MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 7

State whether given statement is True or False
In order to get maximum power transfer from a capacitive source, the load must have an impedance that is the complex conjugate of the source impedance.

0%
True
0%
False

An oscillating circuit contains an inductor of inductance ${ 10 }^{ -6 }$ H and two capacitor each of capacitance $5\times { 10 }^{ -6 }$ farad connected in parallel. Then the resonance frequency of the circuit is

0%
$\dfrac { { 10 }^{ 5 } }{ 2\pi }$
0%
$\dfrac { { 10 }^{ 5 } }{ \pi }$
0%
$\dfrac { { 3\times 10 }^{ 5 } }{ 2\pi }$
0%
$\dfrac { \sqrt { { 2\times 10 }^{ 5 } } }{ \pi }$

Explanation

$\omega =\dfrac { 1 }{ \sqrt { LC } } =\dfrac { 1 }{ \sqrt { { 10 }^{ -6 }\times 2\times 5\times { 10 }^{ -6 } } }$

$=\sqrt { { 10 }^{ 11 } } \approx 3\times { 10 }^{ 5 }\\$

$f=\dfrac { \omega }{ 2\pi } =\dfrac { 3\times { 10 }^{ 5 } }{ 2\pi }$

$=\dfrac { { 1.5\times 10 }^{ 5 } }{ \pi }$

$100\Omega$ resistor is connected to a

$220V,50Hz$ AC supply. Find rms value of current in the circuit and the net power consumed for a complete cycle.

0%
$2.20A,484W$
0%
$3.20A,564W$
0%
$1.60A,278W$
0%
$5.80A,646W$

Explanation

GIven:

$V_{rms}=220V$

$\therefore i_{rms}=\dfrac{V_{rms}}{R}=\dfrac{220V}{100\Omega}=2.20A$
The power consumed for a complete cycle is:

$P=(i_{rms})^2R=(2.20A)^2\times 100\Omega=484W$

When $100\ volt\ DC$ is applied across a solenoid, a current of $1.0\ amp$ flows in it. When $100\ volt\ AC$ is applied across the same coil, the current drops to $0.5\ amp.$ If the frequency of the AC source is $50\ Hz$ the impedance and inductance of the solenoid are:

0%
$200\ \Omega$ and $0.55\ H$
0%
$100\ \Omega$ and $0.86\ H$
0%
$200\ \Omega$ and $1.0\ H$
0%
$100\ \Omega$ and $0.93\ H$

Explanation

We know that

$i=\frac { V }{ R } ,\quad for\quad DC\Rightarrow R=\frac { 100 }{ 1 } =100\Omega \\ i=\frac { V }{ L\omega } ,\quad when\quad AC\quad is\quad applied\Rightarrow L\omega =\frac { V }{ i } =\frac { 100 }{ 0.5 } =200\Omega \\ L=\frac { 200 }{ 2\pi \times 50 } =0.055H$

In resonance, frequency of LC circuit is :

0%
$\dfrac { 1 }{ 2\pi } \sqrt { LC }$
0%
$\dfrac { 1 }{ 2\pi LC }$
0%
$\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { L }{ C } }$
0%
$\dfrac { 1 }{ 2\pi \sqrt { LC } }$

Explanation

$At\quad resonance\quad inductive\quad reactance\quad and\quad capacitive\quad reactance\\ are\quad equal\quad so\quad \omega L=\dfrac { 1 }{ \omega C } \Rightarrow \omega =\dfrac { 1 }{ \sqrt { LC } } \\ and\quad f=\dfrac { \omega }{ 2\pi } =\dfrac { 1 }{ 2\pi \sqrt { LC } }$

In an LCR circuit the potential difference between the terminal of the inductance is

$60\ V$ , between the terminals of the capacitor is

$30\ V$ and that between the terminals of the resistance is

$40\ V$ . The supply voltage will be equal to:

0%
$130\ V$
0%
$10\ V$
0%
$50\ V$
0%
$70\ V$

Explanation

Supply voltage of an LCR circuit

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

since inductor and capacitor potentials are out of phase with each other)

$=\sqrt{40^2+(60-30)^2}\ V$

$=50\ V$

The peak value of an alternating e.m.f. E given by $E={ E }_{ 0 } \cos\omega t$ is 10 volt and frequency is 50 Hz. At time $t = (1/600)$ sec, the instantaneous value of e.m.f. is

0%
$10$ volt
0%
$5\sqrt { 3 }$ volt
0%
$5$ volt
0%
$1$ volt

Explanation

$E={ E }_{ 0 }cos\omega t=10cos50\times 2\pi \times t$

$=E(t=\dfrac { 1 }{ 600 } s)=10cos\dfrac { 100\pi }{ 600 }$

$=10cos\dfrac { \pi }{ 6 } =5\sqrt { 3 }$ Volt

A series LCR circuit is tuned to resonance. If the angular frequency of the applied AC voltage at resonance is $\omega$ , the impedance of the circuit then is:

0%
$R+\omega L+(\frac { 1 }{ \omega C } )$
0%
R
0%
$\sqrt { { R }^{ 2 }+\omega L+{ \left( \frac { 1 }{ \omega C } \right) }^{ 2 } }$
0%
$\sqrt { { R }^{ 2 }+{ \left( \omega L-\frac { 1 }{ \omega C } \right) }^{ 2 } }$

Explanation

At resonance

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

$\therefore impedance =R$

A series

$LCR$ circuit is connected to a source of alternating emf

$50 \ V$ and if the potential differences across inductor and capacitor are

$90\ V$ and

$60\ V$ respectively, the potential difference across resistor is:

0%
$400\ V$
0%
$40\ V$
0%
$80\ V$
0%
$1600\ V$

Explanation

The potential difference across inductor and capacitor will be opposite to each other and that of the resistor will be perpendicular to them. The resultant of three must be

$50V$ .
As we know,

$V=\sqrt{{(V_L-V_C)}^2 + {V_R}^2}$
Hence,

$\sqrt { { (90-60) }^{ 2 }+{ V }_{ R }^{ 2 } } =50$

$900+{ V }_{ R }^{ 2 }=2500$

${ V }_{ R }^{ 2 }=1600$

${ V }_{ R }=40\ V$

In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the change in inductance, so that the circuit remains in resonance?

0%
4 times
0%
(1/4) times
0%
8 times
0%
2 times

Explanation

$at\quad resonance\quad inductive\quad reactance\quad and\quad capacitive\quad reactance\\ are\quad equal\quad so\quad \omega L=\frac { 1 }{ \omega C } \Rightarrow \omega =\frac { 1 }{ \sqrt { LC } } \\ so\quad if\quad the\quad capacitance\quad is\quad made\quad one-forth,\quad then\quad the\quad inductance\\ must\quad be\quad four\quad times\quad so\quad that\quad \omega \quad doesn't\quad change.$

A light bulb is rated at

$100W$ for a

$220V$ supply. Find the peak voltage of the source:

0%
$111V$
0%
$211V$
0%
$311V$
0%
$411V$

Explanation

$Answer:-$ C

The rated voltage in bulb is rms voltage.

$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } \\ { V }_{ 0 }=\sqrt { 2 } \times 220=311.08V$

In R-L-C series circuit, the potential differences across each element is

$20V$ . Now the value of the resistance alone is doubled, then P.D. across R, L and C respectively.

0%
$20V, 10V, 10V$
0%
$20V, 20V, 20V$
0%
$20V, 40V, 40V$
0%
$10V, 20V, 20V$

Explanation

Circuit is at resonance (

$VL = VC$ )

∴ circuit is purely resistance

Resistance is doubled, current in the circuit is half the initial value

∴ New current

$I'=I/2$

∴

$VR = 20 V$ (equal to applied voltage earlier)

$VL = 10V$

$VC = 10V$

A sine wave with an rms value of

$12 V$ is riding on a dc level of

$18 V$ . The maximum value of the resulting waveform is

0%
$6 V$
0%
$30$
0%
$35 V$
0%
$0 V$

Explanation

Rms value of voltage

$V_{rms} = 12$ volts

Thus peak voltage

$V_o = \sqrt{2} V_{rms}$

$\therefore$

$V_o = 1.414 \times 12 = 17$ volts

Maximum value of resulting waveform

$V_{max} = V_o + 18$

$\implies$

$V_{max} = 17+18 = 35$ volts

For a certain load, the true power is

$150 W$ and the reactive power is

$125 W$ . The apparent power is

0%
$19.52 W$
0%
$195.2 W$
0%
$275 W$
0%
$25 W$

Explanation

True power

$P = 150 W$

Reactive power

$P_r = 125 W$

Thus apparent power

$P_a = \sqrt{P^2 + P_r^2}$

$\therefore$

$P_a = \sqrt{(150)^2+ (125)^2}$

$\implies$

$P_a = 195.2 W$

A sinusoidal voltage

$V=200sin314t$ is applied to a

$10\Omega$ resistor. Find rms current.

0%
$14.14A$
0%
$28.28A$
0%
$56.56A$
0%
$100A$

Explanation

$Answer:-$ A

General sinusoidal voltage variation is given by:

$V=V_0sin(\omega t)$

Here, in this equation

$V_0$ is peak voltage.

So, on comparing both equation we get:

$V_0=200V ,\omega=314$

Relation between rms voltage and peak voltage is:

$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 200 }{ \sqrt { 2 } } =141.4V$

Relation between rms current and peak current is:

$I_{rms}=\dfrac { { V }_{ rms } }{ R } =\dfrac { 141.4 }{ 10 } =14.14A$

A sinusoidal voltage

$V=200sin314t$ is applied to a

$10\Omega$ resistor. Find rms voltage.

0%
$141.4V$
0%
$314.2V$
0%
$519.6V$
0%
$278.9V$

Explanation

$Answer:-$ A

General sinusoidal voltage variation is given by:

$V=V_0sin(\omega t)$

Here, in this equation

$V_0$ is peak voltage.

So, on comparing both equation we get:

$V_0=200V ,\omega=314$

Relation between rms voltage and peak voltage is:

$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 200 }{ \sqrt { 2 } } =141.4V$

A sinusoidal voltage

$V=200sin314t$ is applied to a

$10\Omega$ resistor. Find the frequency of the supply.

0%
$40 Hz$
0%
$50 Hz$
0%
$60 Hz$
0%
$80 Hz$

Explanation

$Answer:-$ B

General sinusoidal voltage variation is given by:

$V=V_0sin(\omega t)$

On comparing both equation we get:

$\omega =314=2\pi f\\ f=\dfrac { 314 }{ 2\pi } =50Hz$

A sinusoidal voltage

$V=200sin314t$ is applied to a

$10\Omega$ resistor. Find the peak voltage.

0%
$200V$
0%
$400V$
0%
$600V$
0%
$800V$

Explanation

$Answer:-$ A

General sinusoidal voltage variation is given by:

$V=V_0sin(\omega t)$

Here, in this equation

$V_0$ is peak voltage.

So, on comparing both equation we get:

$V_0=200V ,\omega=314$

$100\Omega$ resistor is connected to a

$220V, 50Hz$ AC supply. Find

$rms$ value of current in the circuit :

0%
$1.10A$
0%
$2.20A$
0%
$3.30A$
0%
$4.40A$

Explanation

$Answer:-$ B

The rated voltage in bulb is rms voltage so using:

$V_{rms}=I_{rms}R$

$I_{rms}=\dfrac{220}{100}=2.20A$

A light bulb is rated at

$100W$ for a

$220V$ supply. Find the rms current through the bulb:

0%
$0.25A$
0%
$0.45A$
0%
$0.65A$
0%
$0.85A$

Explanation

$Answer:-$ B

Power is given by:

$P=\dfrac { { V }^{ 2 } }{ R } \\ 100=\dfrac { { (220 })^{ 2 } }{ R } \\ R=484\Omega$

The rated voltage in bulb is rms voltage so using:

$V_{rms}=I_{rms}R$

$I_{rms}=\dfrac{220}{484}=0.45A$

A condenser of

$250 \mu F$ is connected in parallel to a coil of inductance 0.16 mH, while its effective resistance is

$20 \Omega$ . Determine the resonant frequency.

0%
$9\times 10^4 Hz$
0%
$16\times 10^7 Hz$
0%
$8\times 10^5 Hz$
0%
$9\times 10^3 Hz$

Explanation

For the given case we have standard formula for resonant frequency as:

$f\:=\dfrac{1}{2\pi}\:\sqrt{\dfrac{1}{LC}-\dfrac{R^2}{L^2}}$

$f\:=\dfrac{1}{2\pi}\:\sqrt{\dfrac{1}{0.16\times 10^{-3}\times 250\times 10^{-6}}-\dfrac{400}{({0.16\times10^{-3}})^{2}}}$

On directly putting the given values in he equation we get:

$\omega\:=\:7.98\times10^5$

If a capacitance C is connected in series with an inductor of inductance L, then the angular frequency will be

0%
$\sqrt{\dfrac{1}{LC}}$
0%
$\sqrt{\dfrac{L}{C}}$
0%
LC
0%
$\sqrt{LC}$

Explanation

The peak current of an

$RLC$ circuit is

$\frac{1}{\sqrt{R^{2}+(\frac{1}{{\omega C}}-\omega L)^{2}}}$ , where

$\omega$ is angular frequency

It varies maximum when

We can get the angular frequency by equating

$\frac{1}{\omega C}-\omega L=0$

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

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

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

In A.C. circuit in which inductance and capacitance are joined in series, current is found to be maximum when the value of inductance is

$0.5H$ and the value of capacitance is

$8\mu F$ . The angular frequency of applied alternating voltage will be

0%
400 Hz
0%
500 Hz
0%
4000 Hz
0%
5000 Hz

Explanation

Given,

Inductance and capacitance are joined in series, and inductance

$L=0.5H$ ,

And capacitance

$C= 8\mu F=8\times10^{-6}F$ .

We know that,

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

$\omega=\dfrac{1}{\sqrt{8\times 10^{-6}\times 0.5}}=\dfrac{1}{2\times 10^{-3}}=0.5\times 10^3HZ=500HZ$

The reactance of coil when used in an A.C. power supply ( 220volts, 50 cycles/sec) is 50ohms. The inductance of the coil is nearly

0%
0.16 H
0%
0.22 H
0%
2.2 H
0%
1.6 H

Explanation

Given :

$f = 50$ cycles/sec
Reactance

$X_L = 50\Omega$
Inductance of the coil

$L = \dfrac{X_L}{w} = \dfrac{50}{2\pi \ f } = \dfrac{50}{2\pi\times 50} = 0.16 \ H$

In a series RC circuit, when the frequency and the resistance are halved, the impedance

0%
doubles
0%
is halved
0%
is reduced to one-fourth
0%
cannot be determined without values

Explanation

Capacitive reactance is given by

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

$f$ is the frequency.
Impedance of the RC circuit

$Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + \dfrac{1}{4\pi^2 f^2 C^2}}$
If frequency and resistance are halved, we cannot determined unless the values are given.

A waveform has a baseline of

$3 V$ , a duty cycle of

$20$ %, and an amplitude of

$8 V$ . The average voltage value is

0%
$4 V$
0%
$4.6 V$
0%
$1.6 V$
0%
$11 V$

Explanation

Baseline voltage is

$3V$ and amplitude is

$8V$ .

As we know duty cycle

$DC=\dfrac{T_H}{T_H+T_L}$ .

In this problem it is given as

$0.20$ , thus we get

$0.20 = \dfrac{T_H}{T_H+T_L}$

$\implies T_L=4T_H$

Average value of voltage

$V_a=\dfrac{T_H\times8+T_L\times3}{T_H+T_L}=\dfrac{20}{5}=4V$

$\therefore$ Answer is A.

A series combination of resistor (

$R$ ) and capacitor (

$C$ ) is connected to an A.C. source of angular frequency '

$\omega$ '. Keeping the voltage same, if the frequency is changed to

$\omega/3$ , the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is:

0%
$\sqrt{0.6}$
0%
$\sqrt{3}$
0%
$\sqrt{2}$
0%
$\sqrt{6}$

Explanation

Let r.m.s. value of supply voltage be

$v$ .

Impedance of RC circuit is given by:

$Z = \sqrt {R^2 + X_C^2} = \sqrt {R^2 + \dfrac{1}{\omega^2 C^2}}$

Then, r.m.s. current in the circuit is:

$\displaystyle i = \cfrac{v}{Z} = \cfrac{ v } { \sqrt { R^2 + \frac {1}{\omega^2C^2} }}$

$\implies \cfrac{i_2}{i_1} = \cfrac{\sqrt {R^2 + \frac{1}{\omega^2 C^2}} } { \sqrt {R^2 + \frac{1}{(\omega/3)^2 C^2}} }$

Given :

$\dfrac{i_2}{i_1} = \dfrac{1}{2}$ and

$\omega_2 = \omega/3$

On substituting values and simplifying, we get:

$\cfrac{1}{4} = \cfrac { 1+\omega^2 R^2 C^2 }{9 + \omega^2 R^2 C^2}$

$\implies \omega RC = \sqrt{5/3}$

Ratio at capacitive reactance and resistance in original frequency is:

$\cfrac{X_C}{R} = \cfrac{1}{\omega RC} = \sqrt {3/5} = \sqrt {0.6}$

Which one of the following graphs in following figure represents variation of reactance

$'X_c'$ of a capacitor with frequency 'f' of an ac supply?

Explanation

Capacitor reactance is given by:

$X_c = \dfrac{1}{2\pi f C}, C$ is the capacitance.

$X_c\ \&\ f$ are inversely proportional.

In a series LCR circuit, resonance occurring at 105Hz. At that time, the potential difference across the 100 resistance is 40V while the potential difference across the pure inductor is 30v. The inductance L of the inductor is equal to

0%
$2.0 \times 10^{-4}$
0%
$3.0 \times 10^{-4}$
0%
$1.2 \times 10^{-4}$
0%
$2.4 \times 10^{-4}$

Explanation

$Current(i)=\dfrac{V_r}{R}=\dfrac{100}{40}=2.5A$

the inductor value is

$V_L=i*(wL)$

$L=\dfrac{30}{2.5*2*\pi*105}=1.2*10^{-4}H$

An inductor (L = 20 H), a resistor (R = 100

$\Omega$ ) and a battery (E = 10 V) are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1 ms after short circuiting.

0%
$4.5 \times 10^5 A$
0%
$3.2 \times 10^{-5} A$
0%
$9.9\times 10^{-2} A$
0%
$6.7 \times 10^{-4} A$

Explanation

Given data :

$L = 20 H$

$R = 100 \Omega$

$E = 10 V$

Time after short circuiting the circuit,

$t = 1 ms.$

The initial current,

$I'$ in the circuit is,

$I' = \dfrac{E}{R}$

$I' = \dfrac{10}{100}$

$I' = 0.1 A$ ........(1)

The time constant,

$\tau$ is,

$\tau = \dfrac{L}{R}$

$\tau = \dfrac{20}{100}$

$\tau = 0.2 s$ ........(2)

Now, the current at

$t = 1\ ms = 0.001\ s$ , is:

$I = I' e^{-\dfrac{t}{\tau}}$

Using equations (1) and (2) here, we get:

$I = (0.1)(e^{-\dfrac{0.001}{0.2}})$

$I = (0.1)(e^{-0.005})$

$I = (0.1)(0.995)$

$I = 0.0995 A$

$I = 9.95 \times 10^{-2} A$

If the value of C in a series RLC circuit is decreased, the resonant frequency

0%
is not affected
0%
increases
0%
is reduced to zero
0%
decreases

Explanation

Resonant frequency in the series RLC circuit

$\nu_r = \dfrac{1}{2\pi\sqrt{LC}}$

$\implies \ \nu_r\propto \dfrac{1}{\sqrt{C}}$
Thus resonant frequency of the circuit increases if the value of

$C$ decreases.

In a series RLC circuit that is operating above the resonant frequency, the current

0%
lags the applied voltage
0%
leads the applied voltage
0%
is in phase with the applied voltage
0%
is zero

Explanation

Capacitive reactance is given by

$X_C = \dfrac{1}{wC}$
Inductive reactance is given by

$X_L = wL$
At resonance,

$X_L = X_C$

$\implies \ wL = \dfrac{1}{wC}$
But a frequency higher than resonance frequency,

$X_L > X_C$
So the circuit behaves as a inductive circuit at a frequency higher that resonant frequency and the current lags behind the voltage in an inductive circuit.
Thus option A is correct.

$90 \Omega$ resistor, a coil with

$30 \Omega$ of reactance, and a capacitor with

$50 \Omega$ of reactance are in series across a

$12 V$ ac source. The current through the resistor is

0%
$9 mA$
0%
$90 mA$
0%
$13 mA$
0%
$130 mA$

Explanation

Resistance=

$R=90\Omega$

Reactance of coil

$=X_L=30\Omega$

Reactance of capacitor

$=X_C=50\Omega$

Impedance of circuit

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

$\implies X=\sqrt{90^2+(50-30)^2}=\sqrt{8100+400}$

Current

$=I=\dfrac{E}{X}=\dfrac{12}{\sqrt{8500}}$

$\implies I=0.13A=130mA$

Answer-(D)

A resistor of

$3 k \Omega$ , a

$0.05 \mu F$ capacitor and a

$120 mH$ coil are in series across a

$5 kHz, 20 V$ ac source.What is the impedance, expressed in polar form?

0%
$636\Omega$
0%
$3,769\Omega$
0%
$433\Omega$
0%
$4,337\Omega$

Explanation

Given,

Resistance, $R = 3 \, k\Omega = 3000 \Omega$

Capacitance, $C = 0.05 \, \mu F = 0.05 \times 10^{-6} \, F$

Inductance, $L = 120 \, mH = 120 \times 10^{-3} \, H$

$f = 5 \, kHz = 5000 \, Hz$

It is asked to find the impedance in polar form

We have ,

Impedence, $Z = \sqrt{R^2 + (X_L – X_C)^2}$

Also

Inductive Reactance, $X_L = L \omega , X_C = \dfrac{1}{C \omega}$

We have $\omega = 2 \pi F = 2 \times 3.14 \times 5000 = 31416$

$X_L = L \omega = 120 \times 10^{-3} \times 31416 = 3769.9$

Capacitive Reactance, $X_C = \dfrac{1}{C \omega} = \dfrac{1}{31416 \times 0.05 \times 10^{-6}} = 636.61$

$Z = \sqrt{3000^2 + (3769.9 – 636.61)^2} = \sqrt{9000,000 + 9819090.9} = 4337.2 \approx 4337 \Omega$

$12 \Omega$ resistor, a

$40 \mu F$ capacitor, and an

$8 mH$ coil are in series across an ac source. The resonant frequency is

0%
$28.1 Hz$
0%
$281 Hz$
0%
$2,810 Hz$
0%
$10 Hz$

Explanation

$\omega=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{8\times 10^{-3}\times 40\times 10^{-6}}}$

$\implies \omega=\dfrac{10^4}{\sqrt{32}}$

$\omega=2\pi\nu$ where

$\nu$ =frequency

frequency

$=\nu=\dfrac{\omega}{2\pi}=\dfrac{10^4}{2\times 3.143\times \sqrt{32}}$

$\implies \nu=281Hz$

Answer-(B)

$6 kHz$ sinusoidal voltage is applied to a series RC circuit. The frequency of the voltage across the resistor is

0%
$0 Hz$
0%
$12 kHz$
0%
$6 kHz$
0%
$18 kHz$

Explanation

Frequency of applied voltage and voltage or current accross any component are same. As we know,

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

$Z$ has nothing to do with frequency

$\implies$ $$I

$and$ V$$ have same frequency.

And also according to KVL

$V=V_1+V_2+V_3$

$\implies$ voltage and current across any component have same frequency i.e

$6kHz$ .

A resistor and a capacitor are in series across a

$20 V$ ac source. Circuit impedance is

$4.33 k \Omega$ . Current flow in the circuit is

0%
$9.2 mA$
0%
$92 mA$
0%
$4.6 mA$
0%
$460 mA$

Explanation

Voltage of the source

$V = 20$ volts

Circuit impedance

$Z = 4.33 k\Omega = 4.33\times 10^{3}\Omega$

Thus current in the circuit

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

$\therefore$

$I = \dfrac{20}{4.33\times 10^3} = 4.6 \times 10^{-3} A$

$\implies$

$I = 4.6 mA$

In a parallel RC circuit, there is

$100 mA$ through the resistive branch and

$100 mA$ through the capacitive branch. The total rms current is

0%
$200 mA$
0%
$100 mA$
0%
$282 mA$
0%
$141 mA$

Explanation

Current in resistive branch

$I_r = 100 mA$

Current in capacitive branch

$I_c = 100 mA$

Thus total rms current

$I_{rms} = \sqrt{I_r^2 + I_c^2}$

$\therefore$

$I_{rms} = \sqrt{100^2 + 100^2}$

$\implies$

$I_{rms} = \sqrt{2} \times 100 = 141 mA$

State whether given statement is True or False
In an ac circuit, power to the load peaks at the frequency at which the load impedance is the complex conjugate of the output impedance.

0%
True
0%
False

Explanation

According to maximum power transfer theorem. In pure DC circuit maximum power transfer to the load when load resistance is equal to internal resistance.

But in case of AC circuit consider the circuit as shown in figure

$i=\frac { E }{ Z+{ Z }_{ L } }$

Power delivered to load

${ P }_{ L }={ I }^{ 2 }X$

${ P }_{ L }=\frac { { E }^{ 2 } }{ { \left( Z+{ Z }_{ L } \right) }^{ 2 } } { X }_{ L }=\frac { { E }^{ 2 }{ X }_{ L } }{ \left[ { \left( X+{ X }_{ L } \right) ^{ 2 } }+\left( Y+{ Y }_{ L } \right) ^{ 2 } \right] }$

For maximum power transfer theorem

$Y+{ Y }_{ L }=0\\ { Y }_{ L }=-Y$

$X+{ X }_{ L }=2{ X }_{ L }\\ { X }_{ L }=X\\ Therefore\quad { Z }_{ L }=X-IY$

A resistor

$R$ , an inductor

$L$ and capacitor

$C$ are connected in series to an oscillator of frequency

$n$ . If the resonant frequency is

${ n }_{ r }$ , then the current lags behind voltage, when

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$n=0$
0%
$n < { n }_{ r }$
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$n={ n }_{ r }$
0%
$n > { n }_{ r }$

Explanation

When reactance of inductance is more than the reactance of condenser, the current lag behind the voltage.
Thus

$\omega L < \dfrac { 1 }{ \omega c }$ or

$\omega > \dfrac { 1 }{ \sqrt { LC } }$
or

$n > \dfrac { 1 }{ 2\pi \sqrt { LC } }$ or

$n > { n }_{ r }$

${ n }_{ r }=$ resonant frequency

The natural frequency of the circuit shown in adjoining figure is

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$\dfrac{1}{2\pi \sqrt{LC}}$
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$\dfrac{1}{2\pi \sqrt{2LC}}$
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$\dfrac{2}{2\pi \sqrt{LC}}$
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Zero

Explanation

In the given circuit, two capacitor and the two inductor are in series.

$\therefore =L_s=L+L=2L$
and

$\dfrac{1}{C_s}=\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{2}{C}\Rightarrow C_s=\dfrac{C}{2}$

$\therefore$ Natural frequency of the circuit

$v=\dfrac{1}{2\pi \sqrt{L_sC_s}}=\dfrac{1}{2\pi \sqrt{2L}\times \sqrt{C/2}}$

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

The rms value of a.c. voltage with peack of

$311 V$ is

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$220 V$
0%
$311 V$
0%
$180 V$
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$320 V$

Explanation

Peak value of a.c. voltage,

$V_o = 311$ volts

The rms value of voltage,

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

$\therefore$

$V_{rms} = \dfrac{311}{\sqrt{2}} = \dfrac{311}{1.414} = 220$ volts

The average half-cycle value of a sine wave with a

$40 V$ peak is

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$25.48 V$
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$6.37 V$
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$14.14 V$
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$50.96 V$

Explanation

The source voltage

$V = V_o \sin wt$

Average half-cycle value of sine wave

$V_{avg} = \dfrac{\int_o^{T/2} V_o \sin (wt) dt} {T/2}$

$V_{avg} = \dfrac{V_o} {T/2} \times \int_o^{T/2} \sin wt dt$

$V_{avg} = \dfrac{V_o} {T/2} \times \dfrac{\cos wt}{w} \bigg|^{T/2}_o$

$V_{avg} = \dfrac{V_o} {T/2} \times \dfrac{\cos\frac{wT}{2} - 1}{w}$

$V_{avg} = \dfrac{V_o} {T/2} \times \dfrac{-1 - 1}{w}$

$V_{avg} = \dfrac{V_o} {\pi } \times (-2)$

$\implies$

$|V_{avg} | = \dfrac{2V_o}{ \pi}$

We get

$|V_{avg}| = \dfrac{2\times 40}{\pi} = 25.48$ volts

The instantaneous emf and current equations of an RLC series circuit are

$e=200\sin { \left( \omega t-\dfrac { \pi }{ 6 } \right) }$

$i=20\sin { \left( \omega t+\dfrac { \pi }{ 6 } \right) }$
The average power consumed per cycle is

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Zero
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$2000 W$
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$1000 W$
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$500 W$

Explanation

Instantaneous power of the circuit,

$P=ei$

Hence the average power of the circuit

$=\dfrac{\int_0^{2\pi/\omega} Pdt}{2\pi/\omega}$

$=\dfrac{\int_0^{2\pi/\omega}4000sin(\omega t-\dfrac{\pi}{6})sin(\omega t+\dfrac{\pi}{6})}{2\pi/\omega}W$

Using:

$cosA-cosB=-2sin(\dfrac{A+B}{2})sin(\dfrac{A-B}{2})$

Average power

$=\dfrac{\int_0^{2\pi/\omega}4000\dfrac{1}{2}(cos\dfrac{\pi}{3}-cos2\omega t)dt}{2\pi/\omega}W$

$=1000\ W$

An LC resonant circuit contains a capacitor

$400 pF$ and an inductor

$100 \mu H$ . It is set into oscillations coupled to an antenna. Calculate the wavelength of the radiated electromagnetic wave.

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$377 \ mm$
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$377 \ cm$
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$377 \ m$
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$3.77 \ cm$

Explanation

Wavelength

$=\lambda = \dfrac{Velocity}{Frequency}$

Frequency of the Resonating LC circuit =

$\dfrac{1}{2\pi \times \sqrt {LC}}$

$\implies Wavelength = C\times 2 \pi \times \sqrt { LC }$

$=3\times { 10 }^{ 8 }\times 2\times 3.14\times \sqrt { 100\times { 10 }^{ -6 }\times 400\times { 10 }^{ -12 } }$

$=376.8 m$

In the adjoining circuit, if the reading of voltmeter

$V_1$ and

$V_2$ are 300 volts each, then the reading voltmeter

$V_3$ and ammeter A are respectively

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220 V, 2.2 A
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100 V, 2.0 A
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220 V, 2.0 A
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100 V, 2.2 A

Explanation

Given,

$V_1=V_2=300V;V_3=?,i=?$
As, We know,

$V=\sqrt{V^2_3+(V_1-V_2)^2}$

We now

$(V_1-V_2)^2=(300-300)^2=0$

$\therefore 220=\sqrt{V_3^2}=V_3\Rightarrow V_3=220V$

$\therefore I=\dfrac{V_3}{R}=\dfrac{220}{100}=2.2A$

So the reading of voltmeter

$V_3=220V$ and Ammeter

$A=2.2A$ .

An L-C-R circuit contains

$R=50\Omega$ ,

$L=1 mH$ and

$C=0.1\mu F$ . The impedence of the circuit will be minimum for a frequency of

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$\dfrac { { 10 }^{ 5 } }{ 2\pi } Hz$
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$\dfrac { { 10 }^{ 6 } }{ 2\pi } Hz$
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$2\pi \times { 10 }^{ 5 }Hz$
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$2\pi \times { 10 }^{ 6 }Hz$

Explanation

Impedance of L-C-R circuit will be minimum for a resonant frequency so,

${ v }_{ 0 }=\dfrac { 1 }{ 2\pi \sqrt { LC } }$

$=\dfrac { 1 }{ 2\pi \sqrt { 1\times { 10 }^{ -3 }\times 0.1\times { 10 }^{ -6 } } } =\dfrac { { 10 }^{ 5 } }{ 2\pi } Hz$

The instantaneous values of current and voltage in an AC circuit are

$i=100\sin 314 t$ amp and

$e=200\sin (314t+\dfrac{\pi}{3})V$ respectively. If the resistance is

$1\Omega$ , then the reactance of the circuit will be:

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$\sqrt{3}\Omega$
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$100\sqrt{3}\Omega$
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$-200\sqrt{3}\Omega$
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$-200/ \sqrt{3}\Omega$

Explanation

We have,

$V_0=i_0Z$

$\Rightarrow 200=100Z$

$Z=2\Omega$
So, the impedance of the circuit

$Z^2=R^2+X^2_L$

$\Rightarrow (2)^2=(1)^2+X^2_L$

$X_L=\sqrt{3}\Omega$

The value of alternating emf

$E$ in the given circuit will be

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$220 V$
0%
$140 V$
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$100 V$
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$20 V$

Explanation

Value of alternating emf

$E = \sqrt{V_R^2 + (V_c-V_L)^2}$

$\therefore$

$E= \sqrt{(80)^2 + (100 - 40)^2}$

$E = \sqrt{80^2 + 60^2}$

$\implies$

$E = \sqrt{10000} = 100V$

If L-R circuit connected to a battery of constant emf

$E$ switch

$S$ is closed at time

$t=0$ . If

$e$ denotes the induced emf across inductor and

$I$ the current in the circuit at any time

$t$ . Then which of the following graphs shows the variation of

$e$ with

$I$ ?

Explanation

$V_L$ or

$e=E-V_R=E-iR$

Thus,

$e-i$ graph is a straight line with negative slope and positive intercept.

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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