Home
Class 12
PHYSICS
An alternating voltage v(t)=220sin100 pi...

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.

Promotional Banner

Similar Questions

Explore conceptually related problems

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

The rms value of an ac of 50Hz is 10A. The time taken be an alternating current in reaching from zero to maximum value and the peak value will be

The time required for a 50Hz alternating current to increase from zero to 70.7% of its peak value is-

A resostance of 20 Omega is connected to a source of an alternating potential V=220 sin (100 pi t) . The time taken by the corrent to change from the peak value to rms value is

A resistance of 20ohms is connected to a source of an alternating potential V=220sin(100pit) . The time taken by the current to change from its peak value to r.m.s. value is