An ac voltage is applied to a resistance Rand an inductor Lin series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is

π/6
π/4
π/2
zero

Solution

A 220 volts input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

3.6 ampere
2.8 ampere
2.5 ampere
5.0 ampere

Solution

In the given circuit the reading of voltmeter V₁ and V₂ are 300 volt each. The reading of the voltmeter V₃ and ammeter A are respectively

150 V and 2.2 A
220 V and 2.0 A
220 V and 2.0 A
100 V and 2.0 A

Solution

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°. The value of C is

\(\frac{1}{\pi f(2\pi fL)R}\)
\(\frac{1}{2\pi f(2\pi fL-R)}\)
\(\frac{1}{\pi f(2\pi fL+R)}\)
\(\frac{1}{2\pi f(2\pi fL+R)}\)

Solution

In a series resonant circuit, having L,C and R as its elements,the resonant current is i. The power dissipated in circuit at resonance is

\(\frac{i^{2}R}{(\omega L-1/\omega C)}\)
zero
i²ωL
i²R

Solution

At resonance wL= 1/wC and i = E/R , So power dissipated in circuit is P = i²R.

A capacitor has capacitance C and reactance X, if capacitance and frequency become double, then reactance will be

Solution

The reactance of capacitor X =¹⁄_ωC where w is frequency and C is the capacitance of capacitor.

A step up transformer operates on a 230 V line and supplies a current of 2 ampere. The ratio of primary and secondary winding is 1:25 . The current in primary is

25 A
50 A
15 A
12.5 A

Solution

The primary winding of transformers has 500 turns where as its secondary has 5000 turns. The primary is connected to an A.C. supply of 20 V,50 Hz.The secondary will have anout put of

2 V,5 Hz
200 V,500 Hz
2 V,50 Hz
200 V,50 Hz

Resonance frequency of LCR series a.c. circuit is f₀. Now the capacitance is made 4 times, then the new resonance frequency will become

f₀/4
2f₀
f₀
f₀/2

Solution

The instantaneous voltage through a device of impedance 20 Ω is e =80 sin 100 πt. The effective value of the current is

3 A
2.828 A
1.732 A
4 A

Solution

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