Electrostatics 5 Question 29
31. For the circuit shown, which of the following statements is true?
(1999, 2M)
(a) With $S_{1}$ closed, $V_{1}=15 \mathrm{~V}, V_{2}=20 \mathrm{~V}$
(b) With $S_{3}$ closed, $V_{1}=V_{2}=25 \mathrm{~V}$
(c) With $S_{1}$ and $S_{2}$ closed, $V_{1}=V_{2}=0$
(d) With $S_{3}$ closed, $V_{1}=30 \mathrm{~V}, V_{2}=20 \mathrm{~V}$
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Answer:
Correct Answer: 31. (d)
Solution:
- When $S_{3}$ is closed, due to attraction with opposite charge, no flow of charge takes place through $S_{3}$. Therefore, potential difference across capacitor plates remains unchanged or $V_{1}=30 \mathrm{~V}$ and $V_{2}=20 \mathrm{~V}$.
Alternate Solution
Charges on the capacitors are
$ q_{1}=(30)(2)=60 \mathrm{pC} $
and $q_{2}=(20)(3)=60 \mathrm{pC}$ or $q_{1}=q_{2}=q$ (say)
The situation is similar as the two capacitors in series are first charged with a battery of emf $50 \mathrm{~V}$ and then disconnected.
$\therefore$ When $S_{3}$ is closed,
$ \text { and } \quad V_{2}=20 \mathrm{~V} \text {. } $
