In the circuit shown in the figure- there are two parallel plate capacitors each of capacitance C- The switch S-1- is pressed first to fully charge the capacitor C-1- and then released- The switch S-2- is then pressed to charge the capacitor C-2- After some time- S-2- is released and then S-3- is pressed- After some time-the charge on the upper plate of C-1- is 2CV-0-the charge on the upper plate of C-1- is CV-0-the charge on the upper plate of C-1- is 0-the charge on the upper plate of C-2- is -CV-0-

Question

Toppr Admin · Accepted Answer

Sol-xA0-B-xA0-D-After switch S1 is closed- C1 is charged by 2CV0- when switch S2 is closed- C1 and C2 both have upper plate charge CV0-When S3 is closed- then upper plate of C2 becomes charged by -x2212-CV0 and lower plate by -CV0

Sol. (B, D)After switch S1 is closed, C1 is charged by 2CV0, when switch S2 is closed, C1 and C2 both have upper plate charge CV0.When S3 is closed, then upper plate of C2 becomes charged by −CV0 and lower plate by +CV0

$S o l . (B, D)$
After switch $S_{1}$ is closed, $C_{1}$ is charged by $2 C V_{0}$ , when switch $S_{2}$ is closed, $C_{1}$ and $C_{2}$ both have upper plate charge $C V_{0}$ .
When $S_{3}$ is closed, then upper plate of $C_{2}$ becomes charged by $- C V_{0}$ and lower plate by $+ C V_{0}$