Alternating Current Test

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Alternating Current Test - 43

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Question 1
1 / -0

The simplest type of AC voltage or current is the one which

A
Varies exponentially

B
Varies sinusoidally

C
Varies linearly

D
Does not vary uniformly

Solution

Solution:-
AC voltage or current can be of any form, but the simplest type is a sine wave because any periodic wave can be represented as a combination of sine waves.
Option B is correct.
Question 2
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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

A
50 Hz

B
60 Hz

C
25 Hz

D
40 Hz

Solution

$$Y=100V ,C=2 \mu F=20 \times 10^{-6} $$
$$I = 0.628 A, v = ?$$
$$X_C = \displaystyle \frac{V}{I} = \frac{100}{0.628}$$
$$\Rightarrow \dfrac{ 1}{2 \pi v C} = \displaystyle \frac{100}{0.628}$$
$$v = \displaystyle \frac{0.628 }{100 \times 2 \pi C}$$
$$= 50 Hz$$
Question 3
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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

A
$$\displaystyle \frac{1}{2 \pi f (2 \pi f L - R)}$$

B
$$\displaystyle \frac{1}{2 \pi f ( 2\pi f L + R)} $$

C
$$\displaystyle \frac{1}{ \pi f (2 \pi f L - R)}$$

D
$$\displaystyle \frac{1}{ \pi f (2 \pi f L + R)}$$

Solution

Here $$X_C - X_L = R$$
$$\Rightarrow \displaystyle \frac{1}{2 \pi f C} = (R + 2 \pi f L)$$
$$\Rightarrow \displaystyle C = \frac{1}{2 \pi f (2 \pi f L + R)}$$
Question 4
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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

A
1000 watt

B
40 watt

C
20 watt

D
Zero

Solution

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
$$\displaystyle P={ I }_{ r.m.s }\times { V }_{ r.m.s }\times \cos { \left( { \pi }/{ 2 } \right) } =0\\ $$
Question 5
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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.

A
If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

B
If both Assertion and Reason are true but the Reason is not the correct explanation of the Assertion.

C
If Assertion is true statement but Reason is false.

D
If both Assertion and Reason are false statements.

Solution

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.
Question 6
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Why the current produced by the generator is called an alternating current?

A
continuous decrease and increase of current occurs.

B
after every half cycle direction of flow of current is reversed.

C
both

D
none

Solution

The current produced by a generator has the following property - after every half cycle, the direction of the flow of current is reversed. The given image depicts the situation correctly.
Because of this reason, the current produced by the generator is also called alternating current.
Question 7
1 / -0

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

A
Becomes zero instantaneously

B
Grows monotonically

C
Decays monotonically

D
Oscillates instantaneously

Solution

In an LC circuit current oscillates between,maximum and minimum value. So, LC circuit needs oscillations (electrical). It occurs due to discharging and charging of capacitor and magnetisation and demagnetisation of inductor
Question 8
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A lamp is connected in series with a capacitor and an AC source. What happens if the capacity of the capacitor is reduced?

A
The lamp shines more brightly

B
The lamp shines less brigthly

C
There is no change in the brightness of the lamp

D
Brightness may increase or decrease depending on the frequency of the AC

Solution

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.
Question 9
1 / -0

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

A
$$\displaystyle { 1 }/{ RC }$$

B
$$\displaystyle { R }/{ L }$$

C
$$\displaystyle { 1 }/{ \sqrt { LC } }$$

D
$$\displaystyle { C }/{ L }$$

Solution
Question 10
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The Current in resistance R at resonance is

A
Zero

B
Minimum but finite

C
Maximum but finite

D
Infinite

Solution

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.