductor and $35\mu F$ capacitors are connected in series with $200$ V, $50$ Hz ac supply. Calculate the impedance of the circuit and also find the current through the circuit?

If an LC circuit has $C= 1\mu F$ & $L = 0.25$ henry. neglecting any resistance in the circuit, what is the frequency of oscillation ?

The figure shows the graphical variation of the reactance of a capacitor with frequency of ac source.
Find the capacitance of the capacitor.

An ac source of emf $V = V_0 \, \sin \omega t$ is connected to a capacitor of capacitance C. Deduce the expression for the current (I) flowing in it. Plot the graph of (i) V vs $\omega t$ , and (ii) I vs. $\Omega t$ .

Show that in an $AC$ circuit containing a pure inductor, the voltage is ahead of current by $\pi /2$ in phase

Answer in brief:
What is the natural frequency of the LC circuit? What is the reactance of this circuit at this frequency?

Alternating $emf$ of $e = 220\sin 100 \pi t$ is applied to a circuit containing an inductance of $(1/\pi)henry$ . Write an equation for instantaneous current through the circuit. What will be the reading of the $AC$ galvanometer connected in the circuit?

How soon does the current amplitude in an oscillating circuit with quality factor $Q=5000$ decrease $\eta =2.0$ times if the oscillation frequency is $\nu=2.2MHz$ ?

Conceptual Questions
(a) Why does a capacitor act as a short circuit at high frequencies?
(b) Why does a capacitor act as an open circuit at low frequencies?

it is observed that they resonate at frequency of 325 $\times$ s Hz. Find the value of s.

An electric lamp which runs at $100V$ dc and consumes $10A$ current is connected to ac mains at $150V, 50 Hz$ cycles with a choke coil in series. The inductance of voltage across the choke is given as $x\times 10^{-3} H$ .Find $x$ Neglect the resistance of choke.

What is the reactance in ohms of a $2.00H$ inductor at a frequency of $50.0Hz$ ?

A series LCR circuit with R = 20 $\Omega$ , L = 1.5 H and C = 35 $\mu$ F is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

What will be the effect on Inductive reactance $X_{L}$ and capacitive reactance $X_{C}$ if frequency of ac source is increased?

In the given circuit, the potential difference across the inductor $L$ and resistor $R$ are $200 V$ and $150 V$ respectively and the rms value of current is $5 A$ . Calculate the impedance of the circuit.

Calculate the resonant frequency and Q-factor (Quality factor) of a series L-C-R circuit containing a pure inductor of inductance 4 H, capacitor of capacitance $27\mu F$ and resistor of resistance $8\cdot 4\Omega$

Calculate resonant frequency and Q-factor of a series L-C-R circuit containing a pure inductor of inductance 3 H, Capacitor of capacitance $27\mu F$ and resistor of resistance $7.4\Omega$ .

Answer the following:
What is the impedance of a capacitor of capacitance C in an AC circuit using source of frequency n Hz?

An alternating current of frequency $\omega =314s^{-1}$ is fed to a circuit consisting of a capacitor of capacitance $C=73\mu F$ and an active resistance $R=100\Omega$ connected in parallel. Find the impedance of the circuit.

If a circuit made up of a resistance $1 \Omega$ and inductance 0.01H, and alternating emf 200 V at 50 H is connected, then the phase difference between the current and the emf in the circuit is :

The current is given as $\dfrac{x}{10} A$ . Find $x$

An $L-C-R$ series circuit with $L=0.120H, R=240\Omega$ , and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$ . What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor?

A charged 30 $\mu$ F capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

The reactance of a $2.00\mu F$ capacitor at a frequency of $50.0Hz$ is $\dfrac{x}{10}k\Omega$ . Find $x$

The capacitance of a capacitor whose reactance is $2.00\Omega$ at $50.0Hz$ is $\dfrac{x}{10}mF$ Find $x$

The inductance of an inductor whose reactance is $2.00\Omega$ at $50.0Hz$ is $\dfrac{x}{100} mH$ . Find $x$

What is the phase difference between AC e.m.f and current in the following?
Pure resistor and pure inductor.

Alternating current flowing through a certain electrical device leads over the potential difference across it by $90^{\circ}$ . State whether this device is a resistor, capacitor or an inductor

A light bulb is rated at $100W$ for a $220V$ ac supply. Calculate the resistance of the bulb.

Calculate the current (in $A$ ) in the circuit.

If the inductance of the choke coil is $x\times10^{-4}H$ . Find $x$

The maximum current in the coil is given as $\dfrac{x}{100} A$ . Find $x$ .

The maximum current in the circuit is $x\times 10^{-2} A$ . Find $x$ .

Leading is $\dfrac{x}{1000}H$

In the circuit shown in figure, match the following two columns. In column $II$ quantities are given in $SI$ units.

A $300\Omega$ resistor is connected in series with a $0.800H$ inductor. The voltage across the resistor as a function of time is $V_R=(2.50V)\cos[(950 rad/s)t]$ . Determine the inductive reactance of the inductor.

An $L-C-R$ series circuit with $L=0.120H, R=240\Omega$ , and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$ . What is the impedance of the circuit?

The source voltage $V$ ,

An alternative voltage given by $V=140( \sin{314} t)$ is connected across a pure resistor of $50 \Omega$ . Find
(i) The frequency of the source.
(ii) The rms current through the resistor.

Adjoining figure shows a series $RCL$ circuit connected to an ac source which generates an alternating emf of frequency $50\ Hz$ . The reading of the voltmeters $V_{1}$ and $V_{2}$ are $80\ V$ and $60\ V$ respectively.
Find:
The current in the circuit

At resonance, what is the relation between impedance of a series $LCR$ circuit and its resistance $R$ ?

Adjoining figure shows a series $RCL$ circuit connected to an ac source which generates an alternating emf of frequency $50\ Hz$ . The reading of the voltmeters $V_{1}$ and $V_{2}$ are $80\ V$ and $60\ V$ respectively.
Find:
The capacitance $C$ of the capacitor

In an alternating circuit applied voltage is $220V$ . If $R=8\Omega$ , ${X}_{L}={X}_{C}=6\Omega$ then write the values of the following:
(a) Root mean square value of voltage.'
(b) Impedance of circuit.

Arrive at the expression for the impedance of a series $LCR$ circuit using phasor diagram method and hence write the expression for the current through the circuit.

A $20$ volt $5$ watt lamp is used on a.c mains of $200$ volts $50$ c.p.s. Calculate the value of
(a) capacitance (b) inductance to be put in series to run the lamp
(c) how much pure resistance should be included in place of the above device so that lamp can run on its voltage?

A $220\ volt$ input is supplied to a transformer. The output circuit of $2.0\ ampere$ at $440\ volts$ . If the efficiency of the transformer is $80 \%$ , the current drawn by the primary windings of the transformer is

A capacitor of $20 \mu F$ is connected in series with a $25 \Omega$ resistance to a peak e.m.f $240v$ , $50Hz$ A.c. Calculate
(i)the capacitive reactance of the circuit.
(ii)an impedance of the circuit

In the circuit shown, initially the switch is in position 1 for a long . Then the switch is shifted to position 2 for a long time. Find the total heat produced in $R_2$ .

The peak voltage of an ac supply is $300V$ . What is the rms voltage?

When a coil is connected to a $100\ V\ dc$ supply. The current is $2\ A$ . When the same coil is connected to a.c. source $E=100\sqrt { 2 } \sin { \omega t }$ , the current is $1\ A$ . Find the inductive reactance used :

When a coil is connected to a $100$ $V$ DC supply, the current is $2$ $A$ . When the same coil is connected to AC source $E=100\sqrt{2}\sin\omega t$ , the current is $1$ $A$ . Find the inductive reactance used in the circuit.

The r.m.s current in an ac circuit is $1\ A$ . If the watt less current be $\sqrt {3}A$ , Then what is the power factor

Plot a graph showing variation of capacitive reactance with the change in the frequency of the $AC$ source.

In the given circuit find out :
(i) maximum value of current in the circuit,
(ii) root mean square value of current in the circuit,

Consider the $LCR$ circuit shown in Fig. Find the net current $i$ and the phase of $i$ . Show that $i = \dfrac {V}{Z}$ . Find the impedance $Z$ for this circuit.

Find the time dependence of the current flowing through the inductance $L$ of the circuit shown in Fig. after the switch $Sw$ is shorted at the moment $t = 0$ .

A circuit is set up by connecting inductance L=100mH resistor R=100 $\Omega$ and a capacitor of reactance 200 $\Omega$ in series An alternating emf of 150 $\sqrt{2}V,500m^{-1}$ Hz is applied across this series combination. Calculate the power dissipated.

The figure shows the graphical variation of the reactance of a capacitor with frequency of ac source.
An ideal inductor has the same reactance at $100 Hz$ frequency as the capacitor has at the same frequency. Find the value of inductance of the inductor.

In the circuit shown in figure, the battery has negligible internal resistance. Show that the current in the circuit through the battery rises instantly to its steady state value $E/R$ when the switch is closed, provided that resistance R is $\sqrt{L/C}$

When under load and operating at an rms voltage of $220V$ , a certain electric motor draws an rms current of $3.00 A$ . It has a resistance of $24.0\Omega$ and no capacitive reactance. What is its inductive reactance?

(a) At what frequency would a $6.0 mH$ inductor and a $10\mu F$ capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same $L$ and $C$ .

An inductor $200 mH$ , a capacitor $100 \mu F$ and a resistor $10 \Omega$ are connected in series to an a.c. source of $100 V$ , having variable frequency. At what frequency of the applied voltage will the power factor of the circuit be $1$ ?

A $1.50\mu F$ capacitor has a capacitive reactance of $12.0\Omega$ . (a) What must be its operating frequency? (b) What will be the capacitive reactance if the frequency is doubled?

A generator with an adjustable frequency of oscillation is wired in series to an inductor of $L=2.50 mH$ and a capacitor of $C=3.00\mu F$ . At what frequency does the generator produce the largest possible current amplitude in the circuit?