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 [2013]

Question

Smart Achievers · Accepted Answer

(2, 4)

When switch $S_{1}$ is pressed, the capacitor $C_{1}$ gets charged such that its upper plate acquires a positive charge $+ 2 C V_{0}$ and the lower plate $- 2 C V_{0} .$

When switch $S_{2}$ is pressed and $S_{1}$ is released, As $C_{1} = C_{2},$ the charge gets distributed equally. The upper plates of $C_{1}$ and $C_{2}$ now take charge $+ C V_{0}$ each, and the lower plates $- C V_{0}$ each.

When $S_{2}$ is released and $S_{3}$ is pressed, the charge on the upper plate of $C_{1}$ is $C V_{0}$ and the charge on the upper plate of $C_{2}$ is $- C V_{0} .$

Solution

1 The charge on the upper plate of $C_{1}$ is $2 C V_{0}$

2 The charge on the upper plate of $C_{1}$ is $C V_{0}$

3 The charge on the upper plate of $C_{2}$ is 0

4 The charge on the upper plate of $C_{2}$ is $- C V_{0}$

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Solution

1 The charge on the upper plate of C1 is 2CV0

2 The charge on the upper plate of C1 is CV0

3 The charge on the upper plate of C2 is 0

4 The charge on the upper plate of C2 is -CV0

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1 The charge on the upper plate of $C_{1}$ is $2 C V_{0}$

2 The charge on the upper plate of $C_{1}$ is $C V_{0}$

3 The charge on the upper plate of $C_{2}$ is 0

4 The charge on the upper plate of $C_{2}$ is $- C V_{0}$