A $6\,\mu F$ capacitor is charged from $10\;volts$ to $20\;volts$. Increase in energy will be
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(b) $\Delta E = {E_{Final}} - {E_{Initial}} = \frac{1}{2}C(V_{Final}^2 - V_{Initial}^2)$
$ = \frac{1}{2} \times 6 \times ({20^2} - {10^2}) \times {10^{ - 6}}$
$ = 3 \times (400 - 100) \times {10^{ - 6}} = 3 \times 300 \times {10^{ - 6}} = 9 \times {10^{ - 4}}\,J$
$ = \frac{1}{2} \times 6 \times ({20^2} - {10^2}) \times {10^{ - 6}}$
$ = 3 \times (400 - 100) \times {10^{ - 6}} = 3 \times 300 \times {10^{ - 6}} = 9 \times {10^{ - 4}}\,J$
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