MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 5

fixed and $C$ is varied for tuning then minimum and maximum value of $C$ is

0%
C, 3C
0%
C, 6C
0%
C, 9C
0%
C, 12C

Explanation

$\displaystyle \frac{f_{max}}{f_{min}} = 3$

$\therefore \dfrac{\sqrt{LC_{max}}}{\sqrt{LC_{min}}} = 3$

$\dfrac{C_{max}}{C_{min}}= 9$

$C_{min}= C$

$\therefore C_{max}=9C$

The correct curve representing the variation of capacity reactance

$X_c$ with frequency

$f$ is

Explanation

${ X }_{ C }=\dfrac { 1 }{ C2\pi f }$

${X}_{C}$ and

$f$ are inversely proportional.

A resistance

$R$ and a capacitor

$C$ are joined to a source of

$AC$ of constant e.m.f and variable frequency. The potential difference across

$C$ is

$V$ . If the frequency of

$AC$ is gradually increased,

$V$ will

0%
increase
0%
decrease
0%
remain constant
0%
first increase and then decrease

Explanation

In complex plane,

$V = \dfrac { {V}_{AC} }{ 1+ jRC\omega }$
Therefore as

$\omega ($ frequency

$)$ increases,

$V$ decreases.

A series

$AC$ circuit has a resistance of

$4 \Omega$ and a reactance of

$3 \Omega$ . The impedance of the circuit is

0%
$5 \Omega$
0%
$7 \Omega$
0%
$12/7 \Omega$
0%
$7/12 \Omega$

Explanation

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

where,

$R = resistance$

$X = reactance$

Given

$R = 4\Omega$ and

$X = 3\Omega$

Substitute values back in equation

$Z = \sqrt{{4}^{2}+{3}^{2}}$

$Z = 5\Omega$

The natural frequency of a

$L-C$ circuit is

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

Explanation

When circuit is at natural frequency,

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

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

$AC$ source producing emf

$\displaystyle E = E_0 [cos (100 \pi s^{-1}) t] + E_0 cos [(500 \pi s^{-1})t ]$ is connected in series with a capacitor and a resistance, the steady state current in the circuit is found to be

$i = i_1 [cos (100 \pi s^{-1}) t + \varphi_1] + i_2 cos [(500 \pi s^{-1})t + \varphi_2]$

0%
$\displaystyle i_1>i_2$
0%
$\displaystyle i_1 = i_2$
0%
$\displaystyle i_1 < i_2$
0%
Insufficient information

Which of the following plots may represent the reactance of a series

$LC$ combination?

0%
(a)
0%
(b)
0%
(c)
0%
(d)

Explanation

Reactance of capacitor,

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

Reactance of inductor,

$X_L=\omega L$

Since in series combination, net reactance

$=X_C+X_L=j(\omega L-\dfrac{1}{\omega C})$

So reactance magnitude tends to infinity as frequency tends to 0 as well as infinity

In the adjacent circuit diagram, initially switch S is opened and the circuit is in steady state. At time

$t=0$ , the switch S is closed and the new steady state is reached. Choose the correct option (s)

0%
Current in the inductor when the circuit reaches the new steady state is 4A.
0%
The net change in the magnetic flux in the inductor is 1.5 Wb
0%
The potential difference across the inductor is 9 volt when the circuit reaches the new steady state
0%
The change stored in the capacitor in the new steady state is 1.2 mC

An inductor, a resistor and a capacitor are joined in series with an

$AC$ source. As the frequency of the source is slightly increased from a very low value, the reactance

0%
of the inductor increases
0%
of the resistor increases
0%
of the capacitor increases
0%
of the circuit increases

Explanation

Reactance in electriacal and electronic systems is the opposition of a circuit element to a change of electric current or voltage.
Ideally a resistor has zero reactance.

Reactance of inductor of inductance

$L,\quad X_L= L \omega$ . It increases with frequency.

Reactance of capacitor of capacitance

$C,\quad X_C =\dfrac{ 1}{C \omega }$ . It decreases with frequency.

Reactance of circuit

$= L \omega - \dfrac{1}{C \omega}$ . It follows a curve both increasing and decreasing when the frequency increases.

In an

$L-C-R$ circuit the value of

$X_L,X_C$ and

$R$ are 300

$\Omega$ , 200

$\Omega$ and 100

$\Omega$ respectively.The total impedance of the circuit will be

0%
600 $\Omega$
0%
200 $\Omega$
0%
141 $\Omega$
0%
310 $\Omega$

Explanation

Total impedance

$= R + j({X}_{L} -{X}_{C} )$

$= 100 +j100$

$=100\sqrt { 2 } \measuredangle 45$

$=141 \measuredangle 45$

The resultant reactance in an

$L-C-R$ circuit is

0%
$\displaystyle X_L+X_C$
0%
$\displaystyle X_L-X_C$
0%
$\displaystyle \sqrt{X^2_L+X^2_C}$
0%
$\displaystyle \sqrt{X^2_L-X^2_C}$

Explanation

Reactance is the nonresistive component of impedance in an AC circuit, arising from the effect of inductance or capacitance or both and causing the current to be out of phase with the electromotive force causing it.
Therefore, reactance of the

$L-C-R$ circuit is

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

The angular frequency of an AC source is 10 radian/sec. The reactance of

$1 \mu F$ capacitor will be

0%
$\displaystyle 10^4 \Omega$
0%
$\displaystyle 10^2 \Omega$
0%
$\displaystyle 10^1 \Omega$
0%
$\displaystyle 10^5 \Omega$

Explanation

Reactance

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

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

$={ 10 }^{ 5 }\Omega$

A solenoid of length 10 cm, diameter 1 cm, number of turns 500 with relative permeability of the core 2000, is connected to an ac source of frequency 50 Hz. Then, the reactance is

0%
zero
0%
55 $\Omega$
0%
105 $\Omega$
0%
155 $\Omega$

Explanation

Inductance of the solenoid

$L =\dfrac { \mu { N }^{ 2 }A }{ l } = 0.493 H$

reactance of this solenoid

$=L\omega=L 2\pi f=155\Omega$

An inductance coil of

$1$

$H$ and a condenser of capacity

$1$

$pF$ produce resonance. The resonant frequency will be

0%
$\displaystyle \dfrac{10^6}{\pi} Hz$
0%
$\displaystyle 27 \pi \times 10^6 Hz$
0%
$\displaystyle \dfrac{2\pi}{10^6} Hz$
0%
$\displaystyle \dfrac{10^6}{2\pi} Hz$

Explanation

$f$ is resonant frequency then

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

$L = 1H$

$C = 1pF = {10}^{-12}$

$f = \dfrac{1}{2\pi\sqrt{ 1 \times {10}^{-12}}}$

$f = \dfrac{{10}^{6}}{2\pi} Hz$

If the values of

$L, C$ and

$R$ in a series

$L-C-R$ circuit are 10 mH, 100

$\mu F$ and

$100 \Omega$ respectively then the value of resonant frequency will be

0%
$\displaystyle \frac{10^3}{2 \pi} Hz$
0%
$\displaystyle 2\times 10^3 Hz$
0%
$\displaystyle 2\times \frac{10^3}{pi}Hz$
0%
$\displaystyle 10^3 Hz$

Explanation

Resonant frequency

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

$=\dfrac { 1 }{ 2\pi \sqrt { { 10 }^{ -6 } } }$

$=\dfrac { { 10 }^{ 3 } }{ 2\pi } Hz$

$AC$ voltmeter in an

$L-C-R$ circuit reads 30 volt across resistance, 80 volt across inductance and 40 volt across capacitance. The value of applied voltage will be

0%
50 Volt
0%
25 Volt
0%
150 Volt
0%
70 Volt

Explanation

$V=\sqrt { { V }_{ R }^{ 2 }+{ ({ V }_{ L }^{ }-{ V }_{ C }^{ }) }^{ 2 } }$

$V=\sqrt { { 30}^{ 2 }+{ (80^{ }-40^{ }) }^{ 2 } }$

$V=50$ volt

A coil of 10 mH and 10

$\Omega$ resistance is connected in parallel to a capacitance of

$0.1 \mu F$ . The impedance of the

0%
$\displaystyle 10^2 \Omega$
0%
$\displaystyle 10^4 \Omega$
0%
$\displaystyle 10^6 \Omega$
0%
$\displaystyle 10^8 \Omega$

The reactance of a condenser of capacity 50

$\mu$ F for an

$AC$ of frequency

$2 \times 10^3$ Hertz will be

0%
5 ohm
0%
$\displaystyle \frac{2}{\pi}$
0%
$\displaystyle \frac{3}{\pi}$
0%
$\displaystyle \frac{5}{\pi}$

Explanation

Given :

$\nu = 2000$ Hz ,

$C = 50\mu F$

Capacitive reactance

$X_c = \dfrac{1}{2\pi \nu C}$

$= \dfrac{1}{2\pi (2000)\times 50\times 10^{-6}}$

$= \dfrac{5}{\pi}$

$\Omega$

In the following circuit the values of

$L,C,R$ and

$E_0$ are 0.01 H,

$\displaystyle 10^{-5} F, 25 \Omega$ and 220 volt respectively. The value of current flowing in the circuit at

$f=0$ and

$f=\infty$ will respectively be

0%
8 $A$ and 0 $A$
0%
0 $A$ and 0 $A$
0%
8 $A$ and 8 $A$
0%
0 $A$ and 8 $A$

Explanation

$I= \dfrac { 220 }{ 25+j\left(.01\times 2\pi f\quad -\quad \dfrac { 1 }{ { 10 }^{ -5 }2\pi f } \right) }$

$I = 0$ at

$f= 0$

$I=0$ at

$f= \infty$

The capacitive reactance of a condenser of capacity 25

$\mu$ F for an AC of frequency 4000 Hz will be

0%
$\displaystyle \frac{5}{\pi} \Omega$
0%
$\displaystyle \frac{10}{\pi} \Omega$
0%
$\displaystyle 5 \pi \Omega$
0%
$\displaystyle \frac{\pi}{5} \Omega$

Explanation

${ X }_{ C }=\dfrac { 1 }{ \omega C }$

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

${ X }_{ C }=\dfrac { 1 }{2\pi \times 4000\times 25\times { 10 }^{ -6 } }$

${ X }_{ C }=\dfrac { 5 }{ \pi }$ .

In the figure two identical bulbs, each with filament resistance

$100\ \Omega$ are connected to a resistor

$R = 100\ \Omega$ , and an inductor

$\displaystyle (X_L = 100 \Omega)$ as shown in the Figure. Then, which bulb glows more

0%
$\displaystyle B_1$
0%
$\displaystyle B_2$
0%
both glow equally
0%
cannot be predicted

Explanation

Impedance in branch containing bulb1

$Z_1=200\Omega$

impedance in branch containing bulb 2

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

${ Z }_{ 2 }=\sqrt { { 100 }^{ 2 }+{ 100 }^{ 2 } }$

${ Z }_{ 2 }=100\sqrt { 2 }$

Since,

$Z_1>Z_2$

$B_2$ will glow more than

$B_1$ .

The values of

$X_L, X_C$ and

$R$ in an AC circuit are 8

$\Omega$ , 6

$\Omega$ and 10

$\Omega$ respectively. The total impedance of the circuit

0%
$\displaystyle 10.2 \Omega$
0%
$\displaystyle 12.2 \Omega$
0%
$\displaystyle 10 \Omega$
0%
$\displaystyle 24.4 \Omega$

Explanation

Total impedance

$= R +j({X}_{L} - {X}_{C} )$

$=10 +j(8-6) = 10+j2$

$= 10.2\quad \measuredangle { tan }^{ -1 }\left(\dfrac { 2 }{ 10 } \right)$

The turning circuit of a radio receiver has a resistance of

$50\ \Omega$ , an inductor of

$10\ \text{mH}$ and a variable capacitor. A

$1\ \text{MHz}$ radio wave produces a potential difference of

$0.1\ \text{mV}$ . The values of the capacitor to produce resonance is (Take

${\pi}^{2}=10$ ):

0%
$2.5\ \text{pF}$
0%
$5.0\ \text{pF}$
0%
$25\ \text{pF}$
0%
$50\ pF$

Explanation

$L=10\ \text{mH}={10}^{-2}\ \text{H}$

$f=1\ \text{MHz}={10}^{6}\ \text{Hz}$

$f=\cfrac {1}{2\pi \sqrt {LC}}$

$f^2=\cfrac {1}{ 4 \pi ^2LC }$

$\Rightarrow C=\dfrac { 1 }{ 4\pi ^2 f ^2 L } =\dfrac { 1 }{ 4\times 10\times 10^{12}\times 10^{-2}} =\dfrac {10^{- 11}}{4} =2.5\ \text{pF}$

The ac generator in the Figure supplies 150

$\displaystyle V_{(max)}$ at 50 Hz. With the switch open as shown in the diagram, the resulting current leads the generator emf by

$60^{\circ}$ . With the switch in position 1, the current lags the generator emf by

$30^{\circ}$ . When the switch is in position 2, the maximum current is 3A. Then, the value of R is:

0%
50/ $\sqrt 3 \Omega$
0%
83.3 $\Omega$
0%
133.3 $\Omega$
0%
50 $\Omega$

Explanation

Answer : A

$switch\quad in\quad position\quad 2:\\ { I }_{ max }=\dfrac { 150 }{ \left| L\omega -\dfrac { 1 }{ C\omega } \right| } =3\quad \Rightarrow \left| L\omega -\dfrac { 1 }{ C\omega } \right| =50\\ open\quad switch\quad condition\quad :\\ \hat { Z } =R+j(L\omega -\dfrac { 1 }{ C\omega } )=Z\angle -60\quad \Rightarrow \dfrac { { R }^{ 2 } }{ { R }^{ 2 }+{ \left| L\omega -\dfrac { 1 }{ C\omega } \right| }^{ 2 } } ={ cos }^{ 2 }(-60)\quad \Rightarrow \dfrac { { R }^{ 2 } }{ { R }^{ 2 }+{ 50 }^{ 2 } } =\left( \dfrac { 1 }{ 2 } \right) ^{ 2 }\Rightarrow R=\dfrac { 50 }{ \sqrt { 3 } } A\\$

An ideal choke takes a current of

$10\ Amperes$ when connected to an ac supply of

$125\ volt$ and

$50\ Hz$ . A pure resistor under the same condition takes a current of

$12.5\ Ampere$ . If the two are connected to an ac supply of

$100\sqrt {2}$ volt and

$40\ Hz,$ then the current in a series combination of the above resistor and inductor is:

0%
$10\ A$
0%
$12.5\ A$
0%
$20\ A$
0%
$25\ A$

Explanation

In ideal choke coil all voltage drop occur at inductor. Resistance is neglected in case of ideal choke coil. so there is only inductance.
Let inductance of choke coil

$= L$
given value of AC voltage

$(V) = 125\ volt$ , frequency

$(f) = 50\ Hz$
Current in inductor

$=\dfrac { V }{ \omega L }$

${\omega L} =$ Resistance due to inductor,

$\omega$ =

$2\pi f$
Current in inductor

$=\dfrac { V }{ \omega L } = \dfrac { V }{ 2\pi fL } = \dfrac { 125 }{ 2\pi \times 50\times L }$
Value of current is given in question

$=10\ A$ .

$\dfrac { 125 }{ 2\pi \times 50\times L } = 10\Longrightarrow \quad L= \dfrac { 125 }{ 1000\pi } \ H$

Let the value of resistance is

$R$
So in pure resistance case
Current

$=\dfrac { V }{ R }$ , current given

$= 12.5\ A$

$\dfrac { 125 }{ R } = 12.5 \Longrightarrow \ R =10\ \Omega$

Now this

$R\ \&\ L$ is connected in series to another AC source of

$V = 100\sqrt { 2 }$ volt,
frequency

$(f) = 40\ Hz$
so impedance of combination

$(Z) = \sqrt { { R }^{ 2 }+{ \omega L }^{ 2 } } = \sqrt { { R }^{ 2 }+{ 2\pi fL }^{ 2 } }\ \Omega$

$Z=\sqrt { { 10 }^{ 2 }+\left( { 2\pi \times 40\times \dfrac { 125 }{ 1000\pi } } \right) ^{ 2 } }$ , putting value of

$R,L,$ and new

$f$ .

$Z=10\sqrt { 2 } \ \Omega$
Current in new combination

$= \dfrac { V }{ Z }= \dfrac { 100\sqrt { 2 } }{ 10\sqrt { 2 } } = 10\ A$
So, A is correct Answer.

For the circuit shown in the Figure, the current through the inductor is 0.9 A while the current through the condenser is 0.4 A

0%
current drawn from generator I = 1.13 A
0%
$\displaystyle \omega = \frac{1}{\left(1.5 LC\right)}$
0%
I = 0.5 A
0%
I = 0.6 A

Explanation

The current in inductor and capacitor is always at an phase difference of

$180^o$
for

$V=V_osin\omega t$
Capacitor,

$i=i_osin(\omega t - \dfrac{\pi}{2})$
Inductor,

$i=i_osin(\omega t + \dfrac{\pi}{2})$
So, the current from both branches will be

$0.9 + (-0.4) = 0.5A$

A broadcasting centre broadcasts at 300 metre band. A capacitor of capacitance 2.5

$\mu$ F is available. The value of the inductance required for resonant circuit is nearly

0%
$\displaystyle 1 \times 10^{-4} H$
0%
$\displaystyle 1 \times 10^{-8} H$
0%
$\displaystyle 1 \times 10^{-6} H$
0%
$\displaystyle 1 \times 10^{-2} H$

Explanation

$c=\lambda \nu \quad at resonance$

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

$\sqrt { LC } =\dfrac { 300 }{ 3\times 10^8\times 2\pi } =\dfrac{ 10^{-6 }}{2\pi}$

$LC=\dfrac{10^{-12}}{4 \pi^2}$

$L=\dfrac {10^{ -12 } }{ 2.5\times 10^{ -6 }\times 4\times 9.87 } =\dfrac {10^{ -6 } }{ 98.7 } \cong {10}^{ -8 }\ H$

A circuit containing a

$80\ mH$ inductor and

$60 \ \mu F$ capacitor in series is connected to a

$230\ V, 50\ Hz$ supply. If the resistance in the circuit is negligible then the current amplitude will be

0%
11.63 A
0%
8.23 A
0%
9.2 A
0%
13.67 A

Explanation

$\omega =2 \pi f =100 \pi$ ;

$f= 50Hz$

Impedance of circuit

$Z=j\left( L \omega -\dfrac{1}{C \omega}\right) =j\left(80 \times 10^{-3} \times 100 \pi -\dfrac{1}{60\times 10^{-6} \times 100 \pi}\right)$

Current amplitude

$=\left| \dfrac{V}{Z} \right | =\dfrac { 230 }{ \left| L\omega -\dfrac { 1 }{ C\omega } \right| } =8.226A\quad \quad \quad (\because \omega =2\pi 50)$

If in a series L-C-R ac circuit, the voltages across L, C and R are

$V_1, V_2$ and

$V_3$ respectively, then the voltage of the source is always

0%
equal to $\displaystyle V_1+V_2+V_3$
0%
equal to $\displaystyle V_1-V_2+V_3$
0%
more than $\displaystyle V_1+V_2+V_3$
0%
none of the above is true

Explanation

The voltages across different elements in AC circuit add vectorially.

The voltages across inductor and capacitor are out of phase and at

$90^{\circ}$ to that across resistor.

Hence net voltage across source

$=\sqrt{(V_1-V_2)^2+V_3^2}$ .

A high-impedance ac voltmeter is connected in turn across the inductor, the capacitor, and the resistance in a series circuit having an ac source of 100 V(rms) and gives the same reading in volts in each case. Then, this reading is:

0%
100/3 volts
0%
300 volts
0%
100 volts
0%
incomplete data

Explanation

Let the same reading be $x V$ ,
then let, ${ Z }_{ total }=R+j({ X }_{ L }-{ X }_{ C })=Z\angle \theta \\ I=\dfrac { { V }_{ S } }{ Z } \angle -\theta \\ therefore,\quad { V }_{ R }=\left| \dfrac { { V }_{ S } }{ Z } R\angle -\theta \right| \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ L }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ L }\angle 90-\theta \right| \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ C }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ C }\angle -90-\theta \right| \\ given,\quad { V }_{ R }={ V }_{ L }={ V }_{ V }=x\\ \quad \Rightarrow \quad R={ X }_{ L }={ X }_{ C }\\ \Rightarrow { Z }_{ total }=R=Z\angle 0=Z\quad \quad \\ \Rightarrow \quad \quad { V }_{ R }=\left| \dfrac { { V }_{ S } }{ Z } R\angle -\theta \right| ={ V }_{ S }\\ \quad \quad \quad \quad { V }_{ L }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ L }\angle 90-\theta \right| =\left| j{ V }_{ S } \right| ={ V }_{ S }\\ \quad \quad \quad \quad { V }_{ C }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ C }\angle -90-\theta \right|=\left| -j{ V }_{ S } \right| ={ V }_{ S }\\ \Rightarrow x={ V }_{ S }=100V$

In an ideal parallel LC circuit, the capacitor is charged by connecting it to a D.C. source which is then disconnected. The current in the circuit

0%
Becomes zero instantaneously
0%
Grows monotonically
0%
Decays monotonically
0%
Oscillates instantaneously

An ac source producing emf

$\displaystyle e = E_0[cos (100 \pi t) + cos (500 \pi t)]$ is connected in series with a capacitor and a resistor. The steady state current in the circuit is found to be

$\displaystyle i = I_1 cos (100 \pi t + \varphi_1) + I_2 cos (500 \pi t + \varphi_2)$

0%
$\displaystyle I_1 > I_2$
0%
$\displaystyle I_1 = I_2$
0%
$\displaystyle I_1 < I_2$
0%
nothing can be said

Explanation

We can assume two different sources connected in series with same orientation.

${ E }_{ 1 }={ E }_{ 0 }\cos { 100\pi t }$

${ E }_{ 2 }={ E }_{ 0 }\cos { 500\pi t }$

such that

$e={ E }_{ 0 }\cos { 100\pi t }+{ E }_{ 0 }\cos { 500\pi t }$

becomes,

$e={ E }_{ 1 }+{ E }_{ 2 }$

Now, solving using principle of superposition we get,

$i_{ 1 }=100\pi C\cos { (100 } \pi t+\phi_1)$

$i_{ 2 }=500\pi C\cos { (500 } \pi t+\phi_2)$

Final current will be,

$i=i_{ 1 }+i_{ 2 }$

From this we get

$I_{ 1 }=100\pi C$

and

$I_{ 2 }=500\pi C$

From this we get,

$I_{ 1 }<I_2$ .

In an electric circuit the applied alternating emf is given by

$\displaystyle E = 100\sin(314t)$ volts and current flowing

$\displaystyle I = \sin \left(314t + \dfrac{\pi}{3}\right)$ , Then, the impedance of the circuit is

$(in\ \Omega)$

0%
$\dfrac{100}{ \sqrt{2}}$
0%
$100$
0%
$100 \sqrt{2}$
0%
$None\ of\ the\ above$

Explanation

Phasor of

$E=100 \angle{(\pi /2)}$

phasor of

$I=1 \angle{(\pi /2 + \pi /3)}$

phasor of impedance

$=\dfrac{E}{I} = 100 \angle{(-\pi/3)}$

impedance magnitude

$= 100\ \Omega$

A coil of negligible resistance is connected in series with a 90

$\Omega$ resistor across a 120 V, 60 Hz line. An ac voltmeter reads 90 V across the resistance, then the inductance of the coil is approximately

0%
0.2 H
0%
0.3 H
0%
0.4 H
0%
0.7 H

Explanation

Using Ohm's law

$V=IR$

where V is potential difference across resistance and I is current in it.

As both inductor and resistor are connected in series the current will be same in both the inductor and resistor, which is given as

$I=\dfrac { 120 }{ \sqrt {R^2+L^2 \omega^2}}$

From given data

$V=90 \ V$ and

$R= 90 \ \Omega$

$90=\dfrac { R }{ \sqrt {R^2+L^2 \omega^2}} 120$

$\dfrac { 1 }{ 120 } =\dfrac { 1 }{ \sqrt { { 90 }^{ 2 }+{ L }^{ 2 }{ (2\pi 60) }^{ 2 } } }$

$(120)^2=(90)^2+120^2\pi^2 L^2\Rightarrow L^2=\dfrac{120^2-90^2}{120^2 \pi^2}$

$L=\sqrt{\dfrac{30\times 210}{120^2\pi^2}}$

$L=0.21\quad H$

In an ac circuit, the resistance of R

$\Omega$ is connected in series with an inductance

$I$ . If phase angle between voltage and current be

$\displaystyle 45^{\circ}$ , the value of inductive reactance will be:

0%
$R/2$
0%
$\displaystyle R/\sqrt 2$
0%
$R$
0%
Insufficient data

Explanation

$\dfrac { R }{ \sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } } =cos(45)=\dfrac { 1 }{ \sqrt { 2 } } \\ \Rightarrow 2{ R }^{ 2 }={ R }^{ 2 }+{ X }_{ L }^{ 2 }\quad \\ \Rightarrow \left| { X }_{ L } \right| =R$

You apply

$\displaystyle f = \frac{1}{2\pi \sqrt{LC}}$ to find L. Which frequency will you select?

0%
when $V_R$ is minimum
0%
when $V_R$ is maximum
0%
when $V_C$ is maximum
0%
when $V_C$ is minimum

Explanation

Since resonant frequency,

$f=\dfrac { 1 }{ 2\pi \sqrt { LC } }$ is being applied, current from the variable AC source will be maximum when potential across resistor is maximum, because

${ V }_{ R }=iR$ and

$R$ is fixed.

Thus when

${ V }_{ R }$ is maximum, we can measure the value of inductance without requiring an ammeter.

In the circuit shown in fig., the resonant frequency is

0%
$200Hz$
0%
$220Hz$
0%
$225.08Hz$
0%
$230Hz$

Explanation

The resonant frequecy of a circuit is given by f=

$\frac{1}{2\pi\sqrt{LC}}$ .

Given L= 0.1H C = 5

$\mu$ F R = 5

$\Omega$ .

Substituting them in formula for f gives

f= 225.08 Hz.

If resistance of

$100\Omega$ and the inductance of

$0.5$ henry and capacitance of

$10\times {10}^{6}$ farad are connected in series through

$50$ Hz A.C supp;y, then impedence is

0%
$1.8765\Omega$
0%
$18.76\Omega$
0%
$187.6\Omega$
0%
$101.3\Omega$

Explanation

$z=\sqrt { { R }^{ 2 }+{ \left( \omega L-\cfrac { 1 }{ \omega C } \right) }^{ 2 } }$
Here

$R=100W, L=0.5 henry, C=10\times {10}^{6}farad$

$\omega=2p \pi=100\pi$ $

In which of the following circuit, there may be no change in current with increase in the frequency when same ac current is passed through them:

Explanation

The impedance inductor and capacitor are

$2\pi fL\quad \& \quad 1/2\pi fC$ respectively, since both of them have the quantity 'f' in their expression the change in frequency will effect their impedance and hence the current in the circuit, so the circuit only with Resistance can be independent of frequency in terms of current.

In an

$L-C-R$ circuit the capacitance is changed from

$C$ to

$4C$ . For the same resonant frequency, the inductance should be changed from

$L$ to

0%
L/4
0%
L/2
0%
2L
0%
4L

Explanation

$\omega =\dfrac { 1 }{ \sqrt { LC } }$
For,

$\omega = constant$

$\rightarrow \sqrt { LC } = constant$
Therefore,

$C$ is inversely proportional to

$L$
So, if

$C$ is made

$4C$ then

$L$ should be reduced to

$L/4$ to keep

$\omega$ constant.

In the given circuit, the current drawn from the source is

0%
$20A$
0%
$10A$
0%
$5A$
0%
$5\sqrt {2}A$

Explanation

$\text{All the three are connected in parallel so voltage are same across them}$

$X_L=10\Omega ,\quad X_c=20\Omega ,\quad R = 20\ \Omega ,\quad V= 100\ \text{volt}$

$\text{Current in resistor }(i_1)= \dfrac { V }{ R } =\dfrac { 100 }{ 20 } =5\ \text{A}$

$\text{Current in inductor }({ i }_{ 2 })= j V X_l = j\dfrac {100}{10} = 10j\ \text{A}$

$\text{current in capacitor }(i _3) = -j\dfrac { V }{ X_C } = -j\dfrac { 100 }{ 20 } = -5j\ \text{A}$

$\text{So, total current } i = i_1 + \left(i_2+i_3\right) j$

$i = 5 + (10 -5)j = 5 + 5j$

$\text{Therefore magnitude of current} = \sqrt {5^2+ 5^2} = 5\sqrt {2}\ \text{A}$

The most stable frequency can be generated using

0%
LC series circuit
0%
LC parallel circuit
0%
crystal oscillator
0%
RC phase shift oscillator

With increase in frequency of an ac supply, the impedance of an

$L-C-R$ series circuit

0%
remains constant
0%
increases
0%
decreases
0%
decrease at first becomes minimum and then increases.

Explanation

We know that in an L-C-R circuit impedance is given by,

$Z=\sqrt { { R }^{ 2 }+{ ({ X }_{ L }-{ X }_{ C }) }^{ 2 } }$
When

${ X }_{ L }=\omega L$ is less than

${ X }_{ C }=1/\omega C$
With increase in

$\omega$ impedance decrease and become minimum when

${ X }_{ L }={ X }_{ C }$
After that

${ X }_{ L }=\omega L$ will be greater than

${ X }_{ C }=1/\omega C$

Two coils

$A$ and

$B$ are connected in series across a

$240\ \text{V}, 50\ \text{Hz}$ supply. The resistance of

$A$ is

$5\ \Omega$ and the inductance of

$B$ is

$0.02\ \text{H}$ . The power consumed is

$3\ \text{kW}$ and the power factor is

$0.75$ . The impedence of the circuit is

0%
$0.144\ \Omega$
0%
$1.44\ \Omega$
0%
$14.4\ \Omega$
0%
$144\ \Omega$

Explanation

Given :

$E_v = 240 \ V$ ,

$\cos\phi = 0.75$ and

$p = 3000 \ W$

Power consumed

$P=\dfrac {E_v^2\cos \phi }{ Z } =3000$

where

$Z$ is the impedence of the circuit.

$\therefore \dfrac { (240)^2(0.75)}{Z} = 3000$

$\Rightarrow Z=14.4\ \Omega$

In a circuit inductance

$L$ and capacitance

$C$ are connected as shown in figure.

${A}_{1}$ and

${A}_{2}$ are ammeters.
When key

$K$ is pressed to complete the circuit, then just after closing key

$(K)$ , the readings of

${A}_{1}$ and

${A}_{2}$ will be

0%
zero in both ${A}_{1}$ and ${A}_{2}$
0%
maximum in both ${A}_{1}$ and ${A}_{2}$
0%
zero in ${A}_{1}$ and maximum in ${A}_{2}$
0%
maximum in ${A}_{1}$ and zero in ${A}_{2}$

Explanation

$\text{Initially there is no D.C current in inductive circuit and maximum D.C}$

$\text{current is in capacitive current. Hence, the current is zero in}$

${A}_{2}\ \text{and maximum in}\ {A}_{1}$

Capacitance

$(C)$ of the capacitor is

0%
$10\mu F$
0%
$15\mu F$
0%
$20\mu F$
0%
None of these

An inductive coil has a resistance of

$100\ \Omega$ . When an a.c. signal of frequency

$1000\ Hz$ is fed to the coil, the applied voltage leads the current by

${45}^{o}$ . What is the inductance of the coil?

0%
$10mH$
0%
$12mH$
0%
$16mH$
0%
$20mH$

Explanation

The applied voltage leads the current by

$45^o$
i.e.,

$\dfrac{X_L}{R}=\tan 45^o$

$X_L=R$

$\Rightarrow 2\times3.14\times 1000 L=100$

$\Rightarrow 6280L=100$

$\Rightarrow L=\dfrac{25}{1570}=0.01592\cong 16 mH$

An alternating voltage of

$220V,50Hz$ frequency is applied across a capacitor of capacitance

$2\mu F$ . The impedence of the circuit is:

0%
$\cfrac{\pi}{5000}$
0%
$\cfrac{1000}{\pi}$
0%
${500}{\pi}$
0%
$\cfrac{5000}{\pi}$

Explanation

Impedence of a capacitor is

${X}_{C}=1/\omega C$

${ X }_{ C }=\cfrac { 1 }{ 2\pi fC } =\cfrac { 1 }{ 2\pi \times 50\times 2\times { 10 }^{ -6 } } =\cfrac { 5000 }{ \pi }$

The instantaneous voltage through a device of impedence

$20\Omega$ is

$e=80\sin { 100\pi t }$ . The effective value of the current is

0%
$3A$
0%
$2.828A$
0%
$1.732A$
0%
$4A$

Explanation

Given equation,

$e=80\sin { \left(100 \pi t \right) } .......(i)$
Standard equation of instantaneous voltage given by

$E={e}_{m}\sin { \left( \omega t \right) } ..........(ii)$
Compare

$(i)$ and

$(ii)$ , we get

${e}_{m}=80V$
where

${e}_{m}$ is the voltage amplitude.
Current amplitude

${I}_{m}=\cfrac{{e}_{m}}{Z}$ where

$Z=$ impedence

$=\dfrac{80}{20}=4A$

${ I }_{ r.m.s }=\cfrac { 4 }{ \sqrt { 2 } } =\cfrac { 4\sqrt { 2 } }{ 2 } =2\sqrt { 2 } =2.828A$

For watt-less power in an

$AC$ circuit the phase angle between the current and voltage is

0%
$0^o$
0%
$90^o$
0%
$45^o$
0%
Not possible

Explanation

Watt-less power in an AC circuit is basically power supposed to be generated by inductive and capacitive reactance and since they are not resistor they generate any heat , and these power wasted is called watt-less power and its phase angle is always

$90^o$ as it has only capacitor and inductor.

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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