MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 1

Which one of the following represents capacitive reatance versus angular frequency graph?

The frequency at which

$1\ H$ inductor will have a reactance of

$2500\ \Omega$ is:

0%
$418\ Hz$
0%
$378\ Hz$
0%
$398\ Hz$
0%
$406\ Hz$

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%
$7/12\,\Omega$

The ratio of peak value and r.m.s value of an alternating current is

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

Explanation

As we know,

$V_0=V_{rms} \times \sqrt 2$

so,

$\dfrac{V_0}{V_{rms}}=\sqrt 2$

In a circuit

$LC = 10^{-3}H, C = 10^{-3} F$ . Find angular frequency.

0%
$1000 \,rad/sec$
0%
$100 \,rad/sec$
0%
$10 \,rad/sec$
0%
$10^{-3} \,rad/sec$

Explanation

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

$w = \dfrac{1}{\sqrt{10^{-3} \times 10^{-3}}} = 1000 \,rad/sec$

Alternating current is flowing in inductance

$L$ and resistance

$R$ . The frequency of source is

$\omega/2\pi$ Which of the following statement in correct:

0%
For low frequency the limiting value of impedance is $L$ .
0%
For high frequency the limiting value of impedance is $\omega L$ .
0%
For high frequency the limiting value of impedance is $R$ .
0%
For low frequency the limiting value of impedance is $\omega L$ .

The power factor of L-R circuit is :

0%
$\dfrac{R}{\omega L}$
0%
$\dfrac{R}{(\omega L)^2+R^2}$
0%
$\omega LR$
0%
$\sqrt{\omega LR}$

Current in the circuit is wattless, if

0%
Inductance in the circuit is zero
0%
Resistance in the circuit is zero
0%
Current is alternating
0%
Resistance and inductance both are zero

Explanation

Current in the circuit is wattless,

Because power =

$i^2R$ , if

$R = 0$ , then

$P = 0$ .

An inductance of

$2H$ , capacitance of

$8\mu F$ and resistance of

$10K\Omega$ are connected in series to an AC source of

$100V$ with adjustable frequency. The angular frequency at which current in the circuit is maximum is :

0%
$50 \ rad/s$
0%
$100 \ rad/s$
0%
$125 \pi \ rad/s$
0%
$250 \ rad/s$

${W_1}and\;{W_2}$ are half power frequencies of series LCR circuit, Then the reasonant frequency of the circuit is-

0%
$\sqrt {{W_1}\;{W_2}}$
0%
$\frac{{{W_1} + {W_2}}}{2}$
0%
$\sqrt {\frac{{{W_1} + {W_2}}}{2}}$
0%
$\frac{{{W_1} -{W_2}}}{2}$

In an a.c. circuit consisting of resistance

$R$ and inductance

$L$ , the voltage across

$R$ is

$60$ volt and that across

$L$ is

$80$ Volt.The total Voltage across the combination is

0%
$140 V$
0%
$20 V$
0%
$100 V$
0%
$70 V$

In a series LCR circuit, the voltage across R is 100 volts and R = 1

$k\Omega$ with C =

$2\mu F$ . The resonant frequency

$\omega$ is 200 rad/s. At resonance the voltage across L is-

0%
$2.5 \times 10^{-2}V$
0%
$40 V$
0%
$250 V$
0%
$4 \times 10^{-3}V$

The instantaneous value of emf and current in an A.C. circuit are;

$E=1.414sin\left ( 100\pi t-\frac{\pi }{4} \right )$ ,

$I=0.707sin(100\pi t)$ . The admittance of the circuit will be _______ mho.

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

Explanation

Admittance =

$\dfrac{1}{impedance}= \dfrac{I_{max}}{V_{max}}$

$= \dfrac{0.707}{1.414}$

$= \dfrac{1}{2}$ mho

The instantaneous value of emf and current in an A.C. circuit are

$E=1.414$ sin(

$100\pi t-\frac{\pi }{4}$ ) ,

$I=0.707$ sin(

$100\pi t$ ) . The resistance of the circuit is

0%
$2\Omega$
0%
$\sqrt{2}\Omega$
0%
$\frac{1}{\sqrt{2}}\Omega$
0%
$\frac{1}{2}\Omega$

Explanation

resistance = impedance

$\times cos(\frac{\pi}{4})$

$= \frac{1.414}{0.707}\times \frac{1}{\sqrt{2}}$

$= \sqrt{2} \Omega$

$A$ series

$R-C$ circuit is connected to

$AC$ voltage source.

$C$ onsider two cases; (

$A$ ) when

$C$ is without a dielectric medium and (B) when

$C$ is filled with dielectric of constant

$4$ . The current

$I_{R}$ through the resistor and voltage

$V_{C}$ across the capacitor are compared in the two cases. Which of the following

$is/are$ true?

0%
$I_{R}^{A}>I_{R}^{B}$
0%
$I_{R}^{A} < I_{R}^{B}$
0%
$V_{c}^{A}>V_{c}^{B}$
0%
$V_{C}^{A} < V_{R}^{C}$

Explanation

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

$V^{2}= (V_{R})^{2} + (V_{C}^{2})$

$= (IR^{2}) + (\dfrac{I}{\omega C^{2}})$

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

The instantaneous value of emf and current in an A.C. circuit are;

$E=1.414sin\left ( 100\pi t-\frac{\pi }{4} \right )$ ,

$I=0.707sin(100\pi t)$ . The impedance of the circuit will be

0%
$1\Omega$
0%
$2\Omega$
0%
$\sqrt{2}\Omega$
0%
$\dfrac{1}{2}\Omega$

Explanation

Given :

$I_{max} = 0.707 \ A$ and

$E_{max} = 1.414 \$ volts
Impedance

$Z = \dfrac{E_{max}}{I_{max}}$

$Z= \dfrac{1.414}{0.707}$

$Z= 2 \Omega$

The capacitive reactance of 50

$\mu$ F capacitance at a frequency of

$2 \times 10^{3}$ Hz will be ____

$\Omega$

0%
$\frac{2}{\pi }$
0%
$\frac{3}{\pi }$
0%
$\frac{4}{\pi }$
0%
$\frac{5}{\pi }$

Explanation

Capacitive reactance =

$\frac{1}{\omega C}$

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

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

$= \frac{5}{\pi} \Omega$

The phase angle between current and voltage in a purely inductive circuit is :

0%
$zero$
0%
$\pi$
0%
$\pi$ /4
0%
$\pi$ /2

Explanation

In the above image waveform of current and voltage in puerly inductive circuit with time is shown.

It is clear from the image that current lags voltage by

$90^o$ .

Hence phase angle between current and voltage in purely inductive circuit is

$\pi /2$

In an LCR circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to :

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

Explanation

$\text{Resonant frequency,}\ f_{r}=\dfrac{1}{2\pi\sqrt{LC}}$

$\text{As the frequency is unchanged so}\ f_r=f'_r$

$LC=L'C'=L'(2C)$

$L'=\dfrac{L}{2}$

When the rms voltages

$V_L, V_C$ and

$V_R$ are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio

$V_L : V_C : V_R=1 : 2 : 3$ . If the rms voltage of the AC source is 100 V, then

$V_R$ is close to :

0%
50 V
0%
70 V
0%
90 V
0%
100 V

In a series resonant

$LCR$ circuit, the voltage across

$R$ is

$100\ volts$ and

$R = 1k\ \Omega$ with

$C=2\ \mu F$ . The resonant frequency is

$200\ rad/s$ . At resonance the voltage across

$L$ is

0%
$4$ x $10^{-3}\ V$
0%
$2.5$ x $10^{-2}\ V$
0%
$40\ V$
0%
$250\ V$

Explanation

The resonant angular frequency ,

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

$L=\dfrac{1}{\omega^2 C}=\dfrac{1}{200^2 \times 2\times 10^{-6}}=12.5 H$

At resonance , impedance

$Z=R$ so the current in the circuit is

$I=\dfrac{V}{Z}=\dfrac{V}{R}=\dfrac{100}{1000}=0.1 A$

Voltage across inductor,

$V_L=IX_L=I\omega L=0.1\times 200\times 12.5=250 V$

In an LCR circuit as shown below both switches are open initially. Now switch

$S_{1}$ is closed,

$S_{2}$ kept open. (

$q$ is charge on the capacitor and

$\tau=RC$ is capacitive time constant). Which of the following statement is correct?

0%
At $t=\tau,\ q=CV/2$
0%
At $t=2\tau,\ q=CV(1-e^{-2})$
0%
At $t=\frac{T}{2},\ q=CV(1-e^{-1})$
0%
Work done by the battery is half of the energy dissipated in the resistor.

Explanation

$C$ harge on the capacitor at any time `

$t$ ' is

$q=CV(1-e^{-t/\tau})$

$t=2\tau$

$q=\ CV(1-e^{-2})$

In a series LCR circuit

$\mathrm{R}=200\ Omega$ and the voltage and the frequency of the main supply is

$220$

$V$ and

$50$

$Hz$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by

$30^0$ . On taking out the inductor from the circuit the current leads the voltage by

$30^0$ . The power dissipated in the LCR circuit is

0%
$305$ $\mathrm{W}$
0%
$210$ $\mathrm{W}$
0%
$Zero \mathrm{W}$
0%
$242$ $\mathrm{W}$

Explanation

The given circuit is under resonance as

$\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$

Hence power dissipated in the circuit is

$\displaystyle \frac{V^2}{R}$

$\mathrm{P}=242\mathrm{W}$

An arc lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

0%
80H
0%
0.08H
0%
0.044H
0%
0.065H

Explanation

Resistance of lamp by ohm's law,

$R=\dfrac{V}{I}=\dfrac{80}{10}=8 \Omega$

When inductor is connected in series and source is applied,

$V=IZ$

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

Given,

$I=10\ A$ for lamp to work,

$\therefore 220=10\sqrt{8^2+(2 \times \pi \times 50 \times L)^2}$

Solving,

$L=0.0648\ H$

In an a.c. circuit, the instantaneous e.m.f. and current are given by

$e=100 \sin 30t$ ,

$i=20\sin\left(30t -\displaystyle\frac{\pi}{4}\right)$ . In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively.

0%
$\displaystyle\frac{50}{\sqrt{2}}, 0$
0%
$50, 0$
0%
$50, 10$
0%
$\displaystyle\frac{1000}{\sqrt{2}}, 10$

Explanation

$P_avg = V_{rms} I_{rms} \cos \theta$

$=(\dfrac{V_o}{\sqrt 2})(\dfrac{I_o}{{\sqrt2}}) cos \theta$

$=(\dfrac{100}{\sqrt 2})(\dfrac{20}{{\sqrt2}}) cos 45$

$=\dfrac{1000}{\sqrt2} watt$

Wattless current =

$I_{rms} sin\theta$

$\dfrac{I_o}{\sqrt2} sin \theta$

$\dfrac{20}{\sqrt2} sin 45$

$10 amp$

In an LCR series a.c. circuit, the voltage across each of the components. L, C and R is 50 V. The voltage across the LC combination will be;

0%
$50\ \text{V}$
0%
$50\sqrt{2}\mathrm{v}$
0%
$100\ \text{V}$
0%
$0\mathrm{V}$ (zero)

Explanation

In a series LCR circuit, the voltage across the inductor (L) and the capacitor (C) are in opposite phase.

So the voltage across LC combination will be

$(50-50)=0$ V

In the given circuit, the AC source has

$\omega =100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are) :

0%
The current through the circuit, $I$ is 0.3 A.
0%
The current through the circuit, $I$ is $0.3\sqrt{2}$ A.
0%
The voltage across $100\Omega$ resistor $=10\sqrt{2}$ V.
0%
The voltage across $50\Omega$ resistor $=10$ V.

Explanation

In the upper branch the net impedance,

$(C=100\mu F,R=100\Omega)$

Therefore, the net impedance will be:

$Z_1=\sqrt{\left(\dfrac{1}{\omega C}\right)^2+R^2}$

$=\sqrt{[100^2+100^2}$

$=100\sqrt2\Omega$

The current flowing the upper branch:

$I_1=\dfrac{V}{Z_1}=\dfrac{20}{100\sqrt2}\ A$

which leads by

$(45^0)$ w.r.t. voltage.

In the lower branch the net impedance,

$(L=0.5\ H\ ,R=50\Omega)$

Therefore, the net impedance will be:

$Z_2=\sqrt{\left({\omega L}\right)^2+R^2}$

$=\sqrt{(0.5\times100)^2+100^2}$

$=50\sqrt2\Omega$

The current flowing the lower branch:

$I_2=\dfrac{V}{Z_2}=\dfrac{20}{50\sqrt2}\ A$

which lags by

$(45^0)$ w.r.t. voltage.

Thus total current

$I$ is given by summation of phasors

$(I_1)$ and

$(I_2)$ which differ by

$(90^0)$ in phase and hence

$I=\sqrt{I^2_1+I^2_2}=\dfrac{1}{\sqrt10}\ A\approx 0.3\ A$

Also voltage across

$100\Omega\space \text{resistor} =I_1R_1=\dfrac{20}{100\sqrt2}\times 100=10\sqrt2V)$

Similarly across

$50\Omega\space \text{resistor}=I_2R_2=\dfrac{20}{50\sqrt2}\times 50=10\sqrt2V$

So, options

$A$ and

$C$ is correct.

In the circuit shown,

$L = 1 \mu H, C = 1 \mu F$ and

$R = 1 k\Omega$ . They are connected in series with an a.c. source

$V = V_0 \sin \omega t$ as shown. Which of the following options is/are correct?

0%
At $\omega >> 10^6 rad. s^{-1}$ , the circuit behaves like a capacitor
0%
At $\omega \sim 0$ the current flowing through the circuit becomes nearly zero
0%
The current will be in phase with the voltage if $\omega = 10^4rad. s^{-1}$
0%
The frequency at which the current will be in phase with the voltage is independent of $R$

Explanation

$Z= R + \hat j\omega L-\dfrac{j}{\omega C}$

$for \ \omega_0 = \dfrac{1}{\sqrt {LC}} = 10^6 \ rad$

$Z= R$

$for \ \omega< \omega_0, \ (\omega L - \dfrac{1}{\omega C}) <0$

Hence, Circuit is capacitive

$for \ \omega >\omega_0, \ (\omega L - \dfrac{1}{\omega C}) >0$

Hence, Circuit is Inductive

Hence, option A is wrong

$B -> \omega \approx0 \implies Z-> \infty \implies i->0$

Hence option B is correct

$C-> Current \ is \ in \ phase \ with \ voltage \ at \ resonance \ (\omega = \omega_0)$

Hence Option C is wrong

D-> Resonant frequency

$\omega_0 = \dfrac{1}{\sqrt{LC}}$

which is independent of R. Hence option D is correct.

A series LR circuit is connected to a voltage source with

$V\left( t \right) ={ V }_{ 0 }\sin { \Omega t }$ . After very large time, current

$I\left( t \right)$ behaves as

$\left( { t }_{ 0 }\gg \frac { L }{ R } \right)$

Explanation

Let

$L$ be the inductance and

$R$ be the resistance then the current

$I$ can be found by using Kirchoff's law as:

$\quad L\dfrac { dI }{ dt } -IR={ V }_{ 0 }\sin { \Omega t } \\ \Rightarrow \dfrac { dI }{ dt } -\dfrac { IR }{ L } =\dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } \\ \Rightarrow \dfrac { d }{ dt } (I{ e }^{ \frac { -Rt }{ L } })=\dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } { e }^{ \frac { -Rt }{ L } }\\ \Rightarrow I{ e }^{ \frac { -Rt }{ L } }=\int _{ 0 }^{ t }{ \dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } { e }^{ \frac { -Rt }{ L } } } dt\quad \\ \Rightarrow I=-\dfrac { { V }_{ 0 }L }{ { R }^{ 2 }+{ \Omega }^{ 2 }{ L }^{ 2 } } (\dfrac { R }{ L } \sin { \Omega t } +\Omega \cos { \Omega t } )\\ \Rightarrow I=-\dfrac { { V }_{ 0 } }{ \sqrt { { R }^{ 2 }+{ \Omega }^{ 2 }{ L }^{ 2 } } } \sin { (\Omega t+\phi ) }$

Where,

$\tan { \phi } =\dfrac { L\Omega }{ R }$

So, it is clearly seen that the solution is dictated by the forcing function which is sinusoidal. Therefore, the steady state solution will be sinusoidal in nature.

So, the correct answer is option B.

$A$ series

$R-C$ combination is connected to an

$AC$ voltage of angular frequency

$(\omega =500radian/s$ . lf the impedance of the

$R-C$ circuit is

$R\sqrt{1.25}$ , the time constant (in millisecond) of the circuit is

0%
1
0%
2
0%
3
0%
4

Explanation

Given :

$\omega = 500$

$radian/s$

Let the capacitance of the capacitor be

$C$ .

Thus resistance of capacitor

$X_C = \dfrac{1}{\omega C} = \dfrac{1}{500 C}$

Impedance of the circuit

$Z = R\sqrt{1.25}$

Using

$Z^2 = R^2 + X_C^2$

$\therefore$

$1.25 R^2 = R^2 + \dfrac{1}{(500)^2 C^2}$

$0.25 R^2 = \dfrac{1}{.25\times 10^6 C^2}$

$\implies$

$R^2 C^2 = \dfrac{10^{-6}}{(0.25)^2}$

We get time constant of the circuit

$RC = \dfrac{10^{-3}}{0.25} =0.004$ s

$= 4$ ms

$40 \mu F$ capacitor is connected to a

$200 V,\ 50 Hz \ ac$ supply. The rms value of the current
in the circuit is, nearly :

0%
$2.05 A$
0%
$2.5 A$
0%
$25.1 A$
0%
$1.7 A$

Explanation

RMS value of applied voltage =

$200\ V$

Impedance of a capacitor is given by:

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

Hence, rms current through it is:

$I_{rms}=\dfrac{V}{X_C}$

$I_{rms}=200 \times 2 \times \pi \times 50 \times 40 \times 10^{-6}$

$I_{rms}=2.51 \ A$

A series R-C circuit is connected to an alternating voltage source. Consider two situation:
(a) When capacitor is filled
(b) When capacitor is mica filled
Current through resister is i and voltage across capacitor is V then :

0%
$V_a=V_b$
0%
$V_a < V_b$
0%
$V_a>V_b$
0%
$i_a = i_b$

Explanation

For series C-R circuit, the impedance

$Z=\sqrt{R^2+X_C^2}$ where

$X_C=\frac{j}{\omega C}$ and current

$I=V/Z$

When the capacitor is filled by mica, the capacitance will be increased. If C increases,

$X_C$ decreases, so the current will increase and

hence voltage across resistance increases and voltage across capacitor decreases. thus,

$V_a>V_b$

An inductor

$20$ mH, a capacitor

$100$

$\mu$ F and a resistor

$50$

$\Omega$ are connected in series across a source of emf, V

$=10$

$\sin 314$ t. The power loss in the circuit is?

0%
$2.74$ W
0%
$0.79$ W
0%
$1.13$ W
0%
$0.43$ W

Explanation

$L = 20 mH$

$C = 100 \mu F$

$R = 50 \Omega$

$V = 10 sin(314 t)$

$V_0= 10$ ,

$\omega = 314$

$X_L = wL= 314 \times 20\times 10^{-3}= 6.28 \Omega$

$X_C = \dfrac{1}{\omega C}=31.8 \Omega$

$Z = \sqrt{R^2 + (X_C- X_L)^2} = 56.1$

$Power \ loss P=\dfrac{V_0^2 R}{2 Z^2}= 0.79 W$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

$\displaystyle\frac{L}{R}$ possesses dimension of time because it is equal to time constant. Time constant is the time which a current takes to reach

$67\%$ of its maximum value in an

$L-R$ circuit. To reduce rate of increase time constant needs to be increased.

A condenser of

$250\mu F$ is connected in parallel to a cell of inductance

$0.16$ mH, while its effective resistance is while its. 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

Given

Capacitance,

$C= 250\mu F= 250\times 10^{-6} F$

Inductance ,

$L=0.16mH= 0.16 \times 10^{-3}H$

Resonant frequency ,

$f=\dfrac1{2\pi \sqrt{LC}}= \dfrac1{2\times 3.14\times \sqrt{250\times 10^{-6} \times 0.16 \times 10^{-3}}}= 8\times 10^5 Hz$

Reciprocal of Impedance is:

0%
Susceptance
0%
Conductance
0%
Admittance
0%
Transconductance

Explanation

Impedance is the opposition a circuit presents to a current when a voltage is applied.

Admittance is a measure of how easily a circuit or device will allow a current to flow.

Admittance is defined as

$Y=\dfrac{1}{Z}$

where

$Z$ is the impedance of the circuit.

$10\ \mu F$ capacitor is connected across a

$200\ V$ ,

$50 \ Hz$ A.C. supply. The peak current through the circuit is :

0%
$0.6\ A$
0%
$0.6\sqrt{2}A$
0%
$(0.06\sqrt{2})A$
0%
$0.6{\pi}\ A$

Explanation

RMS value of applied voltage =

$200\ V$

Impedance of a capacitor is given by:

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

Hence, rms current through it is:

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

$I=200 \times 2 \times \pi \times 50 \times 10 \times 10^{-6}$

$I=0.2 \pi \ A$

Peak current,

$I_{max}=\sqrt 2 I$

$=0.628 \sqrt 2\ A$

If resonant frequency is

$f$ and capacitance become

$4$ times, then resonant frequency will be :

0%
$\dfrac {f}{2}$
0%
$2f$
0%
$f$
0%
$\dfrac {f}{4}$

Explanation

Frequency,

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

$= 4C$ , then

$f' = \dfrac {1}{2\pi \sqrt {4LC}} = \dfrac {f}{2}$ .

In a circuit, the frequency is

$f=\dfrac{1000}{2\pi }$ Hz and the inductance is 2 henry, then the reactance will be

0%
200 $\Omega$
0%
200 $\mu$ $\Omega$
0%
2000 $\Omega$
0%
2000 $\mu$ $\Omega$

Explanation

Given :

$f = \dfrac{1000}{2\pi}$

$L = 2 \ H$
Thus

$w = 2\pi f = 2\pi\times \dfrac{1000}{2\pi} = 1000$
Reactance

$X_L =\omega L$

$X_L = 1000\times 2= 2000\Omega$

The instantaneous value of emf and current in an A.C. circuit are:

$E=1.414sin\left ( 100\pi t-\frac{\pi }{4} \right )$ ,

$I=0.707sin(100\pi t)$ .The reactance of the circuit
will be

0%
$\sqrt{2}$ $\frac{1}{2}\Omega$
0%
2 $\Omega$
0%
$\frac{1}{2}\Omega$
0%
$\frac{1}{\sqrt{2}}\Omega$

Explanation

reactance of the circuit =

$\frac{E_{max}}{I_{max}}$

$= \frac{1.414}{0.707}$

$= 2 \Omega$

The instantaneous value of emf and current in an
A.C. circuit are;

$E=1.414sin\left ( 100\pi t-\dfrac{\pi }{4} \right )$ ,

$I=0.707sin(100\pi t)$ . RMS value of current will be

0%
1A
0%
$\dfrac{1}{\sqrt{2}}A$
0%
$\sqrt{2}A$
0%
$\dfrac{1}{2}A$

Explanation

Given :

$I_{max} = 0.707 \ A$
Rms value of current

$I_{rms} = \dfrac{I_{max}}{\sqrt{2}}$

$I_{rms}= \dfrac{.707}{\sqrt{2}}$

$= \dfrac{1}{2}A$

The instantaneous value of emf and current in an A.C. circuit are;

$E=1.414sin\left ( 100\pi t-\dfrac{\pi }{4} \right )$ ,

$I=0.707sin(100\pi t)$ . The RMS value of emf will be

0%
$2\sqrt{2}V$
0%
$1 V$
0%
$\dfrac{1}{2}V$
0%
$\dfrac{1}{2\sqrt{2}}V$

Explanation

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

$= \dfrac{1.414}{\sqrt{2}}$

$= 1 V$

In an AC circuit containing only capacitance, the current :

0%
leads the voltage by 180
0%
lags the voltage by 90
0%
leads the voltage by 90
0%
remains in phase with the voltage

Explanation

Current leads by $90^{\circ}$

An inductive coil has a resistance of 100

$\Omega$ . When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by

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

0%
$\dfrac{1}{100\pi }$ H
0%
$\dfrac{1}{20\pi }$ H
0%
$\dfrac{1}{40\pi }$ H
0%
$\dfrac{1}{60\pi }$ H

Explanation

$X_L = \omega L$

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

$= \dfrac{100}{2\pi f}$

$= \dfrac{100}{2\pi 1000}$

$= \dfrac{1}{20\pi}H$

If the phase difference between Alternating Voltage and Alternating Current is

$\frac{\pi }{6}$ and the resistance in the circuit is

$\sqrt{300}\Omega$ , then the impedance of the circuit will be

0%
25 $\Omega$
0%
50 $\Omega$
0%
20 $\Omega$
0%
100 $\Omega$

Explanation

impedance

$\times cos \theta =$ resistance
impedance =

$\frac{resistance}{cos \theta}$

$= \frac{\sqrt{300}}{cos \pi /6}$

$= 20 \Omega$

The inductance of a resistanceless coil is 0.5 henry. In the coil, the value of alternating current is 0.2 A, whose frequency is 50 Hz. The reactance of circuit is

0%
15.7 $\Omega$
0%
157 $\Omega$
0%
1.57 $\Omega$
0%
757 $\Omega$

Explanation

Reactance

$= \omega L$

$= 2\pi f L$

$= 2\pi \times 50 \times 0.5$

$= 157 \Omega$

The capacitive reactance at 1600Hz is 81

$\Omega$ .When the frequency is doubled then the capacitive reactance will be:

0%
40.5 $\Omega$
0%
81 $\Omega$
0%
162 $\Omega$
0%
zero

Explanation

Capacitive reactance = $\dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$
$81 = \dfrac{1}{2\pi \times 1600 C}$
$C = \dfrac{1}{2\pi \times 1600 \times 81}$
Capacitive reactance = $\dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$
$81 = \dfrac{1}{2\pi \times 2 \times 1600 \times \dfrac{1}{2\pi \times 1600 \times 81}}$
$C = \dfrac{81}{2} = 40.5 \Omega$

In an LCR circuit, current is I

$=$ 10sin100

$\pi$ t from an A.C. source. The value of average voltage at the ends of the resistance R

$=$ 10

$\Omega$ will be

0%
25V
0%
$\dfrac{25}{\sqrt{2}}V$
0%
zero V
0%
1V

In an A.C. circuit the potential difference across an inductance and resistance joined in series are respectively 16V and 20V. The total potential difference across the circuit is

0%
20.0V
0%
25.6V
0%
31.9V
0%
53.5V

Explanation

Potential difference across the circuit

$V = \sqrt{V_R^2 + V_L^2 } = \sqrt{20^2 + 16^2} = \sqrt{656} = 25.6 V$

A coil is used in a circuit in which an A.C. of frequency 50Hz is flowing. The self-inductance of the coil, in order to produce an impedance of 100

$\Omega$ , will be ___ H

0%
$\pi$
0%
$\frac{1}{\pi }$
0%
$\pi ^{2}$
0%
$\frac{1}{\pi^{2} }$

Explanation

impedance = WL

$= 2\pi f L$

$100 = 2\pi 50\times L$

$L = \frac{1}{\pi}H$

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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