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Question

(D) The energy stored in a capacitor is given by the formula $W = \frac{1}{2} CV^2$, where $C$ is the capacitance and $V$ is the potential difference.

Vedclass · Accepted Answer

$9$ : $1$

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(D) The energy stored in a capacitor is given by the formula $W = \frac{1}{2} CV^2$, where $C$ is the capacitance and $V$ is the potential difference.Initial energy $W_1 = \frac{1}{2} CV_1^2$, where $V_1 = 5 \,V$.Final energy $W_2 = \frac{1}{2} CV_2^2$, where $V_2 = 15 \,V$.The ratio of final energy to initial energy is $\frac{W_2}{W_1} = \frac{\frac{1}{2} CV_2^2}{\frac{1}{2} CV_1^2} = \left(\frac{V_2}{V_1}\right)^2$.Substituting the values, we get $\frac{W_2}{W_1} = \left(\frac{15}{5}\right)^2 = (3)^2 = 9$.Therefore, the ratio is $9 : 1$.

If the potential difference across a capacitor is increased from $5 \,V$ to $15 \,V$, then the ratio of final energy to initial energy stored in the capacitor is

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