In $$LCR$$ series $$AC$$ circuit

If $$R$$ is increased current will decrease.

If $$L$$ is increased current will decrease.

If $$C$$ is increased current will increase.

If $$C$$ is increased current will decrease.

MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 6

In an

$AC$ circuit, the impedance is

$\sqrt 3$ times the reactance, then the phase angle is

0%
$60^o$
0%
$30^o$
0%
zero
0%
none of these

Explanation

$\displaystyle \sin \phi=\frac{X}{Z}$

$\displaystyle =\frac{1}{\sqrt 3}$

$\displaystyle \therefore \phi=\sin^{-1}\left(\frac{1}{\sqrt 3}\right)$

In a L-C-R circuit, as the frequency of an alternating current increases the impedance of the circuit

0%
increases continuously.
0%
decreases continuously.
0%
remains constant.
0%
None of these.

Explanation

Impedance first decreases then increases. At resonance frequency

$Z$ is minimum.

A circuit contains resistance

$R$ and an inductance

$L$ in series. An alternating voltage

$V=V_0\sin \omega t$ is applied across it. The currents in

$R$ and

$L$ respectively will be

0%
$I_R=I_0\cos \omega t, I_L=I_0\cos \omega t$
0%
$I_R=-I_0\sin \omega t, I_L=I_0\cos \omega t$
0%
$I_R=I_0\sin \omega t, I_L=-I_0\cos \omega t$
0%
None of the above.

Explanation

The effective impedance is

$Z= \sqrt{R^2 + (\omega ^2 L^2}$

$I = \frac{V_0}{Z}\sin( \omega t + \phi)$ where

$\tan\phi = \frac{\omega L}{R}$

$AC$ voltage is applied across a series combination of

$L$ and

$R$ . If the voltage drop across the resistor and inductor are

$20V$ and

$15V$ respectively, then applied peak voltage is :

0%
$25V$
0%
$35V$
0%
$25\sqrt 2V$
0%
$5\sqrt 7V$

Explanation

$\displaystyle V=\sqrt{V^2_R+V^2_L}$

$\displaystyle =\sqrt{(20)^2+(15)^2}=25V$
But this is the rms value.

$\therefore$ Peak value

$=\displaystyle \sqrt 2 V_{rms}=25\sqrt 2V$

The impedance of a series

$L-C-R$ circuit in an

$AC$ circuit is

0%
$\displaystyle \sqrt{R+(X_L-X_C)}$
0%
$\displaystyle \sqrt{R^2+(X^2_L-X^2_C)}$
0%
$R$
0%
None of these

Explanation

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

$D$ is the correct Answer

In the

$AC$ network shown in figure, the rms current flowing through the inductor and capacitor are

$0.6A$ and

$0.8A$ respectively. Then the current coming out of the source is

0%
$1.0A$
0%
$1.4A$
0%
$0.2A$
0%
None of the above

Explanation

$I_C$ is

$90^o$ ahead of the applied voltage and

$I_L$ lags behind
the applied voltage by

$90^o$ . So, there is a phase difference of

$180^o$ between

$I_L$ and

$I_C$ .

$\therefore \displaystyle I=I_C-I_L=0.2A$

In a parallel

$L-C-R$ circuit as shown in figure if

$I_R, I_L, I_C$ and

$I$ represents the rms values of current flowing through resistor, inductor, capacitor and the source, then choose the appropriate correct answer

0%
$\displaystyle I=I_R+I_L+I_C$
0%
$\displaystyle I=I_R+I_L-I_C$
0%
$I_L$ or $I_C$ may be greater than $I$
0%
None of the above

Explanation

Voltage

$V$ is same across all the elements.

currents,

$I_R=V/R;I_L=V/\omega L;I_C=\omega C\times I$ , where

$\omega$ is frequency of current source

$I$

now total current is

$I$ is the vector sum of three currents, and cannot be simply added like vectors. Thus option A and B are incorrect.

The vector sum

$I$ can be lower than any of its

$I_R,I_L,I_C$ . So, option C is correct.

In a series

$LC$ circuit, the applied voltage is

$V_0$ . If

$\omega$ is very low, then the voltage drop across the inductor

$V_L$ and capacitor

$V_C$ are

0%
$\displaystyle V_L=\frac{V_0}{2}; V_C=\frac{V_0}{2}$
0%
$\displaystyle V_L=0; V_C=V_0$
0%
$\displaystyle V_L=V_0; V_C=0$
0%
$\displaystyle V_L=-V_C=\frac{V_0}{2}$

Explanation

$X_L=\omega L$
If

$\omega$ is very low, then

$X_L=0$

$\therefore V_L=0$
or

$V=V_C=V_0$

If the rms current through a

$6.8 k \Omega$ resistor is

$8 mA$ , the rms voltage drop across the resistor is

0%
$5.44 V$
0%
$54.4 V$
0%
$7.07 V$
0%
$8 V$

Explanation

Rms current

$I_{rms} = 8 mA$

Resistance of resistor

$R = 6.8k\Omega$

Thus rms voltage

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

$\therefore$

$V_{rms} = (8\times 10^{-3}) \times (6.8\times 10^3)$

$\implies$

$V_{rms} = 54.4 V$

In a series

$L-C-R$ circuit, current in the circuit is

$11A$ when the applied voltage is

$220V$ . Voltage across the capacitor is

$200V$ . If value of resistor

$20\Omega$ , then the voltage across the unknown inductor is

0%
Zero
0%
$200V$
0%
$20V$
0%
None of these

Explanation

The given LCR circuit is in resonance. Inductive reactance magnitude

$X_L$ increases as frequency increases while capacitive reactance magnitude

$X_C$ decreases with the increase in frequency. At one particular frequency, these two reactances are equal in magnitude but opposite in sign; that frequency is called the resonant frequency

$f_O$ for the given circuit.

Hence, at resonance:

$X_L = X_C$

So the voltage across rthe unknown inductor is $200V$ as same as the voltage across the capacitor.

The adjoining figure shows an

$AC$ circuit with resistance

$R$ , inductance

$L$ and source voltage

$V_s$ . Then

0%
the source voltage $V_s=72.8V$ .
0%
the phase angle between current and source voltage is $\tan^{-1}(7/2)$ .
0%
Both $(a)$ and $(b)$ are correct.
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Both $(a)$ and $(b)$ are wrong.

Explanation

$\displaystyle V_S=\sqrt{V^2_R+V^2_L}$

$=\displaystyle \sqrt{(70)^2+(20)^2}=72.8V$

$\displaystyle \tan \phi=\frac{X_L}{R}=\frac{V_L}{V_R}=\frac{20}{70}=\frac{2}{7}$

In series

$L-C-R$ circuit voltage drop across resistance is

$8V$ , across inductor is

$6V$ and across capacitor is

$12V$ . Then

0%
Voltage of the source will be leading in the circuit.
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Voltage drop across each element will be less than the applied voltage.
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Power factor of the circuit will be $3/4$ .
0%
None of the above.

Explanation

$\displaystyle V=\sqrt{V^2_R+(V_C-V_L)^2}=10V$

$V_C > V_L$ , hence current leads the voltage.
Power factor

$\displaystyle =\cos \phi=\frac{8}{10}=0.8$

In the circuit shown in figure, the

$AC$ source gives a voltage

$V=20\cos (2000t)$ . Neglecting source resistance, the voltmeter and ammeter readings will be

0%
$0 V, 2.0A$
0%
$0V, 1.4A$
0%
$5.6V, 1.4A$
0%
$8V, 2.0A$

Explanation

Here,

$X_{L}=\omega L=2000\times 5\times 10^{-3}\Omega =10\Omega$

$X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{2000\times 50\times 10^{-6}}\Omega=10\Omega$

$X_{L}=X_{C}$ . Hence this is a case of resonance.

$\implies i=\dfrac{V}{R}=\dfrac{(20/\sqrt{2})}{10}=1.4A$

Voltmeter reading=

$1.4\times 4 volt=5.6V$

$AC$ voltage source

$V=V_0\sin \omega t$ is connected across resistance

$R$ and capacitance

$C$ as shown in the figure. It is given that

$R=1/\omega C$ and the peak current is

$I_0$ . If the angular frequency of the voltage source is changed to

$\omega/\sqrt 3$ then the new peak current in the circuit is

0%
$\displaystyle \frac{I_0}{2}$
0%
$\displaystyle \frac{I_0}{\sqrt 2}$
0%
$\displaystyle \frac{I_0}{\sqrt 3}$
0%
$\displaystyle \dfrac{I_0}{3}$

Explanation

$\displaystyle R=\frac{1}{\omega C}=X_C$

$\displaystyle \therefore Z=\sqrt{R^2+X^2_C}=\sqrt 2 R$ (as

$X_C=R$ )

$\displaystyle I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt 2 R}$
When

$\omega$ becomes

$\displaystyle \frac{1}{\sqrt 3}$ times,

$X_C$ will become

$\sqrt 3$ times or

$\sqrt 3R$ .

$\displaystyle Z'=\sqrt{(R^2)+(\sqrt 3R)^2}=2R$

$\displaystyle I'_0=\frac{V_0}{Z'}=\frac{V_0}{2R}=\frac{I_0}{\sqrt 2}$

A resistor and an inductor are connected to an ac supply of

$120\space V$ and

$50\space Hz$ . The current in the circuit is

$3\space A$ . If the power consumed in the circuit is

$108\space W$ , then the resistance in the circuit is

0%
$12\space\Omega$
0%
$40\space\Omega$
0%
$\sqrt{(52\times28)}\space\Omega$
0%
$360\space\Omega$

Explanation

$I_{rms}= current in circuit = 3A$

$P = 108 W = {I}_{rms}^{2} R =3^2 R$

$R =108/9= 12 \Omega$

When

$100\space V$ dc is applied across a solenoid, a current of

$1.0\space A$ flows in it. When

$100\space V$ ac is applied across the same coil, the current drops to

$0.5\space A$ . If the frequency of the ac source is

$50\space Hz$ , the impedance and inductance of the solenoid are

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

Explanation

impedance=

$\frac{V_{ac}}{I_{ac}}= \frac {100 V}{0.5 A} = 200 \Omega$
R=

$\frac{V_{dc}}{I_{dc}}=\frac{100}{1}=100 \Omega$

$L \omega$ =

$\sqrt{ {impedance}^{2} - R^2 } = \sqrt { 200^2-100^2 } =173.2$

$L=\frac{173.2}{\omega}=\frac{173.2}{2\pi 50} =0.55 H$

An inductive coil has resistance of

$100\Omega$ . When an ac signal of frequency

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

$45^{\small\circ}$ . What is the inductance of the coil?

0%
$2\space mH$
0%
$3.3\space mH$
0%
$16\space mH$
0%
$\sqrt5\space mH$

Explanation

$tan(45) = 1 = \frac{L \omega}{R}$

$L = R / \omega = R / ( 2 \pi 1000) = .016 H$

$220V$ ,

$50\space Hz$ AC generator is connected to an inductor and a

$50\space \Omega$ resistance in series. The current in the circuit is

$1.0\space A$ . What is the Potential difference across inductor?

0%
$102.2\space V$
0%
$186.4\space V$
0%
$214\space V$
0%
$170\space V$

Explanation

$\left | Z \right| = \dfrac{V_{rms}}{I_{rms} } = \dfrac{220}{1} =220 \Omega$

$X_L = \sqrt{ {\left | Z \right | }^2 - R^2 } = 214 H$

$V_L = \dfrac{ X_L}{ \left | Z \right | } V_{rms} = \dfrac{ 214}{220} 220 = 214 V$

An alternating voltage

$E = 50\sqrt2\sin(100t)\space V$ is connected to a

$1\space \mu F$ capacitor through an ac ammeter. What will be the reading of the ammeter?

0%
$10\space mA$
0%
$5\space mA$
0%
$5\sqrt2\space mA$
0%
$10\sqrt2\space mA$

Explanation

$X_C = \frac{1}{C \omega} = 10000 \Omega$
ammeater reading =

$I_{rms} = \frac{ V_{rms} }{ \left | jX_C \right | } = \frac{ 50 }{10000} = 5 mA$

$50\space W$ ,

$100\space V$ lamp is to be connected to an AC mains of

$200\space V, \space 50\space Hz$ . What capacitor is essential to be put in series with the lamp?

0%
$\displaystyle\frac{25}{\sqrt2}\mu F$
0%
$\displaystyle\frac{50}{\pi\sqrt3}\mu F$
0%
$\displaystyle\frac{50}{\sqrt2}\mu F$
0%
$\displaystyle\frac{100}{\pi\sqrt3}\mu F$

Explanation

for

$100V, 50 W$ lamp ,

$R = \frac{100^2}{50} = 200 \Omega$
capacitor put in series should be such that,

$V_R = \frac{ R} { \sqrt{ R^2 + X_C^2 } } V_s= 100$

$\Rightarrow \frac{1}{C \omega} = X_C= \sqrt{3} R$

$C=\frac{1}{ \omega \sqrt{3} R } =\frac{50}{\pi \sqrt{3} } \mu F$

A coil has an inductance of

$0.7\space H$ and is joined in series with a resistance of

$220\Omega$ . When an alternating emf of

$220\space V$ at

$50\space cps$ is applied to it, then the wattless component of the current in the circuit is (take

$0.7\pi = 2.2)$

0%
$5\space A$
0%
$0.5\space A$
0%
$0.7\space A$
0%
$7\space A$

Explanation

wattless current =

${ I }_{ max }sin({ tan }^{ -1 }\{ L\omega /R\} )=\frac { 220 }{ \sqrt { { R }^{ 2 }+{ L }^{ 2 }{ \omega }^{ 2 } } } sin({ tan }^{ -1 }\{ L\omega /R\} )=0.5A$

A current source sends a current

$I - i_0\cos(\omega t)$ . when connected across an unknown load, it gives a voltage output of

$v = v_0\sin[\omega t + (\pi/4)]$ across that load. then the voltage across the current source may be brought in phase with the current through it by

0%
Connecting an inductor in series with the load
0%
Connecting a capacitor in series with the load
0%
Connecting an inductor in parallel with the load
0%
Connecting a capacitor in parallel with the load

Explanation

$v = v_0\sin[\omega t + (\pi/4)] = v_0\sin[\omega t + (\pi/4) + (\pi/4) - (\pi/4)] = v_0\sin[\omega t +(\pi/2) - (\pi/4)] = v_0\cos[\omega t - (\pi/4)]$
cuurent =

$I - i_0 \cos[\omega t ] = I + i_0 \cos[\omega t + \pi ]$
we therefore see that ac current leads the ac voltage across the load. if an inductor is connected in series with the load, the it induces a lag in the current wrt to the voltage.
Therefore an inductor must be connected in series with the load so as to bring the voltage across the current source in phase with the current.

The maximum value of a.c. voltage in a circuit is 707V. Its r.m.s. value is

0%
70.7 V
0%
100 V
0%
500 V
0%
707 V

Explanation

As we know,

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

$V_{rms}=\dfrac{707}{\sqrt 2}=500\ V$

For an

$LCR$ series circuit with an A.C. source of angular frequency

$\omega$ , which statement is correct?

0%
Circuit will be capacitive if $\omega > \displaystyle\frac{1}{\sqrt{LC}}$
0%
Circuit will be capacitive if $\omega = \displaystyle\frac{1}{\sqrt{LC}}$
0%
Power factor of circuit will be unity if capacitive reactance equals inductive reactance
0%
Current will be leading voltage if $\omega > \displaystyle\frac{1}{\sqrt{LC}}$

Explanation

Circuit will be capacitive if , total reactance =

$X_L - X_C < 0$

$\Rightarrow X_C > X_L \Rightarrow \frac{1}{C \omega} > L \omega \Rightarrow \omega < \frac{1}{ \sqrt{ L C }}$

Current will be leading the voltage if the circuit is capacitive , i.e.,

$\omega < \frac{1}{ \sqrt{ L C }}$

Power factor=

$\dfrac{ R } { Z }$ ; pf factor will be unity if

$R= Z \Rightarrow X_L=X_C$ , i.e., capacitive reactance is equal to inductive reactance.

$LCR$ series

$AC$ circuit

0%
If $R$ is increased current will decrease.
0%
If $L$ is increased current will decrease.
0%
If $C$ is increased current will increase.
0%
If $C$ is increased current will decrease.

Explanation

$\displaystyle I=\frac{V}{Z}$

$\displaystyle =\frac{V}{\sqrt{R^2+\left(\omega L\sim \frac{1}{\omega C}\right)^2}}$

By increasing

$R$ , current will definitely decrease by change in

$L$ or

$C$ , current may increase or decrease.

Current in an ac circuit is given by

$I = 3\sin\omega t + 4\cos\omega t$ , then

0%
RMS value of current is $5\space A$
0%
Mean value of this current in any one half period will be $6\pi$
0%
If voltage applied is $V = V_m\sin\omega t$ , then the circuit may be containing resistance and capacitance only
0%
If voltage applied is $V = V_m\sin\omega t$ , the circuit may contain resistance and inductance only

Explanation

$I = 3sin \omega t + 4 cos \omega t$

$\Rightarrow I= 5( \frac{3}{5} sin \omega t + \frac{4}{5} cos \omega t )$

$\Rightarrow I = 5 sin( \omega t + \alpha )$

$; \alpha = tan^{-1} {\frac{4}{5} }$
therefore,
rms value of current =

$5/ \sqrt{2} A$
Mean value of current one half period =

$\frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ 5sin\theta \quad d\theta } =\frac { 1 }{ \pi } 5\quad { \left| -cos\theta \right| }_{ 0 }^{ \pi }=\frac { 10 }{ \pi } \quad A$
The current lags the voltage by an angle

$\theta$ , therefore the circuit may contain resistance and capacitor only.

In an ac circuit, the potential differences across an inductance and resistance joined in series are, respectively,

$16\space V$ and

$20\space V$ . The total potential difference across the circuit is

0%
$20\space V$
0%
$25.6\space V$
0%
$31.9\space V$
0%
$53.5\space V$

Explanation

In phasor,

$V_R= 20 \angle{0}$ ;

$V_L= 16 \angle{(-90)}$
Total potential difference is =

$V_R+V_L= 25.6 \angle{(-38.66)}$
Magnitude of total potential difference =

$25.6 V$

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

0%
becomes zero instantaneously.
0%
grows monotonically.
0%
decays monotonically.
0%
oscillates instantaneously.

Explanation

$\frac { q }{ c } +L\frac { di }{ dt } =0\Rightarrow \frac { { d }^{ 2 }q }{ d{ t }^{ 2 } } +\frac { q }{ LC } =0\\ which\quad is\quad the\quad equation\quad of\quad harmonic\quad motion.\\ q={ q }_{ 0 }sin\omega t\Rightarrow i={ i }_{ 0 }cos\omega t\\ thus,\quad the\quad current\quad starts\quad oscillation.$

What is the value of inductance L for which the current is a maximum in a series LCR circuit with

$C=10 \mu F$ and

$\omega = 1000s^{-1}$ ?

0%
$10 mH$
0%
$100mH$
0%
$1 mH$
0%
cannot be calculated unless R is known

Explanation

Current is maximum at resonance

$\displaystyle \Rightarrow \omega^2 = \frac{1}{LC} \Rightarrow L = \frac{1}{\omega^2 C}$

$\displaystyle = \frac{1}{(1000)^2 (10 \times 10^{-6})} = 0.1 H =100 mH$

Voltage across each elements of a series LCR circuit are given by

$V_L= 60V, V_C = 20V, V_R= 30V$ Find out source voltage

0%
50V
0%
100V
0%
150V
0%
200V

Explanation

$V=\sqrt { { { V }_{ R } }^{ 2 }+{ \left( { V }_{ L }-{ V }_{ C } \right) }^{ 2 } } =\sqrt { { 30 }^{ 2 }+{ \left( 60-20 \right) }^{ 2 } } =50V$

A coil has resistance

$30 ohm$ and inductive reactance

$20 ohm$ at

$50 Hz$ frequency. If an

$ac$ source of

$200 volt, 100 Hz$ , is connected across the coil, the current in the coil will be

0%
$4.0 A$
0%
$8.0 A$
0%
$\displaystyle\frac{20}{\sqrt{13}} A$
0%
$2.0 A$

Explanation

$\omega = 50 \times 2\pi$ then

$\omega L = 20\Omega$
If

${\omega}^{\prime} = 100 \times 2\pi$ then

${\omega}^{\prime} L = 40\Omega$

$I = \displaystyle\frac{200}{Z} \displaystyle\frac{200}{\sqrt{{R}^{2} + {\left({\omega}^{\prime}L\right)}^{2}}} = \displaystyle\frac{200}{\sqrt{{30}^{2} + {\left(40\right)}^{2}}}$

$I = 4 A$ .

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when

0%
Frequency of the AC source is decreased
0%
Number of turns in the coil is reduced
0%
A capacitance of reactance $X_{C}=X_{L}$ is included in the same circuit
0%
An iron rod is inserted in the coil

An a.c. supply of 100 volts is applied to a capacitor of capacitance 20

$\mu$ F. If the current in the circuit is 0.628 A, the frequency of a.c. must be

0%
50 Hz
0%
60 Hz
0%
25 Hz
0%
40 Hz

Explanation

$Y=100V ,C=2 \mu F=20 \times 10^{-6}$

$I = 0.628 A, v = ?$

$X_C = \displaystyle \frac{V}{I} = \frac{100}{0.628}$

$\Rightarrow \dfrac{ 1}{2 \pi v C} = \displaystyle \frac{100}{0.628}$

$v = \displaystyle \frac{0.628 }{100 \times 2 \pi C}$

$= 50 Hz$

In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45

$^o$ . The value of C is

0%
$\displaystyle \frac{1}{2 \pi f (2 \pi f L - R)}$
0%
$\displaystyle \frac{1}{2 \pi f ( 2\pi f L + R)}$
0%
$\displaystyle \frac{1}{ \pi f (2 \pi f L - R)}$
0%
$\displaystyle \frac{1}{ \pi f (2 \pi f L + R)}$

Explanation

Here

$X_C - X_L = R$

$\Rightarrow \displaystyle \frac{1}{2 \pi f C} = (R + 2 \pi f L)$

$\Rightarrow \displaystyle C = \frac{1}{2 \pi f (2 \pi f L + R)}$

In an A.C. circuit, the current flowing in inductance is

$\displaystyle I=5\sin { \left( 100t-{ \pi }/{ 2 } \right) }$ ampers and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to

0%
1000 watt
0%
40 watt
0%
20 watt
0%
Zero

Explanation

Power,

$\displaystyle P={ I }_{ r.m.s }\times { V }_{ r.m.s }\times \cos { \phi }$
In the given problem, the phase difference between voltage and current is p/2. Hence

$\displaystyle P={ I }_{ r.m.s }\times { V }_{ r.m.s }\times \cos { \left( { \pi }/{ 2 } \right) } =0\\$

Assertion: A capacitor blocks direct current in the steady state.
Reason : The capacitive reactance of the capacitor is inversely proportional to frequency f of the source of emf.

0%
If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
0%
If both Assertion and Reason are true but the Reason is not the correct explanation of the Assertion.
0%
If Assertion is true statement but Reason is false.
0%
If both Assertion and Reason are false statements.

Explanation

Capacitive reactance

$X_C = \dfrac{1}{2\pi f C}$
Frequency of direct current is zero i.e.

$f= 0$

$\implies \ X_C = \infty$
Thus resistance of capacitor for direct current is very large and so it do not allow the direct current to pass through it.
Hence, option A is correct.

If a cell of

$200V$ is connected across an inductor, the current is found to be

$5A$ . When cell is replaced by

$200 V - 100 rad/s$ supply, r.m.s. value of current becomes

$4A$ . The inductance of inductor is-

0%
$0.2$ henry
0%
$0.3$ henry
0%
$0.4$ henry
0%
$0.5$ henry

Assertion: In the purely resistive element of a series LCR, AC circuit the maximum value of rms current increases with increase in the angular frequency of the applied e.m.f.
Reason:

$\displaystyle { I }_{ max }=\frac { { \varepsilon }_{ max } }{ z } ,z=\sqrt { { R }^{ 2 }+{ \left( \omega L-\frac { I }{ \omega C } \right) }^{ 2 } }$
where

$\displaystyle { I }_{ max }$ is the peak current in a cycle.

0%
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
0%
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
0%
If Assertion is correct but Reason is incorrect.
0%
If Assertion is incorrect but Reason is correct.

Explanation

Solution:-

Given Assertion is correct i.e. value of current increases with increase in angular frequency of applied emf since it makes the impedence of circuit reduces to only resistance of resistor.

But the given reasn is incorrect since current becomes maximum due to decrease in impedence of circuit as inversely proportional to it (reduces to resistance).

$I_{max} = \cfrac{\epsilon}{Z}$

But not due to maximum of applied voltage/emf.

L, C, R represent physical quantities inductance, capacitance and resistance respectively. The combinations which have the dimensions of frequency are

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$\displaystyle { 1 }/{ RC }$
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$\displaystyle { R }/{ L }$
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$\displaystyle { 1 }/{ \sqrt { LC } }$
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$\displaystyle { C }/{ L }$

The Current in resistance R at resonance is

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Zero
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Minimum but finite
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Maximum but finite
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Infinite

Explanation

At resonance

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

$\displaystyle \Rightarrow$ R & current is maximum but finite, which is

$\displaystyle { I }_{ max }=\frac { E }{ R }$ where E is applied voltage.

A lamp is connected in series with a capacitor and an AC source. What happens if the capacity of the capacitor is reduced?

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The lamp shines more brightly
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The lamp shines less brigthly
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There is no change in the brightness of the lamp
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Brightness may increase or decrease depending on the frequency of the AC

Explanation

The lamp shines less brightly because, when capacity of the capacitor is reduced, the capacitive reactance

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

increases. Hence, impedance

$Z$ increases resulting in decrease in current.

The power for the resistor

$i^2R$ decreases, hence the lamp shines less brightly.

If the inductance and capacitance are both doubled in L-C-R circuit, the resonant frequency of the circuit will :

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Decrease to one-half of the original value
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Decrease to one-fourth of the original value
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Increase to twice the original value
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Decrease to twice the original value

Explanation

Resonant frequency in series LCR circuit :

$\omega = \sqrt{\dfrac{1}{LC}}$
When L and C are doubled,

$\omega ' = \sqrt{\dfrac{1}{L'C'}} = \sqrt{\dfrac{1}{2L \times 2C}} =\dfrac{ \omega}{2}$

Resonance frequency of a circuit is

$f$ . If the capacitance is made

$4$ times the initial value, then the resonance frequency will become

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

Explanation

As we know,

Resonance frequency of a circuit

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

$f\propto \dfrac { 1 }{ \sqrt { C } }$

$\Rightarrow \dfrac { { f }^{ \prime } }{ f } =\dfrac { \sqrt { C } }{ \sqrt { 4C } } =\dfrac { 1 }{ 2 }$

$\Rightarrow { f }^{ \prime }=\dfrac { f }{ 2 }$

The resonant frequency of an L-C circuit is

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$\dfrac {1}{2\pi \sqrt {LC}}$
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$\dfrac {1}{2\pi} \sqrt {\dfrac {L}{C}}$
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$\dfrac {1}{4\pi} \sqrt {\dfrac {L}{C}}$
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$\dfrac {1}{2\pi} \sqrt {\dfrac {C}{L}}$

Explanation

Resonance frequency

$f$ of an L-C circuit can be written as

Resonance frequency,

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

$L$ = inductance and

$C$ is capacitance

An alternating emf given by equation

$E = 300 sin [(100\pi )t]$

$volt$ is applied to a resistance

$100\ ohms$ . The rms current through the circuit is (in

$amperes$ ):

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$\displaystyle \frac {3}{\sqrt 2}$
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$\displaystyle \frac {9}{\sqrt 2}$
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$3$
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$\displaystyle \frac {6}{\sqrt 2}$

Explanation

Alternating emf equation

$E = 300$

$sin(100\pi t)$

$\therefore$ Peak emf

$E_o = 300$

$volts$

The rms emf

$E_{rms} = \dfrac{E_o}{\sqrt{2}} = \dfrac{300}{\sqrt{2}}$

$volts$

Value of resistance

$R = 100\ \Omega$

$\therefore$ Value of rms current

$I_{rms} = \dfrac{E_{rms}}{R} = \dfrac{300}{\sqrt{2} \times 100} =\dfrac{3}{\sqrt{2}}$

$A$

220 V, 50 Hz, AC source is connected to an inductance of 0.2 H and a resistance of 20

$\Omega$ in series. What is the current in the circuit?

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3.33 A
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33.3 A
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5 A
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10 A

Explanation

The frequency of AC source

$=f=50Hz$

Thus the impedance of the inductor,

$X_L=\omega L=(2\pi f)L=20\pi\Omega$

Resistance of the resistor

$=20\Omega$

Thus, the net impedance of the circuit

$=\sqrt{X_L^2+R^2}=65.94\Omega$

Thus, the current in the circuit

$=\dfrac{V}{X}=\dfrac{220V}{65.94\Omega}=3.33A$

In L-C-R circuit power of 3 mH inductance and

$4\Omega$ resistance, EMF

$E=4\,cos\,1000t$ volt is applied. The amplitude of current is

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$0.8\mathring{A}$
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$\dfrac{4}{7}\mathring{A}$
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$1.0\mathring{A}$
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$\dfrac{4}{\sqrt{7}}\mathring{A}$

Explanation

Impedance of inductor=

$X_L=\omega H=1000\times 3mH=3\Omega$

Thus the impedance of the circuit=

$X=\sqrt{X_{L}^2+R^2}=5\Omega$

Thus amplitude of current

$I_{max}=\dfrac{E_{max}}{X}=\dfrac{4V}{5\Omega}=0.8A$

The impedance of a circuit, when a resistance

$R$ and an inductor of inductance

$L$ are connected in series in an AC circuit of frequency

$f$ , is

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$\sqrt { R+2{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } }$
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$\sqrt { R+4{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } }$
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$\sqrt { { R }^{ 2 }+4{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } }$
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$\sqrt { { R }^{ 2 }+2{ \pi }^{ 2 }{ f }^{ 2 }{ L }^{ 2 } }$

Explanation

The impedance due to the inductor is

$2\pi fL$ and voltage drop across inductor leads the voltage drop across resistor by

$90°$ .

So, the net impedance will be:

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

In a series LCR circuit, the voltage across the resistance, capacitance and inductance is

$10\ V$ each. If the capacitance is short circuited the voltage across the inductance will be :

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$10\ V$
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$10\sqrt{2}\ V$
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$\dfrac{10}{\sqrt{2}} \ V$
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$20\ V$

Explanation

Since the voltages are all equal, the impedance of all the elements is same.

$R=X_L=X_C$

Total voltage in the circuit=

$I\sqrt{R^2+(X_L-X_C)^2}=IR=10V$

When capacitance is shorted,

$I=\dfrac{V}{\sqrt{R^2+X_L^2}}=\dfrac{10}{\sqrt{2}R}$

Potential drop across inductor=

$IX_L=IR=\dfrac{10}{\sqrt{2}}V$

In an LCR circuit inductance is changed from

$L$ to

$\dfrac{L}{2}$ . To keep the same resonance frequency,

$C$ should be changed to

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$2C$
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$\dfrac{C}{2}$
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$4C$
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$\dfrac{C}{4}$

Explanation

Resonance frequency,

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

For

$f$ to be constant the product

$LC$ must be constant.

So, if we half the value of inductance then the value of capacitance must be

$doubled$ .

$C$ should be changed to

$2C$ .

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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