Networks Analysis and Synthesis - Engineering

Q1:

In figure, the voltage V1 = 4∠0 V. The source voltage V2 is

A 8 - j8 volts

B 4 - j4 volts

C 8 - j4 volts

D 8 - j16 volts

ANS:C - 8 - j4 volts

IR = 2∠OA, I1 = 2∠-90° = -j 2A. Total current = 2 - j2. V2 = 2 (2 - j2) + 4∠0 = 8 - j4 volts.