[" An alternating voltage "v(t)=220sin100 pi t],[" is applied to a purely resistive load of "50Omega],[" The time taken for the current to rise from "],[" zero to the peak value is: "]

An alternating voltage v(t)=220sin100 pi t volt is applied to a purely resistive, load of 50Ω .The time taken for the current to rise from half of the peak value to the peak value is in milliampere.

An alternating voltage v(t) = 220 sin 100 pt volt is applied to a purely resistive load of 50Omega . The time taken for the current to rise from half of the peak value to the peak value is :

An alternating voltage v(t)=220 sin 100pit volt is applied to a pureluy resistive load of 50Omega . The time taken for the current to rise from half of the peak value to the peak value 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 alternative voltage V=30 sin 50t+40cos 50t is applied to a resitor of resistance 10 Omega . Time rms value of current through resistor is

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

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

A sinusoidal voltage v=200 sin 314 t is applied to a resistor of 10Omega resistance. Calculate the rms value of the current

An alternating voltage given by V=140sin314t is connected across a pure resistor of 50Omega . Find the frequency of the source

A sinusoidal voltage v=200 sin 314 t is applied to a resistor of 10Omega resistance. Calculate the rms value of the voltage