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(D) The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2$. The work done $W$ in changing the potential difference from $V_i$ to $V_f$ is

Vedclass · Accepted Answer

$1.67 \ W$

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(D) The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2$. The work done $W$ in changing the potential difference from $V_i$ to $V_f$ is the change in potential energy: $W = \Delta U = \frac{1}{2} C(V_f^2 - V_i^2)$. For the first case,$V_1 = 5 \ V$ to $V_2 = 10 \ V$: $W = \frac{1}{2} C(10^2 - 5^2) = \frac{1}{2} C(100 - 25) = \frac{75}{2} C$. For the second case,$V_2 = 10 \ V$ to $V_3 = 15 \ V$: $W' = \frac{1}{2} C(15^2 - 10^2) = \frac{1}{2} C(225 - 100) = \frac{125}{2} C$. Taking the ratio: $\frac{W'}{W} = \frac{125/2 C}{75/2 C} = \frac{125}{75} = \frac{5}{3} \approx 1.67$. Therefore,$W' = 1.67 \ W$.

The amount of work done in increasing the voltage across the plates of a capacitor from $5 \ V$ to $10 \ V$ is $W$. The work done in increasing it from $10 \ V$ to $15 \ V$ will be (nearly):

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