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Standard XII

Physics

Discharging of Capacitors

Circuit is as...

Question

Circuit is as shown in the figure. Initially, the key is connected to 1 and now shifted to 2, before shifting of key, energy stored in capacitor is $U_{1}$ and after shifting of key, energy stored in capacitor is $U_{2}$ , then $\frac{U_{1}}{U_{2}}$ will be

4 : 1

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2 : 1

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3 : 1

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1 : 1

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Solution

The correct option is D $1 : 1$
When key is connected to $1$ :
Potential difference across capacitor is
$V = 10 - 5 = 5 V$

As the energy stored in the capacitor is given by
$U = \frac{1}{2} C V^{2}$
$U_{1} = 1 / 2 \times C (10 - 5)^{2}$
$U_{1} = 1 / 2 \times C (5)^{2}$

When key is connected to $2$ :
Potential difference across capacitor is
$V = 5 V$ , as the $10 V$ is no longer connected across the capacitor.
$U_{2} = 1 / 2 \times C (5)^{2}$

$∴ \frac{U_{1}}{U_{2}} = \frac{1}{1}$

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Circuit is as shown in the figure. Initially, the key is connected to 1 and now shifted to 2, before shifting of key, energy stored in capacitor is U1 and after shifting of key, energy stored in capacitor is U2, then U1U2 will be

Circuit is as shown in the figure. Initially, the key is connected to 1 and now shifted to 2, before shifting of key, energy stored in capacitor is $U_{1}$ and after shifting of key, energy stored in capacitor is $U_{2}$ , then $\frac{U_{1}}{U_{2}}$ will be