A coil of inductive reactance $1/\sqrt{3} \mathrm{\Omega }$ and resistance 1 Ω is connected to a 200 V, 50 Hz AC supply. The time lag between maximum voltage and current is

• $\frac{1}{200} \mathrm{s}$

• $\frac{1}{300} \mathrm{s}$

• $\frac{1}{500} \mathrm{s}$

• $\frac{1}{600} \mathrm{s}$

In the A.C. circuit shown, keeping 'K' pressed, if an iron rod inserted into the coil, the bulb in the circuit

• glows less brightly

• glows with same brightness (as before the rod is inserted)

• gets damaged

• glows more brightly

The output of a step down transformers is measured to be 48 V when connected to a 12 W bulb. The value of peak current is

• $\frac{1}{\sqrt{2}} \mathrm{A}$

• $\sqrt{2} \mathrm{A}$

• $\frac{1}{2\sqrt{2}} \mathrm{A}$

• $\frac{1}{4} \mathrm{A}$

In the circuit shown in the figure, the AC source gives a voltage V = 20 cos (2000t). Neglecting source resistance, the voltmeter and ammeter reading will be

• 1.68 V, 0.47 A

• 0 V, 1.47 A

• 5.6 V, 1.4 A

• 0 V, 1.4 A

In series LCR circuit, the phase difference between applied voltage and current is

• Positive when X_L > X_C

• positive when X_C > X_L

• 90^o

• 0^o

96.

A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance such that the impedance of the circuit becomes 'Z' the power drawn will be

• 
• 
• 
• 

A small signal voltage V(t) = V_o sin  is applied across an ideal capacitor C:

• over a full cycle the capacitor C does not consume any energy from the voltage source

• current I(t) is in phase with voltage V(t)

• Current I(t) leads voltage V(t) by 180^o

• Current I(t) leads voltage V(t) by 180^o

An inductor of 20 mH, a capacitor of 50  and a resistor of 40 ohm are connected in series across a source of emf V = 10 sin 340 t. the power loss in the ac circuit is,

• 0.67 W

• 0.76 W

• 0.89 W

• 0.89 W

The instantaneous values of alternating current and voltages in a circuit are given as


The average power in Watts consumed in the circuit is

• 1/4

• 

• 1/2

• 1/2

If voltage across a bulb rated 220 V -100W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

• 20%

• 2.5%

• 5%

• 5%