[" In the given circuit the average power developed is - "],[qquad [L=0.2Hquad R=50Omega],[" OOOO MM "],[" V "=200sin(250t)" volt "]]

Find capacitance in a series LCR circuit in order to get maximum power. ( X_L = 250 Omega and f = 50 Hz )

In an AC circuit, the current is given by i=5sin(100t-(pi)/2) and the AC potential is V=200sin(100) volt. Then the power consumption is

In the circuit shown in figure R_1=30Omega, R_2=40 Omega, L=0.4H and C=1/3mF . Find seven function of time I, I_1, I_2, V_(R_1), V_L, V_(R_2) and V_C . Also total power consumed in the circuit. In the given potential function V is in volts and omega rad//s V=200 sin(100t+30)

In an AC circuit, the current is given by i=5sin(100t-(pi)/2) and the AC potential is V=200sin(100t) volt. Then the power consumption is

The potential difference V across and the current I flowing through an instrument in an AC circuit are given by what will be the avrage power: V=5 cos omega t volt I=2 sin omega t Amp.

In a Ac circuit aV and I are given below , then find the power dissiapted in the circuit V= 50 sin (50t) V I=50 sin (50+(pi/3))mA

In a series L.C.R. circuit alternating emf (v) and current (i) are given by the equation v=v_(0) sin omega t , i=i_(0)sin(omega t+(pi)/(3)) The average power dissipated in the circuit over a cycle of AC is

In an A.C. circuit the instantaneous values of e.m.f. And current are given by E=100 sin (200 t) volt and I=2 sin(200 t +(pi)/3) ampere. The average power consumed is the circuit is

In an ac circuit, the current is given by I = 5sin ( 100t-pi/2) and the ac potential is V=200sin(100)volt.Thenthepower consumption is