JEE Advance - Physics (2013 - Paper 1 Offline - No. 11)
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 $$2C{V_0}$$
The charge on the upper plate of $${C_1}$$ is $$C{V_0}$$
The charge on the upper plate of $${C_2}$$ is $$0$$
The charge on the upper plate of $${C_2}$$ is $$ - C{V_0}$$
Explanation
$$
\mathrm{C}_1=\mathrm{C}_2=\mathrm{C}
$$
* When $\mathrm{S}_1$ is closed. charge on $\mathrm{C}_1=+2 \mathrm{VC}$ on upper plate and $-2 \mathrm{Vc}$ on lower plate.
* When $S_1$ is released and $S_2$ is closed. charge on $C_1$ and $C_2=\mathrm{VC}$
* When $\mathrm{S}_2$ is released and $\mathrm{C}_3$ is closed. charge on $\mathrm{C}_2=\mathrm{CV}$
* When $\mathrm{S}_2$ is closed ( $\mathrm{S}_1$ open)
As $\mathrm{C}_1=\mathrm{C}_2$ the charge gets distributed equal. The upper plates of $C_1$ and $C_2$ now take charge to $\mathrm{CV}_0$ each and lower plate $-\mathrm{CV}_0$ each.
* When $\mathrm{S}_1$ is closed. charge on $\mathrm{C}_1=+2 \mathrm{VC}$ on upper plate and $-2 \mathrm{Vc}$ on lower plate.
* When $S_1$ is released and $S_2$ is closed. charge on $C_1$ and $C_2=\mathrm{VC}$
* When $\mathrm{S}_2$ is released and $\mathrm{C}_3$ is closed. charge on $\mathrm{C}_2=\mathrm{CV}$
* When $\mathrm{S}_2$ is closed ( $\mathrm{S}_1$ open)
As $\mathrm{C}_1=\mathrm{C}_2$ the charge gets distributed equal. The upper plates of $C_1$ and $C_2$ now take charge to $\mathrm{CV}_0$ each and lower plate $-\mathrm{CV}_0$ each.
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