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A parallel combination of two capacitors of capacities ' $2 \mathrm{C}$ ' and ' $\mathrm{C}$ ' is connected across $5 \mathrm{~V}$ battery. When they are fully charged, the charges and energies stored in them be ' $\mathrm{Q}_1$ ', ' $\mathrm{Q}_2$ ' and ' $\mathrm{E}_1$ ', ' $\mathrm{E}_2$ ' respectively. Then $\frac{\mathrm{E}_1-\mathrm{E}_2}{\mathrm{Q}_1-\mathrm{Q}_2}$ in $\mathrm{J} / \mathrm{C}$ is (capacity is in Farad, charge in Coulomb and energy in $J$ )
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Verified Answer
The correct answer is:
$\frac{5}{2}$
We know, $\mathrm{Q}=\mathrm{C} . \mathrm{V}$
$\therefore \quad \mathrm{Q}_1=10 \mathrm{C} \text { and } \mathrm{Q}_2=5 \mathrm{C}$
Energy stored, $\mathrm{E}=\frac{1}{2} \mathrm{CV}^2$
$\begin{aligned}
\therefore \quad \mathrm{E}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}^2 & =\frac{1}{2} \times 2 \mathrm{C} \times 25 \\
& =25 \mathrm{~J}
\end{aligned}$
Similarly, $\mathrm{E}_2=\frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2=\frac{1}{2} \times \mathrm{C} \times 25$ $=12.5 \mathrm{~J}$
$\therefore \quad \frac{\mathrm{E}_1-\mathrm{E}_2}{\mathrm{Q}_1-\mathrm{Q}_2}=\frac{12.5}{5}=\frac{5}{2}$
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