Alternating Current Test

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Alternating Current Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

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

A
37.1$$\Omega $$, 0.43 A

B
337.1$$\Omega $$, 0.43 A

C
377.1$$\Omega $$, 0.43 A

D
3.7$$\Omega $$, 0.42 A

Solution

$$X_L = \omega L$$
$$= 2\pi f L$$
$$= 2\pi 60\times 1$$
$$= 377.1 \Omega$$
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, $$I = 0.43\ A$$.
Question 2
1 / -0

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,

A
leads the voltage by $$45^{0}$$

B
lags behind the voltage by $$45^{0}$$

C
leads the voltage by $$90^{0}$$

D
lags behind the voltage by $$90^{0}$$

Solution

$$I = 0.707 sin (100\pi t)$$
$$V = 1.414 sin (100\pi t-\frac{\pi}{4})$$
So, current leads voltage by $$45^0$$.
Question 3
1 / -0

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

A
zero

B
$$\pi $$

C
$$-\dfrac{\pi }{2}$$

D
$$\dfrac{\pi }{2}$$

Solution

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]
Question 4
1 / -0

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.

A
A,R are true and R is the correct reason for A

B
A,R are true and R is not the correct reason for A

C
A is true, R is false

D
A is false, R is true

Solution

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.
Question 5
1 / -0

The capacitor offers zero resistance to

A
D.C. only

B
A.C. & D.C.

C
A.C. only

D
neither A.C. nor D.C.

Solution

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.
Question 6
1 / -0

Voltage and current in an ac circuit are given by $$V = 5\sin \left (100\pi t - \dfrac {\pi}{6}\right )$$ and $$I = 4\sin \left (100 \pi t + \dfrac {\pi}{6}\right )$$.

A
Voltage leads the current by $$30^{\circ}$$

B
Current leads the voltage by $$60^{\circ}$$

C
Voltage leads the current by $$60^{\circ}$$

D
Current and voltage are in phase

Solution

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

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

So, current leading voltage by $$\dfrac{\pi}{6}-(-\dfrac{\pi}{6})$$

$$= \dfrac{\pi}{3} = 60^0.$$
Question 7
1 / -0

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

A
$$tan^{-1}\left ( \dfrac{4}{3} \right )$$

B
$$tan^{-1}\left ( \dfrac{3}{4} \right )$$

C
$$tan^{-1}\left ( \dfrac{1}{4} \right )$$

D
$$tan^{-1}\left ( \dfrac{5}{4} \right )$$

Solution

$$X_L = \omega L$$

$$= 2\pi f L$$

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

$$= 400$$

$$R = 300$$

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

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

$$= {tan}^{-1}(\dfrac{4}{3})$$
Question 8
1 / -0

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

A
500 rad/sec

B
4000 rad/sec

C
$$5 \times10^{5}$$ rad/sec

D
5000 rad/sec

Solution

For maximum current impedance should be zero.

So, $$Z = 0$$

$$\Rightarrow X_L-X_C = 0$$

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

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

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

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

$$= 500 rad/sec$$
Question 9
1 / -0

The frequency at which the inductive reactance of $$2\ H$$ inductance will be equal to the capacitive reactance of 2 $$\mu$$F capacitance (nearly)

A
80 Hz

B
40 Hz

C
60 Hz

D
20 Hz

Solution

$$X_L = X_C$$

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

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

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

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

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

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

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

$$= 80Hz$$
Question 10
1 / -0

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

A
2L

B
L/2

C
L/4

D
4L

Solution

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 = $$4C L_1$$

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

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Alternating Current Test - 30

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Alternating Current Test - 30

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SHARING IS CARING

Question 1

37.1$$\Omega $$, 0.43 A

337.1$$\Omega $$, 0.43 A

377.1$$\Omega $$, 0.43 A

3.7$$\Omega $$, 0.42 A

Solution

Question 2

leads the voltage by $$45^{0}$$

lags behind the voltage by $$45^{0}$$

leads the voltage by $$90^{0}$$

lags behind the voltage by $$90^{0}$$

Solution

Question 3

zero

$$\pi $$

$$-\dfrac{\pi }{2}$$

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

Solution

Question 4

A,R are true and R is the correct reason for A

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

A is true, R is false

A is false, R is true

Solution

Question 5

D.C. only

A.C. & D.C.

A.C. only

neither A.C. nor D.C.

Solution

Question 6

Voltage leads the current by $$30^{\circ}$$

Current leads the voltage by $$60^{\circ}$$

Voltage leads the current by $$60^{\circ}$$

Current and voltage are in phase

Solution

Question 7

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

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

$$tan^{-1}\left ( \dfrac{1}{4} \right )$$

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

Solution

Question 8

500 rad/sec

4000 rad/sec

$$5 \times10^{5}$$ rad/sec

5000 rad/sec

Solution

Question 9

80 Hz

40 Hz

60 Hz

20 Hz

Solution

Question 10

2L

L/2

L/4

4L

Solution

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