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In a LCR series circuit, the potential difference between the terminals of the inductance is 60 V between the terminals of the capacitor is 30 V and that across the resistance is 40 V. Then, supply voltage will be equal to

50 V
70 V
130 V
10 V

62.

A choke is preferred to a resistance for limiting current in AC circuit because

choke is cheap
there is no wastage of power
choke is compact in size
choke is a good absorber of heat

63.
An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of

0.1 mH

1 mH

0.1 H

1.1 H

1.1 H

Resistance of bulb

$R = \frac{V^{2}}{P} = \frac{{(100)}^{2}}{50} = 200 Ω$

Current through bulb is,

$I = \frac{V}{R} = \frac{100}{200} = 0.5 A$

In a circuit containing inductive reactance (X_L) and resistance (R), impedance (Z) of the circuit is

$Z = \sqrt{R^{2} + ω^{2} L^{2}} . . . . . . (i) Here, Z = \frac{200}{0.5} = 400 Ω Now, X_{L}^{2} = Z^{2} - R^{2} = {(400)}^{2} - {(200)}^{2} {(2 πfL)}^{2} = 12 \times 10^{4} L = \frac{2 \sqrt{3} \times 100}{2 π \times 50} = \frac{2 \sqrt{3}}{π} L = 1.1 H$

64.

An inductance of $(\frac{200}{π}) mH$ , a capacitance of $(\frac{10^{- 3}}{π}) F$ and a resistance of 10 Ω are connected in series with an AC source 220 V, 50 Hz. The phase angle of the circuit is

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{2}$
$\frac{π}{3}$

65.

A step-down transformer reduces the voltage of a transmission line from 2200 V to 220 V. The power delivered by it is 880 W and its efficiency is 88%. The input current is

4.65 mA
0.045 A
0.45 A
4.65 A

66.

The value of alternating emf E in the given circuit will be

220 V
140 V
100 V
20 V

67.

A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 V and 50% of power is lost, then the current in secondary will be

2.5 A
5 A
0.25 A
0.5 A

68.

For a series LCR circuit at resonance, the statement which is not true is

Peak energy stored by a capacitor = peak energy stored by an inductor.
Average power = apparent power
Wattless current is zero.
Power factor is zero.

69.

An electric heater rated 220 V and 550 W is connected to AC mains. The current drawn by it is

0.8 A
2.5 A
0.4 A
1.25 A

70.

A resistor and a capacitor are connected in series with an AC source. If the potential drop across the capacitor is 5 V and that across resistor is 12 V, then applied voltage is