Alternating Current Test

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Alternating Current Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If the phase difference between Alternating Voltage and Alternating Current is $$\frac{\pi }{6}$$ and the resistance in the circuit is $$\sqrt{300}\Omega $$ , then the impedance of the circuit will be

A
25$$\Omega $$

B
50$$\Omega $$

C
20$$\Omega $$

D
100$$\Omega $$

Solution

impedance $$\times cos \theta = $$resistance
impedance = $$\frac{resistance}{cos \theta}$$
$$= \frac{\sqrt{300}}{cos \pi /6}$$
$$= 20 \Omega$$
Question 2
1 / -0

The inductance of a resistanceless coil is 0.5 henry. In the coil, the value of alternating current is 0.2 A, whose frequency is 50 Hz. The reactance of circuit is

A
15.7$$\Omega $$

B
157$$\Omega $$

C
1.57$$\Omega $$

D
757$$\Omega $$

Solution

Reactance $$= \omega L $$

$$= 2\pi f L$$

$$= 2\pi \times 50 \times 0.5$$

$$= 157 \Omega$$
Question 3
1 / -0

The capacitive reactance at 1600Hz is 81 $$\Omega $$ .When the frequency is doubled then the capacitive reactance will be:

A
40.5$$\Omega $$

B
81$$\Omega $$

C
162$$\Omega $$

D
zero

Solution

Capacitive reactance = $$\dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$$
$$81 = \dfrac{1}{2\pi \times 1600 C}$$
$$C = \dfrac{1}{2\pi \times 1600 \times 81}$$
Capacitive reactance = $$\dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$$
$$81 = \dfrac{1}{2\pi \times 2 \times 1600 \times \dfrac{1}{2\pi \times 1600 \times 81}}$$
$$C = \dfrac{81}{2} = 40.5 \Omega$$
Question 4
1 / -0

In an LCR circuit, current is I $$=$$ 10sin100$$\pi$$ t from an A.C. source. The value of average voltage at the ends of the resistance R $$=$$ 10$$\Omega $$ will be

A
25V

B
$$\dfrac{25}{\sqrt{2}}V$$

C
zero V

D
1V

Solution

The voltage across the resistance keeps on changing its direction with time periodically, since the source is AC.
Thus the time average of the voltage across the resistor would be zero.
Question 5
1 / -0

In an A.C. circuit the potential difference across an inductance and resistance joined in series are respectively 16V and 20V. The total potential difference across the circuit is

A
20.0V

B
25.6V

C
31.9V

D
53.5V

Solution

Potential difference across the circuit

$$V = \sqrt{V_R^2 + V_L^2 } = \sqrt{20^2 + 16^2} = \sqrt{656} = 25.6 V$$
Question 6
1 / -0

A coil is used in a circuit in which an A.C. of frequency 50Hz is flowing. The self-inductance of the coil, in order to produce an impedance of 100 $$\Omega $$ , will be ___ H

A
$$ \pi$$

B
$$\frac{1}{\pi }$$

C
$$ \pi ^{2}$$

D
$$\frac{1}{\pi^{2} }$$

Solution

impedance = WL
$$= 2\pi f L$$
$$100 = 2\pi 50\times L$$
$$L = \frac{1}{\pi}H$$
Question 7
1 / -0

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

A
30 V

B
1$$\mu $$F

C
20V

D
$$10\sqrt{2}V$$

Solution

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

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

$$= 10\sqrt{2}V$$
Question 8
1 / -0

The source frequency for which a $$5 \mu$$ F capacitor has a reactance of 1000$$\Omega $$ is

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

B
$$\dfrac{1000}{\pi }$$ Hz

C
$$200 \ Hz$$

D
$$5000 \ Hz$$

Solution

$$X_C = \dfrac{1}{\omega C}$$
$$= \dfrac{1}{2\pi f C}$$
$$f = \dfrac{1}{2\pi X_c C}$$
$$= \dfrac{1}{2\pi \times 1000 \times 5 \times 10^{-6}}$$
$$= \dfrac{100}{\pi}Hz$$
Question 9
1 / -0

The inductive reactance of a coil is 2500 $$\Omega $$ . On increasing it’s self-inductance to three times, the new inductive reactance will be:

A
7500$$\Omega $$

B
2500$$\Omega $$

C
1225$$\Omega $$

D
zero

Solution

$$X_L = \omega L$$
So, $$2500 = \omega L$$
When L is increased to 3L
then $$X_L = \omega (3L)$$
$$= 3 \omega L$$
$$= 3(2500)$$
$$= 7500 \Omega$$
Question 10
1 / -0

The unit of susceptance is

A
ohm

B
ohm$$^{-1}$$

C
ohm/cm

D
ohm/m

Solution

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.

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 20

Result

Alternating Current Test - 20

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SHARING IS CARING

Question 1

25$$\Omega $$

50$$\Omega $$

20$$\Omega $$

100$$\Omega $$

Solution

Question 2

15.7$$\Omega $$

157$$\Omega $$

1.57$$\Omega $$

757$$\Omega $$

Solution

Question 3

40.5$$\Omega $$

81$$\Omega $$

162$$\Omega $$

zero

Solution

Question 4

25V

$$\dfrac{25}{\sqrt{2}}V$$

zero V

1V

Solution

Question 5

20.0V

25.6V

31.9V

53.5V

Solution

Question 6

$$ \pi$$

$$\frac{1}{\pi }$$

$$ \pi ^{2}$$

$$\frac{1}{\pi^{2} }$$

Solution

Question 7

30 V

1$$\mu $$F

20V

$$10\sqrt{2}V$$

Solution

Question 8

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

$$\dfrac{1000}{\pi }$$ Hz

$$200 \ Hz$$

$$5000 \ Hz$$

Solution

Question 9

7500$$\Omega $$

2500$$\Omega $$

1225$$\Omega $$

zero

Solution

Question 10

ohm

ohm$$^{-1}$$

ohm/cm

ohm/m

Solution

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