Q.
For the circuit shown figure, which of the following statements is true?
A) With $S_{1}$ closed, $V_{1} = 15$ V,${\ V}_{2} = 20\ $V
B) With $S_{3}$ closed, $V_{1} = {\ V}_{2} = 20\ V$
C) With $S_{1}$and, $S_{3}\ closed,\ V_{1} = {\ V}_{2} = 0$
D) With $S_{1}$ and $S_{3}$closed, ${\ V}_{1} = 30\ $V,${\ V}_{2} = 20\ $V
Explanation
(d) With $S_{1}$and $S_{3}$ closed, the capacitors $C_{1}$ and${\ C}_{2}$ are in series arrangement. In series arrangement potential difference developed across capacitors are in the inverse radio of their capacities. Hence,
$\frac{V_{1}^{'}}{V_{2}^{'}} = \frac{C_{2}}{V_{1}} = \frac{3pF}{3pF} = \frac{3}{2}$and
$V_{1}^{'} + V_{2}^{'} = V_{1} + V_{2} = 30 + 20 = 50V$
On simplification, we get $V_{1}^{'} = V_{1} = 30V\ and\ V_{2}^{'} = V_{2} = 20\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.