MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 11

In an electric circuit applied ac emf is

$e = 20 \sin 300 \ t$ volt and the current is

$i = 4 \sin 300 \ t$ ampere.The reactance of the circuit is

0%
$5 \Omega$
0%
$80 \Omega$
0%
$1.8 \mathrm { k } \Omega$
0%
$0$

In an alternating current circuit, the phase difference between current l and voltage is

$\phi$ , then the Wattless component of current will be -

0%
$I\cos \phi$
0%
$I \tan\phi$
0%
$I \sin\phi$
0%
$I \cos^2\phi$

if value of R is changed then

0%
voltage across L remains same
0%
voltage across C remains sam
0%
voltage across L-C combination remains same w
0%
voltage across L-C combination changes

Explanation

The values of

$R$ is changed then voltage across

$L-C$ combination changes.

Hence, Option

$D$ is correct.

A current of 2A flow in an inductance of 10H. for generating 100V EMF in it the rate of change current will be.

0%
1 A/sec
0%
5 A/sec
0%
10 A/sec
0%
100 A/sec

In a RLC series circuit at resonance, the value of the power factor is :

0%
infinity
0%
zero
0%
$\dfrac{1}{\sqrt{2}}$
0%
$1$

Explanation

At resonance,

$X_L = X_c$

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

Power factor

$\cos\phi = \dfrac{R}{Z} = \dfrac{R}{R} = 1$

A current

$I=3+8\sin 100t$ is passing through a resistor of resistance

$10\Omega$ . The effective value of current is?

0%
$\sqrt{41} A$
0%
$6.4 A$
0%
$4\sqrt{2} A$
0%
$3/\sqrt{2} A$

$8\mu F$ capacitor is connected to the terminals of an A.C. source whose

$V_{rms}$ is

$150$ volt and the frequency is

$60$ Hz, the capacitive reactance is?

0%
$0.332\times 10^3\Omega$
0%
$2.08\times 10^3\Omega$
0%
$4.16\times 10^3\Omega$
0%
$12.5\times 10^3\Omega$

The self inductance of a choke coil is mH. when it is connected with a 10 VDC source then the loss of power is 20 watt. When it connected with 10 volt AC source loss of power is 10 watt. The frequency of AC source will bw-

0%
50 Hz
0%
60 Hz
0%
80 Hz
0%
100 Hz

Explanation

$\begin{array}{l} When\, \, \, dc\, \, is\, \, pass\, \, through\, \, the\, \, inductor\, \, no\, \, induv\tan ce\, \, effect\, \, will\, \, be\, \, their\, \, only \\ resis\tan ce\, \, will\, \, be\, their\, \, resis\tan ce\, \, of\, \, the\, \, inductor\, \, be\, \, R \\ 20=\frac { { 100 } }{ R } \\ R=5\Omega \\ Xl=\omega \times 0.001 \\ power\, \, =\frac { { { v^{ 2 } } } }{ z } \cos \phi \\ Z=\sqrt { 25+{ \omega ^{ 2 } }\times { { 10 }^{ -6 } } } \\ \cos \phi =\frac { R }{ Z } =\frac { 5 }{ Z } \\ 10=\frac { { 100 } }{ Z } \frac { 5 }{ Z } =\frac { { 500 } }{ { { Z^{ 2 } } } } =\frac { { 500 } }{ { 25 } } +{ \omega ^{ 2 } }\times { 10^{ -6 } } \\ 25+{ \omega ^{ 2 } }\times { 10^{ -6 } }=50 \\ { \omega ^{ 2 } }=25\times { 10^{ 6 } } \\ \omega =5000 \end{array}$

$=50 Hz$

Hence, Option

$A$ is correct answer.

Impedance of LR circuit is-

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

An ideal inductor takes a current of 10 A when connected to a

$125 {V}, 50 {Hz}$ ac supply.A pure resistor across the same source takes 12.5 A.If the two are connected in series across a

$100 {V}, 40 {Hz}$ -supply,the current through the circuit will be

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7 A
0%
12.5 A
0%
20 A
0%
25 A

Impedance of a coil having inductance

$0.4H$ at frequency of

$50Hz$ will be

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$20\pi \Omega$
0%
$40\pi \Omega$
0%
$2\pi \Omega$
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$4\pi \Omega$

A capacitor C and inductor L are connected in parallel with a battery of emf e and internal resistance r. At time

$t=0$ , current through the cell be

$i_0$ and at t

$\rightarrow \infty$ ,let the current be i, Then

$\frac{i_0}{i}$ is equal to

0%
$1$
0%
$0$
0%
infinity
0%
cannot be determine.

If an ac source is connected across ideal capacitor and current passing through it is dentoed by curve then instantaneous power is denoted by curve

0%
$c$
0%
$b$
0%
$e$
0%
$a$

The resonance frequency changes if .is a change in L-C-R series A.C. circuit.

0%
only R
0%
only L
0%
only C
0%
L and C

Two inductors each of inductance L are connected in parallel . what is their equivalent inductance?

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

Phase relationship between current (I) and applied voltage (E) for a series LCR circuit is shown here.

$\omega_0$ = Reasonat angular frequency of the circuit and

$\omega$ Applied angular frequency.

0%
$\omega = \omega_o$
0%
$\omega > \omega_o,$ exact relation can't be computed
0%
$\omega < \omega_o,$ exact relation can't be computed
0%
$2 \omega = \sqrt {3\omega_o}$

The inductance of a coil is directly proportional to

0%
its length
0%
the number of turns
0%
the resistance of the coil
0%
the square of the number of turns

In the phasors method for the only capacitive A.C. circuit, voltage is in +X direction then current is in direction.

0%
+X
0%
-X
0%
+Y
0%
-Y

In LC oscillation resistance is

$100\Omega$ and inductance and capacitance is

$1$ H and

$10\mu F$ . Find the half power of frequency.

0%
$266.2$
0%
$366.2$
0%
$166.2$
0%
$233.2$

Explanation

$f_n=\omega_0-\dfrac{R}{2L}=\dfrac{1}{\sqrt{LC}}-\dfrac{R}{2L}$

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

$-\dfrac{100}{2\times 1}$

$=\dfrac{1000}{\sqrt{10}}-\dfrac{100}{2}$

$=100\left[\sqrt{10}-0.5\right]=266.27$ .

A series LCR circuit containing a resistance of

$120 \Omega$ has angular resonance frequency

$4 \times 10^2 rad^{-1}$ . At resonance the voltage across resistance and inductance are 60 V and respectively.
(a) The value of L and C are

$0.2 mH. \frac {1}{32} \mu F$
(b)

$8 \times 10^5 rad/s$ . The current lags the voltage by

$45^o$
(c)

$6 \times 10^5 rad/s,$ the current lags the voltage by

$45^o$

0%
(a), (c) are correct
0%
(b) and (c) are correct
0%
(a), (b) are correct
0%
(a), (b) (c) are wrong

The switch shown in figure is closed at

$t = 0$ find the charge through battery till time

$t = \dfrac {L}{R}$ .

0%
$\dfrac {L_{\epsilon}}{eR^{2}}$
0%
$\dfrac {L_{\epsilon}}{2eR^{2}}$
0%
$\dfrac {2L_{\epsilon}}{eR^{2}}$
0%
None

Explanation

$i = \dfrac {\epsilon}{R} (1 - e^{-t/\tau})$ where

$\tau = L/R$

$\dfrac {dq}{dt} = \dfrac {\epsilon}{R} (1 - e^{-t/\tau})$

$\int_{0}^{q} dq = \int_{0}^{\tau} \dfrac {\epsilon}{R} (1 - e^{-t/\tau})dt$

$q = \dfrac {\epsilon}{R}\left [t - \dfrac {e^{-t/\tau}}{-1/\tau}\right ]_{0}^{\tau}$

$q = \dfrac {\epsilon}{R} \left [\tau + \dfrac {\tau}{e} - \tau \right ]$

$q = \dfrac {\epsilon}{R} \dfrac {\tau}{e} = \dfrac {L\epsilon}{eR^{2}}$ .

If in an A.C., L-C series circuit

$X_c > X_L$ . Hence potential _________.

0%
Lags behind the current by $\pi/2$
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Leads the current by $\pi$ in phase
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Leads the current by $\pi/2$ in phase
0%
Lags behind the current by $\pi$ in phase

Explanation

Potential lags by

$\pi/2$ phase.
1338284_1646268_ans_719e0b40df874157bd18ea560da5bea4.png

Find the time after which current in the circuit becomes

$80 \%$ of its maximum value.

0%
$\dfrac{\ell n 2}{100}$
0%
$\dfrac{\ell n 3}{100}$
0%
$\dfrac{\ell n 5}{100}$
0%
$\dfrac{\ell n 6}{100}$

Explanation

$I_s = \dfrac{V}{R} = \dfrac{V}{0.9 + 0.1} = \dfrac{V}{1}$

$I = I_s \left(1 - e^{-\dfrac{Rt}{L}} \right)$

$0.8V = V \left(1 - e^{- \dfrac{1 \times t}{10 \times 10^{-3}}}\right)$

$0.8 = 1 - e^{-100 \,t}$

$0.2 = e^{- 100 \,t}$

$e^{- 100 \,t} = 5$

$t = \dfrac{\ell n5}{100}$

An oscillator circuit contains an inductor

$0.05H$ and a capacitor of capacity

$80\mu F$ .When the maximum voltage across the capacitor is

$200V$ , the maximum current (in amperes) in the circuit is

0%
$2$
0%
$4$
0%
$8$
0%
$10$
0%
$16$

Explanation

Given

$L=0.05H,C=80\mu F$

$V_{max}=200V$

$\because$ Voltage equation

$V(t)=V_m\sin \omega t$

$\therefore$ Current (i)

$=\dfrac{cdV}{dt}=c\dfrac{d(V_m\sin \omega t)}{dt}=CV_m\omega \cos \omega t$

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

$\therefore i=V_m\sqrt{\dfrac{C}{L}}\cos\omega t$

$\therefore$ Maximum current

$(i_m)=V_m\sqrt{\dfrac{C}{L}}=200\times \sqrt{\dfrac{80\mu}{0.05}}$

$\Rightarrow i_m=8A$

In a series LCR circuit

$R=300\Omega, L=0.9H, C=2\mu F, \omega =1000\ rad/s$ . The impedance of the circuit is?

0%
$500\Omega$
0%
$1300\Omega$
0%
$400\Omega$
0%
$900\Omega$

Explanation

The correct option is (a)

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

$=\sqrt{(300)^2+\left(1000\times 0.9-\dfrac{1}{1000\times 2\times 10^{-6}}\right)^2}$

$Z=\sqrt{(300)^2+(400)^2}\Rightarrow 500\Omega$

$L,C,R$ represent the physical quantities inductances, capacitance and resistance respectively. Which of the following combinations have dimensions of freqeuency?

0%
$\cfrac { 1 }{ RC }$
0%
$\cfrac { R }{ L }$
0%
$\cfrac { 1 }{ \sqrt { LC } }$
0%
$C/L$

The natural frequency of the circuit shown in figure is

0%
$\dfrac{1}{\sqrt{LC}}$
0%
$\dfrac{1}{\sqrt{2LC}}$
0%
$\dfrac{2}{\sqrt{LC}}$
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none of these

Explanation

$\omega=\dfrac{1}{\sqrt{L_{eq}L_{eq}}}$

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

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

The frequency of oscillation of current in the inductor is

0%
$\dfrac{1}{3\sqrt{LC}}$
0%
$\dfrac{1}{6 \pi\sqrt{LC}}$
0%
$\dfrac{1}{\sqrt{LC}}$
0%
$\dfrac{1}{2 \pi\sqrt{LC}}$

Explanation

Equivalent inductance

$L_{eq}=L+2 L=3L$

$,C_{eq}=C+2C=3C$

$\therefore$ Frequency of oscillation

$f=\dfrac{1}{2 \pi \sqrt{L_{eq}C_{eq}}}$

$=\dfrac{1}{6 \pi \sqrt{LC}}$

$E_0$ represents the peak value of the voltage in an ac circuit, the r.m.s. value of the voltage will be[

0%
$\dfrac{E_0}{\pi}$
0%
$\dfrac{E_0}{2}$
0%
$\dfrac{E_0}{\sqrt{\pi}}$
0%
$\dfrac{E_0}{\sqrt{2}}$

Explanation

$E_0$ represents the peak value of the voltage in an ac circuit then ,

we know,

$E_{rms}=\dfrac {E_0}{\sqrt 2}$

The resistance of a coil for dc is in ohms. In ac, the resistance

0%
will remains same
0%
will increase
0%
will decrease
0%
will be zero

Explanation

The coil having inductance

$L$ besides the resistance

$R$ . Hence for ac it’s effective resistance

$\sqrt{R^2 + X^{2}_{L}}$ will be larger than it’s resistance

$R$ for dc.

An alternating current of frequency ' f ' is flowing in a circuit containing a resistance R and a choke L in series. The impedance of this circuit is

0%
$R + 2 \pi \, fL$
0%
$\sqrt{R^2 + 4 \, \pi^2 f^2 L^2}$
0%
$\sqrt{R^2 + L^2}$
0%
$\sqrt{R^2 + 2 \pi \, fL }$

Explanation

As we know,

$Z = \sqrt{R^2 + X_{L}^{2}} , X_L = \omega L$ and

$\omega = 2 \pi f$

$\therefore \, Z = \sqrt{R^2 + 4 \pi^2 f^2 L^2}$

Power delivered by the source of the circuit becomes maximum, when

0%
$\omega L = \omega C$
0%
$\omega L = \dfrac{1}{\omega C}$
0%
$\omega L = - \left ( \dfrac{1}{\omega C} \right )^2$
0%
$\omega L = \sqrt{\omega C}$

Explanation

As we know,

power delivered by the source of the circuit becomes maximum when, load resistance equals to source resistance.

we know, in L-C-R circuit,

load resistance is inductive reactance and source resistance is reactance of capacitor.

e.g.,

$X_L=X_C$

or,

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

hence, Power delivered by the source of the circuit becomes maximum, when ,

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

The natural frequency of a

$L-C$ circuit is equal to

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

An alternating e.m.f. is applied to purely capacitive circuit. The phase relation between e.m.f. and current flowing in the circuit is or In a circuit containing capacitance only

0%
e.m.f is ahead of current by $\dfrac{\pi}{2}$
0%
Current is ahead of e.m.f by $\dfrac{\pi}{2}$
0%
Current lags behind e.m.f by $\pi$
0%
Current is ahead of e.m.f by $\pi$

Explanation

As we know,

For purely capacitive circuit

$e = e_0 \, sin \omega t$

$i = i_0 \, sin \left ( \omega t + \dfrac{\pi}{2} \right )$ i.e., current is ahead of emf by

$\dfrac{\pi}{2}$

A series ac circuit consist of an inductor and a capacitor. The inductance and capacitance is respectively

$1$ henry and

$25\mu F.$ If the current is maximum in circuit then angular frequency will be

0%
$200$
0%
$100$
0%
$50$
0%
$200 / 2\pi$

Explanation

The current in the LC circuit becomes maximum when resonance occurs. So

$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{1 \times 25 \times 10^{-6}}} = \dfrac{1000}{5} = 200$ rad/sec

In a series LCR circuit, operated with an ac of angular frequency

$\omega$ ,the total impedance is

0%
$[R^2 + (L \omega - C \omega)^2]^{1/2}$
0%
$\left [ R^2 + \left (L \omega - \dfrac{1}{C \omega} \right )^2 \right ]^{1/2}$
0%
$\left [R^2 + \left (L \omega - \dfrac{1}{C \omega} \right )^2 \right ]^{-1/2}$
0%
$\left [(R \omega)^2 + \left (L \omega - \dfrac{1}{C \omega} \right )^2 \right ]^{1/2}$

Explanation

As we know,

$X_L=L\omega \ and \ X_C=\dfrac1{\omega C}$

now,

From the above figure,

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

$Z=[R^2+(L\omega -\dfrac1{\omega C})^2]^{\dfrac12}$

1746475_1816881_ans_57dc7c80d1d44cc696876ca70b0844c8.png

When

$100$ volt dc is applied across a coil, a current of

$1$ amp flows through it. When

$100$ volt ac at

$50$ cycle

$s^{-1}$ is applied to the same coil, only

$0.5$ ampere current flows. The impedance of the coil is

0%
$100 \Omega$
0%
$200 \Omega$
0%
$300 \Omega$
0%
$400 \Omega$

Explanation

When dc is supplied

$R = \dfrac{V}{i} = \dfrac{100}{1} = 100 \, \Omega$

When ac is supplied

$Z = \dfrac{V}{i} = \dfrac{100}{0.5} = 200 \, \Omega$

If resistance of

$100 \,\Omega$ inductance of

$0.5$ henry and capacitance of

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

$50$ Hz ac supply, then impedance is

0%
$1.876$
0%
$18.76$
0%
$189.72$
0%
$101.3$

Explanation

As we know,

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

$= \sqrt{100^2 + \left ( 0.5 \times 100 \pi - \dfrac{1}{10 \times 10^{-6} \times 100 \pi} \right )^2} = 189.72 \, \Omega$

In a LCR circuit having

$L = 8.0$ henry,

$C = 0.5 \mu F$ and

$R = 100$ ohm in series. The resonance frequency in per second is

0%
$600$ radian
0%
$600$ Hz
0%
$500$ radian
0%
$500$ Hz

Explanation

Resonance frequency in radian/second is

$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{8 \times 0.5 \times 10^{-6}}} = 500$ rad /sec

The frequency for which a

$5 \mu F$ capacitor has a reactance of

$\dfrac{1}{1000}$ ohm is given by

0%
$\dfrac{100}{\pi} \, MHz$
0%
$\dfrac{1000}{\pi} \, Hz$
0%
$\dfrac{1}{1000} \,Hz$
0%
$1000 \, Hz$

Explanation

As we know,

$X_C = \dfrac{1}{2 \pi \nu C} \, \Rightarrow \, \dfrac{1}{1000} = \dfrac{1}{2 \pi \times \nu \times 5 \times 10^{-6}}$

$\Rightarrow \, \nu = \dfrac{100}{\pi}$ MHz

If an

$8\, \Omega$ resistance and

$6 \, \Omega$ reactance are present in an ac series circuit then the impedance of the circuit will be

0%
$20 \, ohm$
0%
$5 \, ohm$
0%
$10 \, ohm$
0%
$14 \sqrt{2} \, ohm$

Explanation

As we know,

Impedance

$Z = \sqrt{R^2 + X^2} = \sqrt{(8)^2 + (6)^2} = 10 \, \Omega$

Which of the following components of a LCR circuit with ac supply dissipates energy

0%
L
0%
R
0%
C
0%
All of these

Explanation

Resistor is the components of a LCR circuit, with ac supply ,dissipates energy as

$E=i^2Rt$

And inductor and capacitor stores energy in form of magnetic and electric field energy.

In series

$LCR$ circuit, in the condition of resonance, if

$C=1\mu F;L=1H$ then frequency will be:

0%
${ 10 }^{ 6 }Hz$
0%
$2\pi \times { 10 }^{ 6 }Hz$
0%
$\cfrac { { 10 }^{ 6 } }{ 2\pi } Hz$
0%
$2\pi \times { 10 }^{ -6 }Hz$

Explanation

$\omega =\cfrac { 1 }{ \sqrt { LC } } \Rightarrow f=\cfrac { 1 }{ 2\pi \sqrt { LC } } =\cfrac { 1 }{ 2\pi \times \sqrt { { 10 }^{ -6 }\times { 10 }^{ -6 } } } =\cfrac { { 10 }^{ 6 } }{ 2\pi } Hz$

An alternating current circuit is in resonance at

$10kHz$ frequency. If frequency increases to

$12kHz$ , then what will be effect on impedance?

0%
remains unchanged
0%
increases $1.2$ times
0%
increases and becomes capacitive
0%
increases and becomes inductive

Explanation

In the condition of resonance

$Z=R$ , so impedance is not dependent on the frequency.

Which of the following is used in a circuit which shows that the current lead the voltage at the phase:

0%
pure resistance
0%
pure inductance
0%
pure capacitor
0%
none of these

In pure inductive circuit, the curves between frequency f and reciprocal of inductive reactance

$1/ X_L$ is

Explanation

$X_L = 2 \pi \, fL \, \Rightarrow \, X_L \propto f \, \Rightarrow \, \dfrac{1}{X_L} \propto \dfrac{1}{f}$
i.e., graph between

$\dfrac{1}{X_L}$ and

$f$ will be a hyperbola.

An oscillator circuit consists of an inductance of

$0.5 \,mH$ and a capacitor of

$20 \mu F$ . The resonant frequency of the circuit is nearly

0%
$15.92 \,Hz$
0%
$159.2 \,Hz$
0%
$1592 \,Hz$
0%
$15910 \,Hz$

Explanation

AS we know,

$\omega =\dfrac1{\sqrt{LC}}$

so,

$f=\dfrac1{2\pi \sqrt{LC}}$

now,

$f=\dfrac1{2\pi \sqrt{0.5\times 10^{-3}\times 20\times10^{-6}}}$

$f=1592\ Hz$

A resistance of

$40$ ohm and an inductance of

$95.5$ millihenry are connected in series in a

$50$ cycles/second ac circuit. The impedance of this combination is very nearly

0%
$30 \,ohm$
0%
$40 \, ohm$
0%
$50 \, ohm$
0%
$60 \, ohm$

Explanation

As we know,

$Z = \sqrt{R^2 + (2 \pi \nu L)^2}$

$= \sqrt{(40)^2 + 4 \pi^2 \times (50)^2 \times (95.5 \times 10^{-3})^2} = 50\, ohm$

Which of the following curves correctly represents the variation of capacitive reactance

$X_C$ with frequency f

Explanation

$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2 \pi f C}$ i.e.,

$X_C \propto \dfrac{1}{f}$ hence a rectangular hyperbola.

What is the rms voltage?

0%
$\sqrt{2}\Delta V_{max}$
0%
$\Delta V_{max}$
0%
$\Delta V_{max}/\sqrt{2}$
0%
$\Delta V_{max}/2$
0%
$\Delta V_{max}/4$

Explanation

The average of the squared voltage is

$([\Delta V]^2)_{avg}=\dfrac{(\Delta V_{max})^2}{2}$ . Then its square root is

$\Delta V_{max}=\dfrac{\Delta V_{max}}{\sqrt{2}}$

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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