SSC JE Electrical Engineering 2011 Paper

Question : 106 of 150

Marks: +1, -0

In the circuit shown in the figure given below, instantaneous current i(t) under steady state is given by

Solution:

For DC current inductor is short circuit and capacitor is open circuit. So

i_{\DC} = 0

for AC V = 10 sin t

ω=1

X_L=ω1=2Ω

X_C=1/ω_C=1Ω

I_{\ac}={10\sint}/(R+j(X_L-X_C))

Z =1 + 1 j =√ 2∠ 45°

I_{\ac}={10\sint}/{√2∠45°

I_{\ac}=7.07\sin(t-45°)

i(t)=I_{\ac}+I_{\DC}

=7.07\sin(t-45°)

Go to Question:

Prev Question

Next Question