An alternating emf, e=200 sin `omegat` is applied to a lamp, whose filament has a resistance of `1000Omega`. The rms valve of current 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

An alternating emf, e=300 sin100 pit volt, is applied to a pure resistance of 100 Omega . The rms current through the circuit is

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

An alternating e.m.f. given by e=200 sin 50 t is applied to a circuit containing only a resistance of 50 Omega . What is the value of r.m.s. current in the circuit?

Alternating voltage V = 400 sin ( 500 pi t ) is applied across a resistance of 0.2 kOmega . The r.m.s. value of current will be equal to

An alternating emf given by equation e=300sin(100pi)t V is applied to a resistance 100 Omega . The rms current through the circuit is (in amperes).

An alternating emf given by equation e=300sin(100pi)t V is applied to a resistance 100 Omega . The rms current through the circuit is (in amperes).

An alternating emf of E=20sin 1000tV is applied on a circuit containing a pure resistance of 25Omega in series with a coil of self-inductance 40 mH and resistance 15Omega . Find out the phase difference between E and the current I.

An alternating emf of E=20sin 1000tV is applied on a circuit containing a pure resistance of 25Omega in series with a coil of self-inductance 40 mH and resistance 15Omega . Find out the power factor of the circuit , Also , find out this capacitance of a capacitor that should be connected in series to raise the power factory to unity.