Question

In the circuit shown below, the potential of $A$ with respect to $B$ of the capacitor $C$ is

• A. 2.00 V
• B. -2.00 V
• C. - 1.50 V
• D. + 1.50 V

Answer 1

Answer: (C)

Let I be the current in the circuit at the steady state. There will be no current in capacitor arm.

Applying Kirchhoff’s second law in the closed loop $PQSTP$, we get
$2.5-10I-20I-1.0=0$
$I=\frac{2.5-1.0}{(10+20)}=0.05A$
$\therefore$ $-1.0-0.05\times 20=-0.5+V_{AB}$
or $V_{AB}=-1.5V$