Alternating Current Test

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Alternating Current Test - 17

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Question 1
1 / -0

When the rms voltages $$V_L, V_C$$ and $$V_R$$ are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio $$V_L : V_C : V_R=1 : 2 : 3$$. If the rms voltage of the AC source is 100 V, then $$V_R$$ is close to :

A
50 V

B
70 V

C
90 V

D
100 V

Solution
Question 2
1 / -0

In a series resonant $$LCR$$ circuit, the voltage across $$R$$ is $$100\ volts$$ and $$R = 1k\ \Omega $$ with $$C=2\ \mu F$$. The resonant frequency is $$200\ rad/s$$. At resonance the voltage across $$L$$ is

A
$$4$$ x $$10^{-3}\ V$$

B
$$2.5$$ x $$10^{-2}\ V$$

C
$$40\ V$$

D
$$250\ V$$

Solution

The resonant angular frequency , $$\omega=\dfrac{1}{\sqrt{LC}}$$

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

At resonance , impedance $$Z=R$$ so the current in the circuit is

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

Voltage across inductor, $$V_L=IX_L=I\omega L=0.1\times 200\times 12.5=250 V$$
Question 3
1 / -0

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

A
80H

B
0.08H

C
0.044H

D
0.065H

Solution

Resistance of lamp by ohm's law,
$$R=\dfrac{V}{I}=\dfrac{80}{10}=8 \Omega$$

When inductor is connected in series and source is applied,
$$V=IZ$$
$$V=I\sqrt{R^2+(\omega L)^2}$$
Given, $$I=10\ A$$ for lamp to work,
$$\therefore 220=10\sqrt{8^2+(2 \times \pi \times 50 \times L)^2}$$
Solving,
$$L=0.0648\ H$$
Question 4
1 / -0

In a series LCR circuit $$\mathrm{R}=200\ Omega$$ and the voltage and the frequency of the main supply is $$220$$ $$V$$ and $$50$$ $$Hz$$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $$30^0$$. On taking out the inductor from the circuit the current leads the voltage by $$30^0$$. The power dissipated in the LCR circuit is

A
$$305$$ $$\mathrm{W}$$

B
$$210$$ $$\mathrm{W}$$

C
$$Zero \mathrm{W}$$

D
$$242$$ $$\mathrm{W}$$

Solution

The given circuit is under resonance as $$\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$$

Hence power dissipated in the circuit is $$\displaystyle \frac{V^2}{R}$$

$$\mathrm{P}=242\mathrm{W}$$
Question 5
1 / -0

In an a.c. circuit, the instantaneous e.m.f. and current are given by $$e=100 \sin 30t$$, $$i=20\sin\left(30t -\displaystyle\frac{\pi}{4}\right)$$. In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively.

A
$$\displaystyle\frac{50}{\sqrt{2}}, 0$$

B
$$50, 0$$

C
$$50, 10$$

D
$$\displaystyle\frac{1000}{\sqrt{2}}, 10$$

Solution

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

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

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

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

Wattless current = $$I_{rms} sin\theta$$

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

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

= $$10 amp $$
Question 6
1 / -0

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

A
4L

B
2L

C
L/2

D
L/4

Solution

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

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

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

$$L'=\dfrac{L}{2}$$
Question 7
1 / -0

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

A
$$50\ \text{V}$$

B
$$50\sqrt{2}\mathrm{v}$$

C
$$100\ \text{V}$$

D
$$0\mathrm{V}$$ (zero)

Solution

In a series LCR circuit, the voltage across the inductor (L) and the capacitor (C) are in opposite phase.
So the voltage across LC combination will be $$(50-50)=0$$ V
Question 8
1 / -0

A series LR circuit is connected to a voltage source with $$V\left( t \right) ={ V }_{ 0 }\sin { \Omega t } $$. After very large time, current $$I\left( t \right) $$ behaves as $$\left( { t }_{ 0 }\gg \frac { L }{ R } \right) $$

A

B

C

D

Solution

Let $$L$$ be the inductance and $$R$$ be the resistance then the current $$I$$ can be found by using Kirchoff's law as:
$$\quad L\dfrac { dI }{ dt } -IR={ V }_{ 0 }\sin { \Omega t } \\ \Rightarrow \dfrac { dI }{ dt } -\dfrac { IR }{ L } =\dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } \\ \Rightarrow \dfrac { d }{ dt } (I{ e }^{ \frac { -Rt }{ L } })=\dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } { e }^{ \frac { -Rt }{ L } }\\ \Rightarrow I{ e }^{ \frac { -Rt }{ L } }=\int _{ 0 }^{ t }{ \dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } { e }^{ \frac { -Rt }{ L } } } dt\quad \\ \Rightarrow I=-\dfrac { { V }_{ 0 }L }{ { R }^{ 2 }+{ \Omega }^{ 2 }{ L }^{ 2 } } (\dfrac { R }{ L } \sin { \Omega t } +\Omega \cos { \Omega t } )\\ \Rightarrow I=-\dfrac { { V }_{ 0 } }{ \sqrt { { R }^{ 2 }+{ \Omega }^{ 2 }{ L }^{ 2 } } } \sin { (\Omega t+\phi ) } $$
Where, $$\tan { \phi } =\dfrac { L\Omega }{ R } $$
So, it is clearly seen that the solution is dictated by the forcing function which is sinusoidal. Therefore, the steady state solution will be sinusoidal in nature.
So, the correct answer is option B.
Question 9
1 / -0

In an LCR circuit as shown below both switches are open initially. Now switch $$S_{1}$$ is closed, $$S_{2}$$ kept open. ( $$q$$ is charge on the capacitor and $$\tau=RC$$ is capacitive time constant). Which of the following statement is correct?

A
At $$ t=\tau,\ q=CV/2$$

B
At $$ t=2\tau,\ q=CV(1-e^{-2})$$

C
At $$ t=\frac{T}{2},\ q=CV(1-e^{-1})$$

D
Work done by the battery is half of the energy dissipated in the resistor.

Solution

$$C$$harge on the capacitor at any time ` $$t$$' is

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

at $$t=2\tau$$

$$
q=\ CV(1-e^{-2})
$$
Question 10
1 / -0

$$A$$ series $$R-C$$ combination is connected to an $$AC$$ voltage of angular frequency $$(\omega =500radian/s$$. lf the impedance of the $$R-C$$ circuit is $$R\sqrt{1.25}$$, the time constant (in millisecond) of the circuit is

A
1

B
2

C
3

D
4

Solution

Given : $$\omega = 500$$ $$radian/s$$
Let the capacitance of the capacitor be $$C$$.
Thus resistance of capacitor $$X_C = \dfrac{1}{\omega C} = \dfrac{1}{500 C}$$

Impedance of the circuit $$Z = R\sqrt{1.25}$$
Using $$Z^2 = R^2 + X_C^2 $$

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

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

$$\implies$$ $$R^2 C^2 = \dfrac{10^{-6}}{(0.25)^2}$$
We get time constant of the circuit $$RC = \dfrac{10^{-3}}{0.25} =0.004$$ s $$ = 4$$ ms