MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 4

Figure shows a series LCR circuit connected to a variable frequency 200V source. The source frequency which drives the circuit at resonance is

0%
50 Hz
0%
(50 / $\pi$ )Hz
0%
25 Hz
0%
(25 / $\pi$ ) Hz

Explanation

For resonance frequency

$X_L = X_C$

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

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

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

$= \dfrac{1}{2\pi}\times \dfrac{1}{5\times 80\times 10^{-6}}$

$= \dfrac{1}{2\pi}\times \dfrac{10^3}{20}$

$= \dfrac{25}{\pi}Hz$

A 100

$\Omega$ resistance is connected in series with a 4H inductor. The voltage across the resistor is

$V_{R}=2sin(1000t)V$ . The voltage across the inductor is:

0%
$80sin\left ( 1000t+\dfrac{\pi }{2} \right )$
0%
$40sin\left ( 1000t+\dfrac{\pi }{2} \right )$
0%
$80sin\left ( 1000t-\dfrac{\pi }{2} \right )$
0%
$40sin\left ( 1000t-\dfrac{\pi }{2} \right )$

Explanation

$V_R = 2sin (1000t)V$ So, $V_{max}(R) = 2V$

So, $w = 1000$

$2\pi f = 1000$

$f = \dfrac{500}{\pi}Hz$

$X_L = wL$

$= 1000\times 4$

$= 4000 \Omega$

$R = 100 \Omega$

$i_R = \dfrac{V_R}{R} = \dfrac{2}{100} = 2\times 10^{-2}A = i_L$

( $\because$ current is same in series)

$V_{max}(L) = i_L\times X_L$

$= 4000\times 2\times 10^{-2}$

$= 80V$

So, $V_L = 80 sin (1000t + \dfrac{\pi}{2})$

Since in inductor voltage lead by $\dfrac{\pi}{2}$

In the following circuit, the values of current flowing in the circuit at f=0 and f=

$\infty$ will respectively be

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

Explanation

In a LCR circuit current at

$t = 0$ is zero because inductor does not allow flow of current,

and at

$t = \infty$ also current is zero because capacitor does not allow flow of current.

The inductive reactance of a coil is

$1000\Omega .$ If its self inductance and frequency both are increased two times, then inductive reactance will be

0%
$1000 \Omega$
0%
$2000 \Omega$
0%
$4000 \Omega$
0%
$16000 \Omega$

A condenser of 10

$\mu$ F and an inductor of 1H are connected in series with an A.C. source of frequency 50Hz. The impedance of the combination will be (take

$\pi^{2}$

$=$ 10 ):

0%
zero
0%
Infinity
0%
44.7 $\Omega$
0%
5.67 $\Omega$

Explanation

Impedance = $X_L - X_C = \omega L-\dfrac{1}{\omega C}$

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

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

$= \pi \left (100 - \dfrac{10^6}{2{\pi}^2\times 50\times 10} \right )$

$= \pi \left(100-\dfrac{10^6}{2\times 10 \times 50 \times10} \right)$

$(\because {\pi}^2 = 10)$

$= 0$

So, impedance $= 0$

In an L-C-R series circuit,

$R=\sqrt{5}\Omega ,X_{L}=9\Omega ,X_{C}=7\Omega$ . If applied voltage in the circuit is 50V then impedance of the circuit in ohm will be

0%
$2$
0%
$3$
0%
$2\sqrt{5}$
0%
$3\sqrt{5}$

Explanation

Impedance =

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

$= \sqrt{5+(9-7)^2}$

$= \sqrt{5+(2)^2}$

$= \sqrt{5+4}$

$= \sqrt{9}$

$= 3\Omega$

A coil of inductance 0.1H is connected to 50V, 100Hz generator and current is found to be 0.5A. The potential difference across resistance of the coil is

0%
15V
0%
20V
0%
25V
0%
39V

Explanation

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

$= \dfrac{50}{0.5}$

$= 100\Omega$

$Z^2 = X_L^+R^2$

$(100)^2 = (2\pi fL)^2+R^2$

$(100)^2-[2\pi 100(0.1)]^2 = R^2$

$6052 = R^2$

$R = 78\Omega$

$V_R = iR$

$= 0.5\times 78$

$= 39V$

In an LR circuit, R

$=$ 10

$\Omega$ and L

$=$ 2H. If an alternating voltage of 120V and 60Hz is connected in this circuit, then the value of current flowing in it will be _______ A (nearly)

0%
0.32
0%
0.16
0%
0.48
0%
0.8

Explanation

Impedance(z) =

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

$= \sqrt{(10)^2+(2\pi fL)^2}$

$= \sqrt{(10)^2+(2\pi \times 60\times 2)^2}$

$= \sqrt{100+{\pi}^2\times 57600}$

$= \sqrt{100+568480}$

$= 754.04$

$i = \dfrac{V}{z}$

$= \dfrac{120}{754.04}$

$= 0.16A$

A 220V, 50Hz a.c. generator is connected to an inductor and a 50

$\Omega$ resistance in series. The current in the circuit is 1.0A. The P.D. across the inductor is:

0%
102.2 V
0%
186.4 V
0%
214.24 V
0%
302 V

Explanation

Current =

$\dfrac{V}{Z}$

$1 = \dfrac{220}{Z}$

$\Rightarrow Z = 220\Omega$

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

$(220)^2 = X_L^2+(50)^2$

$X_L^2 = 45,900$

$X_L = 214.24$

$V_L = iX_L$

$= (1)214.24$

$= 214.24V$

A circuit operating at

$\dfrac{360}{2\pi }$ Hz

$\Omega$ contains a 1 F

$\mu$ capacitor and a 20 resistor. The inductor must be added in series to make the phase angle for the circuit zero is

0%
7.7 H
0%
10 H
0%
3.5 H
0%
15 H

Explanation

To make phase angle zero, circuit should be purely resistive, so, reactance should be zero.

Hence,

$X_L = X_C$

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

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

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

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

$L= \dfrac{10^6}{129600}$

$L= 7.7H$

In the series L-C-R circuit figure, the voltmeter and ammeter readings are

0%
$V$ = $100 volt, I$ = $2 amp$
0%
$V$ = $100 volt, I$ = $5 amp$
0%
$V$ = $1000 volt, I$ = $2 amp$
0%
$V$ = $300 volt, I$ = $1 amp$

Explanation

Here,

$V_L=V_C$ .

$V_L$ and

$V_C$ are in oppositr phases. Hence they will cancel each other. Now the resultant potential difference is equal to the applied potential difference.

i.e.

$V=100V$

Impedence,

$Z=R(X_L \, or \, X_C)$

Therefore,

$I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{R}=\dfrac{100}{50}=2\,A$

So, the voltmeter reading is

$100\,V$ and ammeter reading is

$2\,A$ .

A resistor

$R$ and capacitor

$C$ are connected in series across an AC source of rms voltage

$5 V$ . If the rms voltage across

$C$ is

$3 V$ then that across

$R$ is :

0%
$1V$
0%
$2V$
0%
$3V$
0%
$4V$

Explanation

Source voltage =

$\sqrt{(Capacitor\ voltage)^2+(Resistor \ voltage)^2}$

$[V_{s(rms)}]^2 = [V_{C(rms)}]^2+[V_{R(rms)}]^2$

$(5)^2 = (3)^2+[V_{R(rms)}]^2$

$25-9 = [V_{R(rms)}]^2$

$16 = [V_{R(rms)}]^2$

$V_{R(rms)}=4V$

The potential difference between the ends of a resistance R is

$V_{R}$ , between the ends of capacitor is

$V_{C}$

$=$ 2

$V_{R}$ and between the ends of inductance is

$V_{L}$

$=$ 3

$V_{R}$ then the alternating potential of the source in terms of

$V_{R}$ will be:

0%
$\sqrt{2}V_{R}$
0%
$V_{R}$
0%
$\dfrac{V_{R}}{\sqrt{2}}$
0%
$5V_{R}$

Explanation

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

$= \sqrt{V_R^2+(3V_R-2V_R)^2}$

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

$= V_R \sqrt{2}$ .

The natural frequency of an LC - circuit is

$1,25,000$ cycles per second. Then the capacitor C is replaced by another capacitor with a dielectric medium of dielectric constant k. In this case, the frequency decreases by

$25 kHz$ . The value of k is:

0%
$1.56$
0%
$1.7$
0%
$3.0$
0%
$2.1$

Explanation

Natural frequency $= 125000 Hz$

New frequency $= 125000 - 25000$

$= 100000 Hz$

Frequency of L-C circuit

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

So, $2\pi f_1 = \dfrac{1}{\sqrt{LC_1}}$ ---(1)

$2\pi f_2 = \dfrac{1}{\sqrt{LC_2}}$ ---(2)

Dividing (1) by (2)

$\dfrac{f_1}{f_2} = \sqrt{\dfrac{C_2}{C_1}}$

$\sqrt{\dfrac{C_2}{C_1}} = \dfrac{125000}{100000} = \dfrac{5}{4}$

$\dfrac{C_2}{C_1} = \dfrac{25}{16}$

$k = \dfrac{C_2}{C_1} = \dfrac{25}{16} = 1.56$

At resonance, the angle

$\phi$ is

0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{6}$
0%
zero

Explanation

At resonance:

$L\omega=\dfrac{1}{C\omega}$ by definition.
therefore form the passage,

$\tan \phi = \dfrac{ L\omega - \dfrac{1}{C\omega} }{R} =0 \Rightarrow \phi = \tan^{-1}(0)=0$

The reading of voltmeter and ammeter in the following figure will respectively be :

0%
0 and 2A
0%
2A and 0V
0%
2V and 2A
0%
0V and 0A

Explanation

In the problem $X_C = 4\Omega$
and $X_L = 4\Omega$
So, V across $X_C$ and $X_L$ will be same and in opposite direction, So net voltage will be zero. Since voltmeter is connected parallel to Capacitor and inductor so, it will read 0 volts.

Current $=$ $\dfrac{V}{impedance}$

$Z = R$ as $X_L = X_C$

Current $= \dfrac{90} {45} = 2\ A$

In the following diagram, the value of emf of A.C. source will be :

0%
$40 V$
0%
$40\sqrt{2}V$
0%
$\dfrac{40}{\sqrt{2}}V$
0%
$160 V$

Explanation

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

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

$= \sqrt{ 40^2 + 40^2}$

$= 40 \sqrt{2} V$

An inductance of 0.2 H and a resistance of 100

$\Omega$ are connected in series to an A.C. 180 V, 50 Hz supply. The apparent current flowing in the circuit will be

0%
0.525 A
0%
5.25 A
0%
1.525 A
0%
15.25 A

Explanation

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

$= \sqrt{(2\pi fL)^2+R^2}$

$= \sqrt{(2\pi 50\times 0.2)^2+(100)^2}$

$= \sqrt{13947.84}$

$= 118.1 \Omega$

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

$= \dfrac{180}{118.1}$

$= 1.525A$

$LCR$ circuit has

$L=10\ mH, R=3\Omega$ , and

$C =1\mu$

$F$ connected in series to a source of

$15 \cos\omega t$ volt. The current amplitude at a frequency that is

$10\%$ lower than the resonant frequency is:

0%
$0.5 A$
0%
$0.7 A$
0%
$0.9 A$
0%
$1.1 A$

Explanation

Resonant frequency,

$w_o=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(10\times 10^{-3})(10^{-6})}}=10^4 rad/s$

New frequency,

$\omega=(0.9)\omega_o=9\times 10^3 rad/s$

We have new

${ X }_{ L }=\omega L=9\times { 10 }^{ 3 }\times 10\times { 10 }^{ -3 }=90\Omega$ and

$X_C= \dfrac {1}{\omega C}=111.11$

$ohm$ .

Thus we calculate new

$Z$ as

$\sqrt{3^2+[90-111.11]^2}$

$=21.32 \Omega$

Current amplitude

$=\dfrac{V_o}{Z}=\dfrac{15}{21.32}=0.7 A$

In the circuit shown in the figure, (neglecting source resistance) the voltmeter and ammeter readings will respectively be

0%
0 V, 8 A
0%
150 V, 8 A
0%
150 V, 3 A
0%
0 V, 3 A

Explanation

Voltmeter reading is zero since voltage in both Capacitor and inductor is same in magnitude but opposite in sign because current in series is same and both have same reactance.

So,

$i = \dfrac{V}{R}$

$= \dfrac{240}{30}$

$= 8A$

In a series LCR circuit the voltage across resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be

0%
$10 V$
0%
$(10/\sqrt{2}V)$
0%
$(10/\sqrt{3}V)$
0%
$20 V$

In the following circuit, the potential of source is

$E_0=200$ volts,

$R=200\Omega$ ,

$L=0.1$ henry,

$C=10.6$ farad and frequency is variable, then the current at f=0 and f=

$\infty$ is :

0%
0A, 10A
0%
10A, 0A
0%
10A, 10A
0%
0A, 0A

Explanation

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

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

when

$w\rightarrow0$ then

$I=\dfrac { V\omega C }{ \sqrt { { R }^{ 2 }+{ 1 }^{ 2 } } }$

$\Rightarrow I\rightarrow 0$

or when

$w\rightarrow\infty$ ,

${ R }^{ 2 }+{ ({ \omega }^{ 2 }LC-1) }^{ 2 }\rightarrow \infty \quad \quad or\\$

$\dfrac { 1 }{ \sqrt { { R }^{ 2 }+{ ({ \omega }^{ 2 }LC-1) }^{ 2 } } } \rightarrow 0\\$

$\Rightarrow I\rightarrow 0$

therefore, current is zero in both case.

The values of

$X_{L}, X_{C}$ and R in series with an A.C. circuit are 8

$\Omega$ , 6

$\Omega$ and 10

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

$\Omega$

0%
$10.2$
0%
$12.2$
0%
$10$
0%
$24.4$

Explanation

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

$= \sqrt{(10)^2+(8-6)^2}$

$= \sqrt{10^2+2^2}$

$= \sqrt{100+4}$

$= 10.2 \Omega$

A condenser and a 30

$\Omega$ resistance are connected in series. When they are connected to 120V A.C. source then the current flowing in the circuit is 1A The p.d. across the ends of the condenser will be nearly

0%
1 V
0%
116 V
0%
zero
0%
220 V

Explanation

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

$Z = \dfrac{V}{i} = \dfrac{120}{1} = 120\Omega$

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

$(120)^2 = (30)^2+X_C^2$

$X_C^2 = 13500$

$X_C = 116\Omega$

$V_C = i X_C$

$= 1(116)$

$= 116V$

An LCR series circuit with 100

$\Omega$ resistance is connected to an ac source of 200V and of frequency of 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by

$60^{0}$ . When only the inductance is removed, the current leads the voltage by

$60^{0}$ the current through the circuit is:

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

Explanation

Since current lead and lag are same So, circuit is in resonance, so, circuit is purely resistive circuit.

So, $i = \dfrac{V}{R}$

$= \dfrac{200}{100}$

$= 2A$

In the given figure as shown, the reading of the ammeter in ampere is

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

Explanation

$I = \dfrac{E}{\sqrt{R^2 + (X_L - X_C)^2}}$

$= \dfrac{ 110}{\sqrt{ 55^2 + ( 2-2)^2}}$

$= \dfrac{110}{55} = 2 A$

An alternating voltage

$V =200\sqrt{2} \sin{100t}$ , Where

$V$ is in volt and

$t$ is in seconds, is connected to a series combination of 1

$\mu\text{F}$ capacitor and

$10\ \text{k}\Omega$ resistor through an AC ammeter. The reading of the ammeter will be_____

0%
$\sqrt{2}\ \text{mA}$
0%
$10\sqrt{2}\ \text{mA}$
0%
$2\ \text{mA}$
0%
20mA

Explanation

$i=\dfrac { { V }_{ rms } }{ \sqrt { { R }^{ 2 }+{ \left( \dfrac { 1 }{ \omega C } -\omega L \right) }^{ 2 } } } \\$

$\Rightarrow i=\dfrac { 200\sqrt { 2 } \dfrac { 1 }{ \sqrt { 2 } } }{ \sqrt { { 10000 }^{ 2 }+{ \left( \dfrac { 1 }{ 100\times { 10 }^{ -6 } } -100\times 0 \right) }^{ 2 } } } =\dfrac { 200 }{ \sqrt { { 2\times 10000 }^{ 2 } } } \\$

$=10\sqrt { 2 }\ mA$

If alternating current of rms value 'a' flows through resistance R then power loss in resistance is :

0%
Zero
0%
$a^2R$
0%
$\dfrac{a^2R}{2}$
0%
$2 a^2R$

Explanation

[B]

$P = {I^2}_{RMS}R$

$= {a^2}R$

The capacitor of an oscillatory circuit of negligible resistance is enclosed in a evacuated container. The frequency of the circuit is 150 kHZ and when the container is filled with a gas, the frequency changes by 100 HZ. The dielectric constant of the gas.

0%
2
0%
1.53
0%
1.0012
0%
3

Explanation

We know frequency is given by:

$n_1 = \dfrac { 1 }{ 2\pi \sqrt { LC_1 } }$

And

$n_2=\dfrac{1}{2\pi \sqrt{LKC_1}}$

Thus we get the ratio of frequencies as

$\dfrac{n_1}{n_2}=\dfrac{150000}{149900}=\sqrt{k}$

Solving the above equation we get,

$k \approx 1.0012$

In the AC circuit shown,

$X_{L}=7\Omega ,R=4\Omega\ and\ X_{c}=4\Omega .$ The reading of the ideal voltmeter

$V_{2}\ is\ 8\sqrt{2}$ . The reading of the ideal ammeter will be:

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

Explanation

From the given figure,

Given,

$X_L=7\Omega$

$X_C=4\Omega$

$R=4\Omega$

The reading of ideal voltmeter is given by

$V_2=8\sqrt{2}V$

From Ohm's law,

$V_2=I( {R+jX_C})$ (Capacitor and resistor are in series)

$I=\dfrac{V_2}{R+jX_C}$

$I=\dfrac{V_2}{\sqrt{(R)^2+(X_C)^2}}$

$I=\dfrac{8\sqrt{2}}{\sqrt{(4)^2+(4)^2}}$

$I=\dfrac{8\sqrt{2}}{4\sqrt{2}}$

$I=2A$

The correct option is B.

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be:-

0%
$\frac{f} {4}$
0%
8f
0%
$\dfrac{f} {2\sqrt{2}}$
0%
$\frac{f} {2}$

Explanation

f =

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

$^{'}$ =

$\dfrac{1} {2 \pi \sqrt{2L\left (4C \right )}}$
Therefore f

$^{'}$ =

$\left (\dfrac{1} {2 \sqrt{2}} \right ) \dfrac{1} {2 \pi \sqrt{LC}}$ =

$\dfrac{f} {2 \sqrt{2}}$

For a series LCR circuit the power loss at resonance is : -

0%
$\dfrac{V^{2}}{\left [ \omega L-\dfrac{1}{\omega C} \right ]}$
0%
$I^{2}L\omega$
0%
$I^{2}R$
0%
$\dfrac{V^{2}}{C\omega }$

The natural frequency of the circuit shown in fig. is

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

Explanation

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

$C_{eq}=C/2$

$\therefore natural frequency = \dfrac{1}{\sqrt {L_{eq} C_{eq} }}$

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

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

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

$\therefore$ Frequency of oscillation

$f=\dfrac{1}{2\pi \sqrt{L_{eq}C_{eq}}}=\dfrac{1}{6\pi \sqrt{LC}}$

The capacitance in an oscillatory LC circuit is increased by 1%. The change in inductance required to restore its frequency of oscillation is to

0%
decrease it by 0.5%
0%
increase it by 1%
0%
decrease it by 1%
0%
decrease it by 2%

Explanation

$\omega_r=constant$

$\Rightarrow LC=constant$

$\Rightarrow LdC+CdL=0$

$\Rightarrow \dfrac {dL}{L}=-\dfrac {dC}{C}=-1 %$

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

Resonance of inductor, $X_L = 2 \pi f L$

Here f is the frequency of Ac, L is inductance of inductor. Reactance is directly proportional to frequency. Therefore, with increase in frequency reactance also increases.

When current passes through inductor, magnetic field develop across inductor, magnetic field cut by inductor and emf is induced. This emf is called back emf which opposes the applied emf. Back emf increase with increase in frequency of applied emf. Therefore opposition, i.e. reactance also increases with increase in frequency.

The circuit shown in Fig. acts as a

0%
tuned filter
0%
low pass filter
0%
high pass filter
0%
rectifier

A steady potential difference of 100 V produces heat at a constant rate in a resistor. The alternating voltage which will produce half the heating effect in the same resister will be

0%
100 V
0%
50 V
0%
70.7 V
0%
141.4 V

Explanation

The power supplied to the resistor by DC source

$=\dfrac{V_{dc}^2}{R}$

Energy given by AC source

$=\int_0^T \dfrac{V_0^2}{R}dt$

Hence,

$\int_0^T \dfrac{V_0^2sin^2\omega t}{R}dt=\dfrac{1}{2}\times \dfrac{V_{dc}^2T}{R}$

$\implies V_0=V_{dc}=100V$

In an

$\text{LCR}$ circuit the capacitance is made

$\dfrac{1}{4}^{th}$ then what should be the change in inductance that the circuit remains in resonance again?

0%
$8\ \text{times}$
0%
$\dfrac{1}{4}\ \text{times}$
0%
$2\ \text{times}$
0%
$4\ \text{times}$

Explanation

$\displaystyle f_0 = \dfrac{1}{2\pi \sqrt{LC}}$ .

To keep frequency unchanged

$L$ be made 4 times.

The square root of the product of inductance and capacitance has dimensions of

0%
length
0%
mass
0%
time
0%
dimensionless

Explanation

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

$or \ \ \ \sqrt{LC} = T$

A $50\ \text{Hz},\ 20\ \text{V}\ \text{AC}$ source is connected across $R$ series circuit as shown in Figure If the voltage across $R$ is $12\text{V}$ then voltage across capacitor is

0%
$8\ \text{V}$
0%
$16\ \text{V}$
0%
$10\ \text{V}$
0%
$\text{cannot be predicted as values of R and C are not given.}$

Explanation

$V^2 = V_R^2 + V^2_c$

$20^2 = 12^2 + V^2_c$

$V_c=\sqrt{20^2-12^2}$

$V_c = 16\ \text{V}$

A capacitor has capacitance

$0.5nF$ . A choke of

$5\mu H$ is connected in series. An electromagnetic wave of wavelength

$\lambda$ is found to resonate with it. Find

$\lambda$ (in meter).

0%
$10 \pi$
0%
$20\pi$
0%
$30 \pi$
0%
$5\pi$

Explanation

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

$\lambda = \dfrac{C}{f} = 3\times 10^8 \times 2\pi \sqrt{LC}$

$= 3\times 10^8\times 2\pi \sqrt{5\times 10^{-6}\times 5\times 10^{-10}} = 30\pi\ \text{m}$

In above circuit, what is the potential drop across

$ZY$ ?

0%
160 V
0%
$\displaystyle 80\sqrt{80}$
0%
80 V
0%
zero

Explanation

$V_{ZY} = V_C-V_L = 0$

When

$4\ V$ DC is connected across an inductor, current is

$0.2\ A$ . When AC of

$4\ V$ is applied the current is

$0.1\ A$ . Then self inductance of the coil is:
[Given

$\omega = 1000\ rads^{-1}$ ]

0%
$20\ mH$
0%
$40\ mH$
0%
$20 \sqrt{3}\ mH$
0%
none of these

Explanation

$\displaystyle Z = \frac{4}{0.1} = 40\ \Omega = \sqrt{R^2 + (L\omega)^2}$

and

$\displaystyle R = \frac{4}{0.2} = 20\ \Omega$

$\displaystyle \therefore\ L\omega = 20\sqrt3 \Omega$

$\displaystyle L = 20\sqrt3\ mH$

A leaky capacitor

$10\ \Omega/ 60^{\circ}$ is connected in series with a 10

$\Omega$ resistance. Find the overall impedance.

0%
$\displaystyle 10\ \Omega$
0%
$\displaystyle 10 \sqrt 2\ \Omega$
0%
$\displaystyle 15\ \Omega$
0%
$\displaystyle 10\sqrt 3\ \Omega$

Explanation

$\displaystyle \sqrt{r^2+X^2_c} = 10\ \text{and}\ \dfrac{X_c}{r} = \tan 60$

$\displaystyle \therefore \sqrt{r^2+(\sqrt 3r)^2} = 10$ or

$\displaystyle r = 5 \Omega, X_c = 5\sqrt 3 \Omega$

$\displaystyle |Z| = \sqrt{(10+5)^2+5\sqrt 3} = 10\sqrt 3 \Omega$ ;

$\displaystyle \Phi = \tan^{-1} \frac{X_c}{R} = tan^{-1} \left(\dfrac{5\sqrt 3}{15}\right)$

$\displaystyle \tan^{-1} \left(\dfrac{1}{\sqrt 3}\right)$ or

$\displaystyle \Phi = 30^{\circ}$

If the output is taken across a capacitor in a series RLC circuit then it acts as

0%
band-pass filter
0%
high-pass filter
0%
low-pass filter
0%
band reject filter

In the given circuit what is the potential drop across resistance?

0%
40 V
0%
80 V
0%
120 V
0%
zero

Explanation

At resonance the voltage across L and C are same but opposite .

So, at resonance

$|V_L| = |V_C|$

Therefore

$\displaystyle V_R = V_{app} = 120 V$

$\displaystyle I = 6\cos wt + 8\sin wt$ is applied across a resistor of 40

$\Omega$ . Find the potential difference across the resistor.

0%
660 V
0%
80 V
0%
330 V
0%
400 V

Explanation

Here

$I=10(\dfrac{6}{10}\cos\omega t+\dfrac{8}{10}\sin \omega t)$

$=10cos(\delta- \omega t)$

$|I|=10A$

$V=IR=400V$

An inductor

$10 \Omega/60^{\circ}$ is connected to a

$5 \Omega$ resistance in series. Find net impedance.

0%
$15 \Omega$
0%
$12 \Omega$
0%
$13.2 \Omega$
0%
$18 \Omega$

Explanation

$\displaystyle 10 = \sqrt{r^2+X^2_L} \ and\ \frac{X_L}{r} = tan 60$

$\displaystyle 10 = \sqrt{r^2+(r\sqrt 3)^2} \ or\ r = 5 \Omega, X_L = 5\sqrt 3 \Omega.$

$\displaystyle Z = \sqrt{(5+5)^2 + (5\sqrt3)^2} = \sqrt{175} = 13.2 \Omega$

$\displaystyle tan\Phi = \frac{X_L}{R+r} = \frac{5\sqrt3}{10}$

or

$\displaystyle \Phi = tan^{-1} \left(\frac{\sqrt3}{2}\right)$

A capacitor acts as an infinite resistance for

0%
$DC$
0%
$AC$
0%
$DC$ as well as $AC$
0%
neither $AC$ nor $DC$

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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