MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 2

The source frequency for which a

$5 \mu$ F capacitor has a reactance of 1000

$\Omega$ is

0%
$\dfrac{100}{\pi }$ Hz
0%
$\dfrac{1000}{\pi }$ Hz
0%
$200 \ Hz$
0%
$5000 \ Hz$

Explanation

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

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

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

$= \dfrac{1}{2\pi \times 1000 \times 5 \times 10^{-6}}$

$= \dfrac{100}{\pi}Hz$

The inductive reactance of a coil is 2500

$\Omega$ . On increasing it’s self-inductance to three times, the new inductive reactance will be:

0%
7500 $\Omega$
0%
2500 $\Omega$
0%
1225 $\Omega$
0%
zero

Explanation

$X_L = \omega L$
So,

$2500 = \omega L$
When L is increased to 3L
then

$X_L = \omega (3L)$

$= 3 \omega L$

$= 3(2500)$

$= 7500 \Omega$

In an LCR series circuit, the rms voltages across R, L and C are found to be 10 V, 10 V and 20 V respectively. The rms voltage across the entire combination is

0%
30 V
0%
1 $\mu$ F
0%
20V
0%
$10\sqrt{2}V$

Explanation

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

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

$= 10\sqrt{2}V$

In the circuit shown, a 30 V d.c source gives a current 2.0A as recorded in the ammeter A and 30V a.c. source of frequency 100Hz gives a current 1.2A. The inductive reactance is

0%
10ohm
0%
20 ohm
0%
$5\sqrt{34}$ ohm
0%
40 ohm

Explanation

In DC current

$V = iR$

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

$R = \dfrac{30}{2} = 15\Omega$

In a.c. current

$V = i Impedance$

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

$Impedance = \dfrac{30}{1.2} = 25\omega$

$Impedance = \sqrt{R^2+(X_L)^2}$

$(25)^2 = R^2+(X_L)^2$

$X_L = \sqrt{(25)^2-R^2}$

$= \sqrt{625-225}$

$= \sqrt{400}$

$= 20\Omega$

If the values of inductance and frequency in an AC circuit are 2 henry and

$\displaystyle \dfrac{10^3}{2\pi}$ Hz respectively then the value of inductive reactance will be

0%
$\displaystyle \dfrac{ 2\times 10^3}{\pi} \Omega$
0%
$\displaystyle 2\times 10^2 \Omega$
0%
$\displaystyle 10^3 \Omega$
0%
$\displaystyle 2\times 10^3 \Omega$

Explanation

Inductive reactance

$= L\omega =2\times 2\pi \dfrac { { 10 }^{ 3 } }{ 2\pi } =2\times { 10 }^{ 3 }\Omega$

In an oscillating system, a restoring force is a must. In an L-C circuit, the restoring force is provided by a/an

0%
inductor
0%
capacitor
0%
resistor
0%
both A and B

The phase difference between the applied emf and the line current in an anti resonant circuit at resonance is

0%
$\displaystyle \frac{\pi}{2}$ radian
0%
$\displaystyle \pi$ radian
0%
$\displaystyle \frac{3\pi}{2}$ radian
0%
zero

Explanation

We know that phase difference is given by:

$\phi =\arctan { (\dfrac { ({ X }_{ L }-{ X }_{ c }) }{ R } ) }$

at resonance

${ X }_{ L }-{ X }_{ c }=0$

Therefore

$\phi =0$ .

The unit of susceptance is

0%
ohm
0%
ohm $^{-1}$
0%
ohm/cm
0%
ohm/m

Explanation

Susceptance is the imaginary part of admittance .

The admittance is the inverse of impedance.

Unit of admittance is ohm. Unit of admittance is

${ohm}^{-1}$ or siemens.

Unit of susceptance is same as of admittance.

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

0%
zero
0%
$\displaystyle \pi$
0%
$\displaystyle -\frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{2}$

Explanation

In a purely capacitive circuit, current leads the voltage by

$\pi /2$ .

The reactance of a circuit is zero. It is possible that the circuit contains

0%
an inductor and a capacitor
0%
an inductor but no capacitor
0%
a capacitor but no inductor
0%
neither an inductor nor a capacitor

The resonant frequency in an anti resonant circuit is :

0%
$\displaystyle \frac{1}{2\pi \sqrt{LC}}$
0%
$\displaystyle \frac{1}{2\pi} \sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$
0%
$\displaystyle \frac{1}{2\pi} \sqrt{LC}$
0%
$\displaystyle \frac{1}{2\pi} \sqrt{\frac{C}{L}}$

Explanation

For resonance

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

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

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

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

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

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

In an A.C. circuit, the potential difference across an inductance and a resistance joined in series are respectively

$16 V$ and

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

0%
$20.0V$
0%
$25.6 V$
0%
$31.4 V$
0%
$53.5 V$

Explanation

Total potential =

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

$= \sqrt{(16)^2+(20)^2}$

$= \sqrt{256+400}$

$= \sqrt{656}$

$= 25.6V$

An LCR series circuit contains

$L = 8 H$ , C

$=$ 0.5

$\mu$ F and R

$=$ 100

$\Omega$ .The resonant frequency of the circuit is

0%
$\dfrac{100}{\pi }Hz$
0%
$\dfrac{500}{\pi }Hz$
0%
$\dfrac{250}{\pi }Hz$
0%
$\dfrac{125}{\pi }Hz$

Explanation

For resonant frequency reactance should be zero.

So,

$X_L-X_C = 0$

$X_L = X_C$

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

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

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

Thus resonant frequency

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

$f= \dfrac{1}{2\pi}\times \dfrac{1}{\sqrt{8\times 0.5\times 10^{-6}}}$

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

$f= \dfrac{250}{\pi}Hz$

.In the given circuit, the phase difference between voltages across R and C is

0%
zero
0%
$\pi /2$
0%
$\pi$
0%
$3\pi /2$

Explanation

Voltage phase difference between R and C is

$\dfrac{\pi}{2}$ ,
R and L is

$\dfrac{\pi}{2}$ and, L and C is

$\pi$ for a LCR circuit such that voltage in L leads by

$\dfrac{\pi}{2}$ from voltage in R and voltage in R leads by

$\dfrac{\pi}{2}$ from voltage in C.

An experimentalist has a coil of 3 mH. He wants to make a circuit whose frequency is 106 Hz. The capacity of condenser used will be

0%
$752.22 \mu F$
0%
$852.22 \mu F$
0%
$950.22 \mu F$
0%
zero

Explanation

Circuit has to work properly at resonance therfore,

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

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

$C=\dfrac { 1 }{ 4\pi ^{ 2 }f^2L }$

$C=\dfrac { 1 }{ 4\pi ^{ 2 }\times 106^2\times 3\times { 10 }^{ -3 } }$

$C=752.22\mu F$

The correct curve between inductive reactance (XL) and frequency (f) is

Explanation

Since,

${ X }_{ L }=2\pi fL$
therefore its a linear graph hence option(B) is correct

When the values of inductance and capacitance in an

$L-C$ circuit are

$0.5\ H$ and

$8\ \mu$ F respectively then current in the circuit is maximum. The angular frequency of alternating e.m.f. applied in the circuit will be

0%
$\displaystyle 5 \times 10^3\ Radian/sec$
0%
$50\ Radian/sec$
0%
$\displaystyle 5 \times 10^2\ Radian/sec$
0%
$5\ Radian/sec$

Explanation

We know that, maximum current flows when the circuit is at resonance
Therefore, Resonance frequency is given by

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

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

this gives

$\omega =5\times { 10 }^{ 2 }$ Radian/sec

The inductive reactance of a coil is

$2500\ \Omega$ . On increasing its self-inductance three times, the new inductive reactance will be

0%
$7500\ \Omega$
0%
$2500\ \Omega$
0%
$1225\ \Omega$
0%
$zero$

Explanation

We know that ,

${ X }_{ L }=\omega L$
therefore when

$L$ is made three times inductive reactance will also become three times.
Therefore new reactance will be equal to

$3\times 2500\ \Omega=7500\ \Omega$ .

The inductive reactance of a choke coil of

$\displaystyle 1/4 \pi\ mH$ in an

$AC$ circuit of

$50\ Hz,$ will be

0%
25 ohm
0%
0.25 ohm
0%
0.025 ohm
0%
2.5 ohm

Explanation

${ X }_{ L }=\omega L$

${ X }_{ L }==2\pi fL$

${ X }_{ L }=2\pi \times 50\times \dfrac { { 10 }^{ -3 } }{ 4\pi }$

${ X }_{ L }=0.025\ \Omega$

The value of alternating emf in the following circuit will be

0%
220 volt
0%
140 volt
0%
20 volt
0%
100 volt

Explanation

$V=\sqrt { { V }_{ R }^{ 2 }+{ ({ V }_{ L }^{ }-{ V }_{ C }^{ }) }^{ 2 } }$

$V=\sqrt { { 80}^{ 2 }+{ (40^{ }-100^{ }) }^{ 2 } }$

$V=100$ volt

The correct curve between the resistance of a conductor (R) and frequency (f) is

Explanation

The resistance of a resistor is independent of the frequency of the AC source to which it is connected. Thus it has a constant curve against frequency.

However the impedance of an inductor or a capacitor changes with the frequency of the AC source.

Impedance of an inductor

$=\omega L=2\pi fL$

Impedance of a capacitor

$=\dfrac{1}{\omega C}=\dfrac{1}{2\pi fC}$

The capacitive reactance at

$1600\ Hz$ is

$81\ \Omega$ . When the frequency is doubled then capacitive reactance will be

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

Explanation

capacitive reactance is inversely proportional to the frequency

${ X }_{ C }=\dfrac{1}{2\pi fC}$
therefore when frequency is doubled capacitive reactance will get halved therefore new

${ X }_{ C }$ is

$40.5\ \Omega$

$R-C$ circuit is as shown in the following diagram.The capacity reactance and impedance will be

0%
zero
0%
infinity
0%
$\displaystyle \frac{1}{\omega c}$
0%
$\displaystyle \omega c$

Explanation

Since power source is dc therefore,

$\omega =0$

${ X }_{ C }=\dfrac { 1 }{ \omega C }$

${ X }_{ C }=\dfrac { 1 }{0}$

${ X }_{ C }=\infty$

$Z=\infty$

The self inductance of a coil is 1/2 henry. At what frequency will its inductive reactance be 3140

$\Omega$

0%
$100 Hz$
0%
$10 Hz$
0%
$1000 Hz$
0%
$10000 Hz$

Explanation

We know that,

${ X }_{ L }=\omega L$

$\omega =\dfrac { { X }_{ L } }{ L }$

$\omega = 6280$

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

$f=1000$ Hz.

An alternating voltage frequency

$\omega$ is induced in electric circuit consisting of an inductance

$L$ and capacitance

$C$ , connected in parallel. Then across the inductance coil the
(i) current is maximum when

${\omega}^{2}=1/(LC)$
(ii) current is minimum when

${\omega}^{2}=1/(LC)$
(iii) voltage is minimum when

${\omega}^{2}=1/(LC)$
(iv) voltage is maximum when

${\omega}^{2}=1/(LC)$

0%
(i) and (iii) are correct
0%
(i) and (iv) are correct
0%
(ii) and (iii) are correct
0%
(ii) and (iv) are correct

Explanation

In parallel LC circuits, current is minimum when,

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

Also, current and voltage are at angle of 90 degrees.

Hence, voltage is maximum when current is minimum.

Option D is thus correct.

Of the following about capacitive reactance which is correct?

0%
The reactance of the capacitor is directly proportional to its ability to store charge
0%
Capacitive reactance is inversely proportional to the frequency of the current
0%
Capacitive reactance is measured in farad
0%
The reactance of a capacitor in an AC circuit is similar to the resistance of a capacitor in a DC circuit

Explanation

Capacitative reactance is an opposition to the change of voltage across an element.

It is denoted by

$X_C$ and is inversely proportional to the signal frequency(f) and the capacitance C.

$X_C=\dfrac{1}{2\pi fC}$

The capacitive reactance in an AC circuit is

0%
effective resistance due to capacity
0%
effective wattage
0%
effective voltage
0%
None of these

Explanation

Capacitive reactance in an A.C circuit is:

${X}_{C}=\cfrac{1}{\omega C}$ ohm

where,

$C$ is the capacitance of capacitor and

$\omega=2\pi n$ (

$n$ is the frequency of A.C source)

If the inductance of a coil in 1 henry then its effective resistance in a

$D.C$ circuit will be

0%
$\displaystyle \infty$
0%
zero
0%
1 $\Omega$
0%
2 $\Omega$

Explanation

${ X }_{ L }=\omega L$
In DC,

$\omega=0$
Therefore, effective inductive resistance is

${ X }_{ L }=0$

The power loss in an

$AC$ circuit is

$E_{rms}$

$I_{rms}$ , when in the circuit there is only

0%
$C$
0%
$L$
0%
$R$
0%
$L,\ C$ and $R$

Explanation

Inductors and capacitors bring a phase difference between the voltage and current in the circuit, hence changing the p.f. When only a resistance is present,

$Poer\ factor= 1$ .
The power loss in an AC circuit

$=E_{rms} I_{rms} Power\ factor$

With increase in frequency of an A.C. supply, the inductive reactance

0%
decreases
0%
increases directly with frequency
0%
increases as square of frequency
0%
decreases inversely with frequency

Explanation

The inductive reactance

${ X }_{ l }=\omega L$

Hence,

${ X }_{ l } \propto \omega$

As frequency increases

$\rightarrow \omega$

Therefore, inductive reactance increases with frequency.

The impedance of the tweeter and woofer branches are respectively

0%
$\displaystyle \sqrt{R^2+X^2_C} , \sqrt{R^2+X^2_L}$
0%
$\displaystyle \sqrt{R^2+X^2_L} , \sqrt{R^2+X^2_C}$
0%
$\displaystyle \sqrt{R^2+(X_L - X_C)^2}$ each
0%
$\displaystyle \sqrt{\left(\frac{1}{R^2}+\left(\frac{1}{X_C}-\frac{1}{X_L}\right)^2 \right)^{-1}}$ each

Explanation

Impedance of tweeter

$=\left | R + \frac{1}{jC\omega} \right | =\sqrt{R^2+X_C^2}$
Impedance of woofer

$=\left | R + jL\omega \right | =\sqrt{R^2+X_L^2}$

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 of the

0%
inductor increases
0%
resistor increases
0%
capacitor increases
0%
circuit increases

Explanation

The reactance of inductor,

${X}_{L}=\omega L$
The reactance of capacitor,

${X}_{C}=\cfrac{1}{\omega C}$
where

$\omega=2\pi n$ &

$n$ is the frequency of A.C source.

Therefore, reactance of inductor increases.

With increase in frequency of an AC supply, the impedence of an L-C-R series circuit

0%
remains constant
0%
increases
0%
decreases
0%
decreases at first, becomes minimum and then increases

Explanation

We have the formula for Impedance

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

And from the graph it can be easily seen that is

$Decreases$ first and then

$Increases$ .

If a current

$I$ given by

$I={ I }_{ 0 }\sin { \left( \omega t-\pi /2 \right) }$ flows in inductance in an AC circuit which an A.C potential

$E={ E }_{ 0 }\sin{\omega t}$ has been applied, then power consumption

$P$ in the circuit will be

0%
$P={ E }_{ 0 }{ I }_{ 0 }/\sqrt 2$
0%
$P=EI/\sqrt 2$
0%
$P={ E }_{ 0 }{ I }_{ 0 }/2$
0%
zero

Explanation

$P={V}_{r.m.s}\times {I}_{r.m.s}\times \cos { \phi } =\cfrac{1}{2}{E}_{0}\times {I}_{0}\cos { \pi/2 } =0$

In a series combination of

$R,L$ and

$C$ to an A.C source at resonance. If

$R=20\ ohm$ , then impedence

$Z$ of the combination is

0%
$20$ ohm
0%
Zero
0%
$1$ ohm
0%
$400$ ohm

Explanation

At resonance, impedence

$Z=R$
So,

$Z= R = 20\ ohm$

In an

$AC$ circuit, a resistance of

$R$ ohm is connected in series with an inductance

$L$ . If phase angle between voltage and current be

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

0%
$R/4$
0%
$R/2$
0%
$R$
0%
$cannot\ be\ found\ with\ given\ the\ data$

Explanation

$\tan { \phi } =\cfrac { \omega L }{ R } =\cfrac { { X }_{ L } }{ R }$

Given:

$\phi ={ 45 }^{ o }\quad \Rightarrow { X }_{ L }=R$

An inductance of negligible resistance whose reactance is

$22\Omega$ at

$200Hz$ is connected to

$200$ volt. If the line frequency is known to be

$50$ cycles/second, the equation for the line voltage is :

0%
$0.0175$ Henry
0%
$0.175$ Henry
0%
$1.75$ Henry
0%
$17.5$ Henry

Explanation

${X}_{L}=\omega L=2\pi nL$

$\therefore$

$L=\cfrac{{X}_{L}}{2\pi n}=\cfrac{22\times 7}{2\times 22\times 200}H=0.0175H$

In series combination of

$R,L,C$ with an A.C source at resonanace, if

$R=20$ ohm, then impedence

$Z$ of the combination is

0%
$20$ ohm
0%
zero
0%
$10$ ohm
0%
$400$ ohm

Explanation

We know at resonance, reactance (resistance due to inductor and capacitor) be zero (0).
At resonance, Impedance (Z)

$=$ Resistance (R)
Therefore,

$Z = 20$ ohm

Resonance frequency of LCR series a.c. circuit is

${f}_{0}$ . Now the capacitance is made

$4$ times, then the new resonance frequency will become

0%
${f}_{0}/4$
0%
$2{f}_{0}$
0%
${f}_{0}$
0%
${f}_{0}/2$

Explanation

In LCR series circuit, resonance frequency

${f}_{0}$ is given by:

$L\omega =\cfrac { 1 }{ C\omega } \quad \Rightarrow { \omega }^{ 2 }=\cfrac { 1 }{ LC } \quad$

$\therefore \quad \omega =\sqrt { \cfrac { 1 }{ LC } } =2\pi { f }_{ 0 }$

$\therefore \quad { f }_{ 0 }=\cfrac { 1 }{ 2\pi \sqrt { LC } } \quad \implies { f }_{ 0 }\propto \cfrac { 1 }{ \sqrt { C } }$

When the capacitance of the circuit is made

$4$ times, its resonant frequency become

${f}_{0}'$

$\therefore \quad \cfrac { { f }_{ 0 }' }{ { f }_{ 0 } } =\cfrac { \sqrt { C } }{ \sqrt { 4C } } \quad \implies { f }_{ 0 }'=\cfrac { { f }_{ 0 } }{ 2 }$

For the circuit shown in the fig., the current through the inductor is

$0.9A$ while the current through the condenser is

$0.4A$ . Then

0%
current drawn from generator $I=1.13 A$
0%
$\omega=1/(1.5LC)$
0%
$I=0.5A$
0%
$I=0.6A$

Explanation

The current drawn by inductor and capacitor will be in opposite phase.

Hence net current drawn from generator

$={I}_{L}-{I}_{C}=0.9-0.4=0.5amp$

In a pure capacitive A.C circuit current and voltage differ in phase by

0%
${0}^{o}$
0%
${45}^{o}$
0%
${90}^{o}$
0%
${180}^{o}$

Explanation

$i=i_o Sin(\omega t + \pi /2)$
current leads voltage by

$\pi /2$ i.e., current and voltage differ in phase by

$90^o$

In a RLC 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

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

$C$ is doubled,

$L$ should be halved so that resonant frequency remains unchanged.

If the frequency of an A.C is made 4 times of its initial value, the inductive reactance will

0%
be 4 times
0%
be 2 times
0%
be half
0%
remain the same

Explanation

inductive reactance

$= 2 \pi f L$
therefore when f is made 4 times, inductive reactance also becomes 4 times.

In series

$L-C-R$ circuit, the voltages across

$R, L$ and

$C$ are

${V}_{R}, {V}_{L}$ and

${V}_{C}$ respectively. Then the voltage of applied a.c. source must be

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

Explanation

$\quad V =V_R + \left(V_L-V_C \right) j$
So value of apply voltage

$V=\sqrt { { \quad { V }_{ R } }^{ 2 }+{ \left( { { V }_{ L }-{ V }_{ C } } \right) }^{ 2 } }$

In a circuit, the current lags behind the voltage by a phase difference of

$\displaystyle { \pi }/{ 2 }$ , the circuit will contain which of the following:

0%
Only R
0%
Only C
0%
R and C
0%
Only L

Explanation

In an inductor, current lags behind the input voltage by a phase difference of

$\pi/2$ .
Current and voltage are in same phase in resistor whereas current leads the voltage by

$\pi/2$ in a capacitor.
So, the circuit must contain an inductor only.

An A.C. circuit containing only capacitance, the current :

0%
Lags the voltage by $\displaystyle { 90 }^{ o }$
0%
Leads the voltage by $\displaystyle { 90 }^{ o }$
0%
Remains in phase with voltage
0%
Lags the voltage by $\displaystyle { 180 }^{ o }$

Explanation

In an a.c. circuit containing resistance onIy voltage & current remain in the same phase. If circuit contains inductance only, voltage remains ahead of current by phase difference of

$\displaystyle { 90 }^{ o }$ .
If circuit contains capacitance only, current remains ahead of voltage by a phase difference of

$\displaystyle { 90 }^{ o }$ .

Assertion: The resistance offered by an inductor in a d.c circuit is always constant.
Reason : The resistance of inductor in steady state is non-zero.

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.

The self inductance of the motor of an electric fan is 10 H. In order to impart maximum powr of 50 Hz, it should be connected to a capacitance of

0%
$8\mu F$
0%
$4\mu F$
0%
$2\mu F$
0%
$1\mu F$

Explanation

Maximum power (

$I^2 R )$ is obtained when

$I$ is maximum (

$Z$ is minimum).

For

$Z$ minimum,

$X_L=X_C$ , which yields

$C=\dfrac {1}{(2\pi n)^2L}=\dfrac {1}{4\pi^2\times 50\times 50\times 10}$

$\therefore C=0.1\times 10^{-5}F=1\ \mu F$

In a purely inductive circuit, the current:

0%
is in phase with the voltage
0%
is out of phase with the voltage
0%
leads the voltage by $\pi / 2$
0%
lags behind the voltage by $\pi /2$

Explanation

In a purely inductive circuit (an AC circuit containing inductance only) the current lags behind the voltage by

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

Inductive reactance of a coil is expressed in

0%
Amphere
0%
Ohm
0%
Volt
0%
Weber

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 2 - 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 2 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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