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