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 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 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

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

An A.C. source of voltage V=200 "sin" (100 pit)" volts is connected with R=50Omega .The time interval in which of the current goes from its peak value of half of the peak value is: (A) 1/400 sec (B) 1/50 sec (C) 1/300 sec (D) 1/200 sec

An AC source is rated at 220V, 50 Hz. The time taken for voltage to change from its peak value to zero is

An AC source is rated at 220V, 50 Hz. The time taken for voltage to change from its peak value to zero is

An a.c. source is rated at 220 V - 50 Hz. The time taken for voltage to change from its peak value to zero is

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.