In the circuit shown below, the ac source has voltage $V=20\cos \left(\omega t\right)$ volts with ω = 2000 rad/sec.  The amplitude of the current is closest to:
         

1. 2 A

2. 3.3 A

3. $2/\sqrt{5}\mathrm{ \; A}$

4. $\sqrt{5}\mathrm{ \; A}$

In an ac circuit the reactance of a coil is $\sqrt{3}$ times its resistance, the phase difference between the voltage across the coil to the current through the coil will be 

(1) π/3

(2) π/2

(3) π/4

(4) π/6

The capacity of a pure capacitor is 1 farad. In dc circuits, its effective resistance will be 

(1) Zero

(2) Infinite

(3) 1 ohm

(4) 1/2 ohm

The power factor of an ac circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is 

(1) $R/\omega L$

(2) $R/{(R^2+{\omega }^2{L}^2)}^{1/2}$

(3) $\omega L/R$

(4) $R/{(R^2-{\omega }^2{L}^2)}^{1/2}$

An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency ω. The power dissipated in the circuit is:

1. $\frac{(R^2+{\omega }^2{L}^2)}{V}$

2. $\frac{V^{2}R}{(R^2+{\omega }^2{L}^2)}$

3. $\frac{V}{(R^2+{\omega }^2{L}^2)}$

4. $\frac{\sqrt{R^2+{\omega }^2{L}^2}}{V^2}$

In an LCR circuit, the potential difference between the terminals of the inductance is 60 V, between the terminals of the capacitor is 30 V and that between the terminals of the resistance is 40 V. The supply voltage will be equal to:

1. 50 V

2. 70 V

3. 130 V

4. 10 V

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 will be:

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

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

3. $\frac{1}{2\pi f(2\pi fL-R)}$

4. $\frac{1}{\pi f(2\pi fL-R)}$

For the series LCR circuit shown in the figure, what is the resonance frequency and the amplitude of the current at the resonating frequency ?

1. 2500 rad/s and $5\sqrt{2}A$

2. 2500 rad/s and 5A

3. 2500 rad/s and $\frac{5}{\sqrt{2}}A$

4. 25 rad/s and $5\sqrt{2}A$

In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. $E=E_0\cos \left(\omega t\right)$ applied to the circuit. The power consumed in the circuit is:

(1) $\frac{E_0^2}{R}$

(2) $\frac{E_0^2}{2R}$

(3) $\frac{E_0^2}{4R}$

(4) $\frac{E_0^2}{8R}$

One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has a self-inductance of value: (f = 50 Hz) 

(1) 0.052 H

(2) 2.42 H

(3) 16.2 mH

(4) 1.62 mH