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If the potential difference across a capacitor is increased from $5 \mathrm{~V}$ to $15 \mathrm{~V}$, then the ratio of final energy to initial energy stored in the capacitor is
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Verified Answer
The correct answer is:
9 : 1
Initial energy $\mathrm{W}_1=\frac{1}{2} \mathrm{CV}_1^2$
Final energy $\mathrm{W}_2=\frac{1}{2} \mathrm{CV}_2^2$
$$
\therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^2=\left(\frac{15}{5}\right)^2=(3)^2=9
$$
Final energy $\mathrm{W}_2=\frac{1}{2} \mathrm{CV}_2^2$
$$
\therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^2=\left(\frac{15}{5}\right)^2=(3)^2=9
$$
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