An emf `e = e_0 sin omega t` applied to a series L-C-R circuit derives a current `I = I_0 sin omega t` in the circuit. Deduce the expression for the average power dissipated in the circuit.

The alternating e.m.f. of e = e_0 sin omega t is applied across LR series circuit. The impedance and phase angle between current and voltage is

In an AC circuit, e and i are given by e = 150 sin 150 t V and i = 150 sin (150t+pi/3) A. The 1 power dissipated in the circuit is

An alternating e.m.f. given by e = 50 sin 314t is applied to a pure resistance of 100 ohm. Find:- power dissipated in the circuit.

A circuit has a resistance and a reactance each equal to 100 Omega .Find its power factor. If the circuit draws a current of 1.5 A for an applied voltage of 200 V, what is the average power consumed in the circuit?

A sinusoidal voltage of peak value 283V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Omega , L = 25.48 mH and C = 796 muf . Find:-The power dissipated in the surface.

If a current I given by I_o sin (omega-pi/2) flowsin an ac circuit across which an potential of e= e_0sin omegat has been applied, then the power consumption P in the circuit will be.

An alternating e.m.f. is applied to a series L - C - R circuit. It the frequency of the applied e.m.f. is more than the resonant frequency of the circuit then the circuit will act as

An e.m.f. e = 4 cos 1000 t volt is applied to LR circuit of inductance 3 mH and resistance 4 Omega . The amplitude of current in the circuit is

An alternating e.m.f., e=200 sin omega t is applied to a lamp, whose filament has a resistance of 1000 Omega . The r.m.s. value of current is

In an A.C. circuit I =10 cos (100t) ampere and V = 20 sin (100t). The power loss in the circuit will be