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 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

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 :

A sinusoidal voltage of amplitude 5V is applied to resistance of 500 Omega . The r.m.s. current in the circuit is

A sinusoidal voltage V (t) = 100 sin (500 t) is applied across a pure inductance of L = 0.02 H . The current through the coil 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 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 a.c. source of voltage V = 100 sin 100 pi t . Is connected to a resistor of (25)/(sqrt2) ohm . What is the r.m.s value of current (in ampere) through the resistor ?

An alternating voltage given as, V=100sqrt(2)sin100t V is applied to a capacitor of 1muF . The current reading of the ammeter will be equal to ................ mA.