When an alternating e.m.f. `e = 300 sin (120pi t)` volt is applied across a bulb, the peak value of the current is found to be 1.5 A, find the
average power

When an alternating e.m.f. e = 300 sin (120pi t) volt is applied across a bulb, the peak value of the current is found to be 1.5 A, find the resistance of the bulb

An alternating e.m.f., e = 300 sin 100 pi t volt, is applied to a pure resistance of 100 Omega . The r.m.s. current through the cicuit is

A voltage e = 50 sqrt2 sin 314.0 t volt is applied across a 10 ohm resistor. Find:- peak value of emf

When the voltage applied across a bulb is 220 v the current passing through the bulb is 0.1 A . Find the power of the bulb.

An alternating e.m.f. is given by 100 sin 100 pi t o a pure resistance of 50 Omega . calculate:- peak value of e.m.f

Equation of an induced e.m.f. is e = 100 sin 100 pi t volt. Find the peak value of e.m.f. and frequency.

An alternating e.m.f. given by e = 50 sin 314t is applied to a pure resistance of 100 ohm. Find:-peak value of e.m.f.

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

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

An alternating e.m.f. is given by 100 sin 100 pi t to a pure resistance of 50 Omega . calculate:-frequency