MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 3

Symbol of Inductance in electric circuit is-

The peak voltage of an AC supply is

$300V$ . What is the rms voltage?

0%
$121.2V$
0%
$212.1V$
0%
$343.4V$
0%
$434.3V$

Explanation

$Answer:-$ B

$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } \\ { V }_{ rms }=\dfrac{300}{\sqrt { 2 }} =212.16V$

A light bulb is rated at

$100W$ for a

$220V$ supply. Find the resistance of the bulb:

0%
$184\Omega$
0%
$284\Omega$
0%
$384\Omega$
0%
$484\Omega$

Explanation

$Answer:-$ D

Power is given by:

$P=\dfrac { { V }^{ 2 } }{ R } \\ 100=\dfrac { { (220 })^{ 2 } }{ R } \\ R=484\Omega$

The rms value of current in an AC circuit is

$10A$ . What is the peak current?

0%
$14.1A$
0%
$35.2A$
0%
$58.9A$
0%
$23.5A$

Explanation

$Answer:-$ A

Relation between peak current and rms current is:

$I_{rms}=\dfrac { { I }_{ 0 } }{ \sqrt { 2 } } \\ { V }_{ 0 }=\sqrt { 2 } \times 10=14.1A$

In series LCR circuit, the capacitance is changed from

$C$ to

$2C$ . The inductance should be changed from

$L$ to ______ to obtain same resonance frequency.

0%
$4L$
0%
${L}/{2}$
0%
${L}/{4}$
0%
$2L$

Explanation

The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency.This frequency is called the natural frequency of oscillation of the system. If such a system is driven by an energy source, whose frequency is equal to the natural frequency of the system, the amplitude of oscillations becomes large and resonance is said to occur.
At resonance,

${ X }_{ L }={ X }_{ C }$
ie,

${ \omega }_{ r }L=\dfrac { 1 }{ { \omega }_{ r }C }$
or,

${ \omega }_{ r }=\dfrac { 1 }{ \sqrt { LC } }$
or

$2\pi { v }_{ r }=\dfrac { 1 }{ \sqrt { LC } }$
or

${ v }_{ r }=\dfrac { 1 }{ 2\pi \sqrt { LC } }$
or

$LC =$ constant (as

$v$ remains same)

$\therefore \dfrac { { L }_{ 2 } }{ { L }_{ 1 } } =\dfrac { { C }_{ 1 } }{ { C }_{ 2 } }$
or

$\dfrac { { L }_{ 2 } }{ L } =\dfrac { C }{ 2C }$ or

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

The reactance of inductance at

$10^{4}Hz$ is

$10^{4}\Omega$ . Its reactance at

$2\times 10^{4}Hz$ will be :

0%
$10^{4}\Omega$
0%
$2\times 10^{4}\Omega$
0%
$3\times 10^{4}\Omega$
0%
$4\times 10^{4}\Omega$

Explanation

$X_{L} = wL = 2\pi fL$

$X_{L}\propto f$
If frequency is double, then reactance will be double.

If we increase the driving frequency in a circuit with a purely resistive load, then amplitude

$V_R$

0%
remain in the same
0%
increase
0%
decrease
0%
none

Explanation

We know that

$V_R$ does not depend on driving frequency in a purely resistive circuit.So,If we increase the driving frequency in a circuit with a purely resistive load, then amplitude remain in the same.

Therefore, A is correct option.

In a

$LC$ circuit. Angular frequency at resonance is

$w$ . What will be the new angular frequency when inductor's introduces is made

$2$ times and capacitance is made

$4$ times?

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

Explanation

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

Now if

$L^1 = 2L \quad C^1 = 4C$

$w^1 = \dfrac{1}{\sqrt{2L.4C}}$

$w^1 = \dfrac{1}{2\sqrt{2}}\cdot \dfrac{1}{\sqrt{LC}} = \dfrac{w}{2\sqrt{2}}$

The current which does not contribute to the power consumed in an AC circuit is called:

0%
Non-ideal current
0%
Wattless current
0%
Convectional current
0%
Inductance current

Explanation

Wattless current does not contribute to the mean rate of working of the circuit.
As, power factor

$= \frac{\text{true power}}{\text{apparent power}}$

$=cos\phi$

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

$\therefore$ Power factor

$=cos\phi = \frac{R}{Z}$
In a non-inductive circuit,

$X_L=X_C$

$\therefore$ Power factor

$=cos\phi = \frac{R}{\sqrt{R^2}}=\frac{R}{R}=1$

$\therefore \phi = 0^o$
This is the maximum value of power factor. Iris a pure inductor or an ideal capacitor

$\phi = 90^o$

$\therefore$ Power factor

$= cos \phi = cos 90^o= 0$ .
Average power consumed in a pure inductor orb ideal capacitor

$P = E_V \cdot I_V cos\, 90^o = zero$ .
Therefore, current through pure L or pure C; which consumes no power for its maintenance in the circuit is called ideal current or wattless current.

The power loss is less in transmission lines, when :

0%
Voltage is less but current is more
0%
Both voltage and current are more
0%
Voltage is more but current is less
0%
Both voltage and current are less

Explanation

The power cables have some resistance.

Power lost in the wires can be calculated as

$P=I^2R$ with

$R$ as the resistance of the wires and

$I$ as the current that passes through them.
Power at the load is

$P=VI$ .

From this one can see that if voltage is increased by say

$n$ times, then only

$\dfrac{1}{n}$ the current is required to deliver the same power. However, if

$\dfrac{1}{n}$ current is passed on the same wires, only

$\dfrac{1}{n^2}$ of the power will be lost.

State whether given statement is True or False
In an RC circuit, the impedance is determined by both the resistance and the capacitive reactance combined.

0%
True
0%
False

Explanation

Solution:-

In RC circuit,

Impedence is

$Z = R+\cfrac{1}{j \omega C} = R+X_C$

Hence determined by both resistance and capacitive reactance.

When the frequency of the source voltage decreases, the impedance of a parallel RC circuit

0%
increases
0%
decreases
0%
does not change
0%
decreases to zero

$L-C-R$ series circuit, the capacitor is changed from

$C$ to

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

$L$ to.

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

Explanation

The resonant frequency in a L-C-R circuit is given by ,

$f=\frac{1}{2\pi\sqrt{LC}}$ ................eq1

Now , if C is changed to 4C ,and L is changed to L' , to keep the

$f$ constant , then

$f=\frac{1}{2\pi\sqrt{L'\times4C}}$ .................eq2

From eq1 and eq2 ,

$\frac{1}{2\pi\sqrt{LC}}=\frac{1}{2\pi\sqrt{L'\times4C}}$

$L'=L/4$

State whether True or False :

The impedance of a series RC circuit varies directly with frequency.

0%
True
0%
False

Explanation

Solution:-

In series RC circuit

Impedence is

$Z = R+\cfrac{1}{j\omega C}$ i.e.

It varies inversely with frequency.

When the frequency of the AC voltage applied to series LCR circuit is gradually increased from a low value, the impedance of the circuit:

0%
monotonically increases.
0%
first increases and then decreases.
0%
first decreases and then increases.
0%
monotonically decreases.

Explanation

Impedance of series LCR circuit is given by:

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

Drawing the above curve for impedance v/s frequency, we get the plot as shown in the attached figure. Hence, impedance first decreases and then increases.

If A.C. voltage is applied to a pure capacitor, then voltage across the capacitor _________.

0%
Leads the current by phase angle $\left(\displaystyle\frac{\pi}{2}\right)$ rad.
0%
Leads the current by phase angle $(\pi)$ rad.
0%
Lags behind the current by phase angle $\left(\displaystyle\frac{\pi}{2}\right)$ rad.
0%
Lags behind the current by phase angle $(\pi)$ rad.

Explanation

In a capacitor, as the voltage applied is increased, there is current going into the capacitor. Now, as the applied voltage reaches the positive peak, the charging (in flow of current) stops. As the voltage starts to reduce, current flows out of the capacitor and continues until the voltage reaches the negative peak and discharging stops.

If we trace this out on a graph we will see that the current is leading the voltage by 90°, reaching zero current at the peaks of the voltage and maximum current as the voltage is going through zero. Hence, the voltage acrross the capacitor lags behind current by the phase angle

$(\frac{\pi}{2})$ rad.

State whether given statement is True or False
In a series RLC circuit, the larger reactance determines the net reactance of the circuit.

0%
True
0%
False

Explanation

In an RLC circuit the net reactance is calculated by the equation $(X_L – X_C)$ . So the large reactance determines the net reactance of the circuit.

Power dissipated in pure inductance will be

0%
$\dfrac {LI^{2}}{2}$
0%
$2LI^{2}$
0%
$\dfrac {LI^{2}}{4}$
0%
Zero

When the frequency of AC is doubled, the impedance of an LCR series circuit:

0%
is halved
0%
is doubled
0%
increases
0%
decreases

$V=100 \sin 100t$ volt, and

$I=100 \sin(100t+\dfrac {\pi}{6})A$ . then find the watt less power in watt?

0%
$10^{4}$
0%
$10^{3}$
0%
$10^{2}$
0%
$2.5 \times 10^{3}{\sqrt{3}}$

Explanation

$P= V_{rms} \times I_{rms} \times \cos \phi$

$\quad= \large\frac{V_0I_0}{\sqrt{2}\times \sqrt{2}}\times \cos \dfrac{\pi}{6}$

$\quad= \large\frac{100 \times 100}{2} \times \frac{\sqrt{3}}{2}=2.5\times 10^3\sqrt{3}W$

In AC circuit having only capacitor, the current _____.

0%
Leads the voltage by $\dfrac {\pi}{2}$ in phase
0%
Lags behind the voltage by $\dfrac {\pi}{2}$ in phase
0%
Leads the voltage by $\pi$ in phase
0%
Lags behind the voltage by $\pi$ in phase

Explanation

For a purely capacitive circuit current leads the voltage by

$\pi/2$ in phase.

'Z' is not.

0%
Atomic number
0%
Impedance
0%
Zeta potential
0%
Partition function

In a pure capacitive circuit if the frequency of ac source is doubled, then its capacitive reactance will be

0%
remains same
0%
doubled
0%
halved
0%
zero

Explanation

Capacitive reactance,

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

We can see that

$X_c \propto \dfrac{1}{\nu}$ ,

$\nu$ is doubled,

$X_c$ will be halved

In a series

$LCR\, AC$ circuit, the current is maximum when the impedance is equal to:

0%
the reactance
0%
the resistance
0%
zero
0%
twice the reactance
0%
twice the resistance

Explanation

In L-C-R circuit, the current

$I=\displaystyle\frac{E_0}{\sqrt{R^2+(X_L-X_C)^2}}$
When impedance

$Z=R$ , i.e.,

$X_L=X_C$ the current in the L-C-R circuit will be maximum.

Statement A: With an increase in the frequency of AC supply inductive reactance increases.
Statement B: With an increase in the frequency of AC supply capacitive reactance increase.

0%
A is true but B is false
0%
Both A and B are true
0%
A is false but B is true
0%
Both A and B are false

Explanation

$Z_L=WL$

$Z_C=\dfrac{1}{wc}$

$w\uparrow, Z_L\uparrow Z_c\downarrow$

In series L-C-R resonant circuit, to increase the resonant frequency :

0%
L will have to be increased
0%
C will have to be increased
0%
LC will have to be decreased
0%
LC will have to be increased

Explanation

Resonant frequency

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

$LC \downarrow \ if \ \omega_r \uparrow$

At low frequency a condenser offers :

0%
high impedance
0%
low impedance
0%
zero impedance
0%
impedance of condenser is independent of frequency

Explanation

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

$w\downarrow Z\uparrow$

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

$V_{1},V_{2},V_{3}$ respectively. Then the voltage of applied AC source is always equal to :

0%
$V_{1}+V_{2}+V_{3}$
0%
$\sqrt{V_{1}^{2}+(V_{2}+V_{3})^{2}}$
0%
$V_{1}-V_{2}-V_{3}$
0%
$\sqrt{V_{1}^{2}+(V_{2}-V_{3})^{2}}$

Explanation

$V_2$ and

$V_3$ will be

$180^0$ out of phase with each other.

Hence, resultant is

$V_2 -V_3$ .

Each of

$V_2$ and

$V_3$ is having a phase diff of

$90^0$ with

$V_1$ .

Hence, resultant AC voltage which is equal to the applied AC voltage is

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

If the frequency of an alternating e.m.f. is f in L-C-R circuit, then the value of impedance Z

$\underline{\hspace{0.5in}}$ as log(frequency) increases :

0%
increases
0%
increases and then becomes equal to resistance, then it will start decreasing
0%
decreases and when it becomes minimum, equal to the resistance then it will start increasing
0%
go on decreasing

An inductance and resistance are connected in series with an A.C circuit. In this circuit

0%
the current and P.d. across the resistance lead P.d across the inductance by $\dfrac{\pi }{2}$
0%
the current and P.d across the resistance lags behind the P.d. across the inductance by angle $\dfrac{\pi }{2}$
0%
The current across resistance leads and the P.d across resistance lags behind the P.d across the inductance by $\dfrac{\pi }{2}$
0%
the current across resistance lags behind and the P.d across the resistance leads the P.d across the inductance by $\dfrac{\pi }{2}$

Explanation

This is very fundamental. If we apply separate voltages across resistance and inductor, then in resistance, current and voltage both are in same phase whereas in inductor, current across it lags p.d across it by

$\pi /2$ .
Now, when we apply voltage across inductor and resistance connencted in series then current through both of them will be same because of KCL. therefore voltage across resistor will be in same phase with current whereas voltage across inductor will lead the current across it by

$\pi /2$ . therefore current and voltage across resistor lags voltage across inductor by

$\pi /2$ .

If a capacitor is connected to two different A.C. generators, then the value of capacitive reactance is:

0%
directly proportional to frequency
0%
inversely proportional to frequency
0%
independent of frequency
0%
inversely proportional to the square of frequency

Explanation

$X_C=\dfrac{1}{\omega c}$

$\therefore X_C \propto \dfrac{1}{\omega}$

A capacitor is connected to an A.C. circuit, then the phase difference between current and the voltage is :

0%
$\pi$
0%
$\dfrac{\pi }{2}$
0%
$\dfrac{-\pi }{2}$
0%
Zero

Explanation

Current leads voltage by

$\dfrac{\pi}{2}$

$\therefore$ phase difference is

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

Ratio of impedance to capacitive reactance has

0%
no units
0%
ohm
0%
ampere
0%
tesla

Explanation

$Ratio=\dfrac{\omega L}{\dfrac{1}{wc}}$

$=w^2LC$

$WL \ and \ \dfrac{1}{wc}$ has dimensions same as of resistance

Statement ( A ) : The reactance offered by an inductance in A.C. circuit decreases with the increase of AC frequency.
Statement ( B ) : The reactance offered by a capacitor in AC circuit increases with the increase of AC frequency.

0%
A is true but B is false
0%
Both A and B are true
0%
A is false but B is true
0%
Both A and B are false

A mixer of 100

$\Omega$ resistance is connected to an A.C. source of 200V and 50 cycles/sec. The value of average potential difference across the mixer will be:

0%
308 V
0%
264 V
0%
220 V
0%
zero

Choose the wrong statement of the following :

0%
The peak voltage across the inductor can be less than the peak voltage of the source in an LCR circuit
0%
In a circuit containing and a capacitor and an ac source the current is zero at the instant source voltage is maximum
0%
When an AC source is connected to a capacitor, then the rms current in the circuit gets increased if a dielectric slab is inserted into the capacitor
0%
In a pure inductive circuit emf will be in phase with the current

Explanation

In pure inductive circuit,

voltage leads by

$\dfrac{\pi}{2}$ .

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 current,

0%
leads the voltage by $45^{0}$
0%
lags behind the voltage by $45^{0}$
0%
leads the voltage by $90^{0}$
0%
lags behind the voltage by $90^{0}$

Explanation

$I = 0.707 sin (100\pi t)$

$V = 1.414 sin (100\pi t-\frac{\pi}{4})$
So, current leads voltage by

$45^0$ .

The equation of an alternating voltage is

E=220E=220 .Then the impedance of the circuit is:

0%
10 ohm
0%
22 ohm
0%
11 ohm
0%
17 ohm

An inductor has a resistance

$R$ and inductance

$L$ . It is connected to an AC source of emf

$E_{v}$ and angular frequency

$\omega$ ; then the current

$I_{v}$ in the circuit is :

0%
$\dfrac{E_{v}}{\omega L}$
0%
$\dfrac{E_{v}}{R}$
0%
$\dfrac{E_{v}}{\sqrt{R^{2}+\omega ^{2}L^{2}}}$
0%
$\sqrt{\left ( \dfrac{E_{v}}{R} \right )^{2}+\left ( \dfrac{E_{v}}{\omega L} \right )^{2}}$

Explanation

The impedance in R-L circuit is

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

The current,

$\displaystyle I_v=\dfrac{E_v}{Z}$

$=\dfrac{E_v}{\sqrt{R^2+(\omega L)^2}}$

A pure inductor of self inductance 1 H is connected across an alternating voltage of 115 V and frequency 60 Hz. The reactance

$X_{L}$ and peak current respectively are

0%
37.1 $\Omega$ , 0.43 A
0%
337.1 $\Omega$ , 0.43 A
0%
377.1 $\Omega$ , 0.43 A
0%
3.7 $\Omega$ , 0.42 A

Explanation

$X_L = \omega L$

$= 2\pi f L$

$= 2\pi 60\times 1$

$= 377.1 \Omega$
peak current =

$\dfrac{\text{peak voltage}}{X_L}$

$=\dfrac{115 \times \sqrt{2}}{377.1} = 0.43 \ A$

So,

$X_L = 377.1$

peak current,

$I = 0.43\ A$ .

The phase difference between alternating emf and current in a purely capacitive circuit will be

0%
zero
0%
$\pi$
0%
$-\dfrac{\pi }{2}$
0%
$\dfrac{\pi }{2}$

Explanation

In purely capacitive circuit current leads the voltage by

$\pi /2$
So, phase difference between alternating emf and current will be

$-\pi /2$
[Trick : CCL - capacitor current lead]

Assertion : In series LCR circuit, the resonance occurs at one frequency only.
Reason : At resonance the inductive reactance is equal and opposite to the capacitive reactance.

0%
A,R are true and R is the correct reason for A
0%
A,R are true and R is not the correct reason for A
0%
A is true, R is false
0%
A is false, R is true

Explanation

Resonance frequency

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

$Z_C=Z_L$

$\dfrac{1}{wc}=wL$

for a particular set of conductor and capacitor there is only one value of frequency

i.e.

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

Hence both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Voltage and current in an ac circuit are given by

$V = 5\sin \left (100\pi t - \dfrac {\pi}{6}\right )$ and

$I = 4\sin \left (100 \pi t + \dfrac {\pi}{6}\right )$ .

0%
Voltage leads the current by $30^{\circ}$
0%
Current leads the voltage by $60^{\circ}$
0%
Voltage leads the current by $60^{\circ}$
0%
Current and voltage are in phase

Explanation

$V = 5 sin (100\pi t-\dfrac{\pi}{6})$

$I = 4 sin (100\pi t+\dfrac{\pi}{6})$

So, current leading voltage by

$\dfrac{\pi}{6}-(-\dfrac{\pi}{6})$

$= \dfrac{\pi}{3} = 60^0.$

The capacitor offers zero resistance to

0%
D.C. only
0%
A.C. & D.C.
0%
A.C. only
0%
neither A.C. nor D.C.

Explanation

Capacitive reactance is given as

$X_c=\dfrac{1}{\omega C}$
From this relation we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.
Likewise, as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC.

A coil of self - inductance

$\left ( \dfrac{1}{\pi } \right )$ H is connected in series with a 300

$\Omega$ resistance. A voltage of 200V at frequency 200Hz is applied to this combination. The phase difference between the voltage and the current will be

0%
$tan^{-1}\left ( \dfrac{4}{3} \right )$
0%
$tan^{-1}\left ( \dfrac{3}{4} \right )$
0%
$tan^{-1}\left ( \dfrac{1}{4} \right )$
0%
$tan^{-1}\left ( \dfrac{5}{4} \right )$

Explanation

$X_L = \omega L$

$= 2\pi f L$

$= 2\pi (200) \dfrac{1}{\pi}$

$= 400$

$R = 300$

phase difference

$\theta = {tan}^{-1} (\dfrac{X_C}{R})$

$= {tan}^{-1}(\dfrac{400}{300})$

$= {tan}^{-1}(\dfrac{4}{3})$

In an A.C. circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of the inductance is 0.5 Henry and capacitance is 8

$\mu$ F . The angular frequency of the input A.C. voltage must be equal to

0%
500 rad/sec
0%
4000 rad/sec
0%
$5 \times10^{5}$ rad/sec
0%
5000 rad/sec

Explanation

For maximum current impedance should be zero.

So,

$Z = 0$

$\Rightarrow X_L-X_C = 0$

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

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

$= \dfrac{1}{\sqrt{0.5\times 8\times 10^{-6}}}$

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

$= 500 rad/sec$

The frequency at which the inductive reactance of

$2\ H$ inductance will be equal to the capacitive reactance of 2

$\mu$ F capacitance (nearly)

0%
80 Hz
0%
40 Hz
0%
60 Hz
0%
20 Hz

Explanation

$X_L = X_C$

$WL = \dfrac{1}{WC}$

$W^2 = \dfrac{1}{LC}$

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

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

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

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

$= \dfrac{10^3}{4\pi}$

$= 80Hz$

In an LCR series circuit, the capacitor is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to

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

Explanation

For resonance:

$X_L = X_C$

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

$\omega = \dfrac{1}{\sqrt{2C}}$

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

For same resonant frequency:
LC should be constant.

So, LC =

$4C L_1$

$\Rightarrow L_1 = \dfrac{LC}{4C}$

$= \dfrac{L}{4}$

In an A.C circuit, a resistance R is connected in series with an inductance L. If the phase angle between voltage and current be

$45^{0}$ , the value of inductive reactance will be

0%
$\dfrac{R}{4}$
0%
$\dfrac{R}{2}$
0%
$R$
0%
$\dfrac{R}{3}$

Explanation

Phase angle

$(\phi) = 45^0 = {tan}^{-1}(1)$

$= {tan}^{-1}(\dfrac{X_L}{R})$

So,

$\dfrac{X_L}{R} = 1$

$\Rightarrow X_L = R$

When the frequency of applied emf in an LCR series circuit is less than the resonant frequency,then the nature of the circuit will be:
a) Capacitive b) resistive c) inductive

0%
only a is true
0%
only a and b are true
0%
only b and c are true
0%
a, b and c are true

Explanation

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

So, $X_C \propto \dfrac{1}{\omega} = \dfrac{1}{2\pi f}$

$X_L = \omega L$ So, $X_L \propto \omega = 2\pi f$

So, as frequency decreases $X_C$ increases So, circuit becomes capacitive circuit.

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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