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 $V_{AB} = - 1.5 \, V$