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2. Electric Potential and Capacitance
easy
A $6\,\mu F$ capacitor is charged from $10\;volts$ to $20\;volts$. Increase in energy will be
A
$18 \times {10^{ - 4}}\,J$
B
$9 \times {10^{ - 4}}\,J$
C
$4.5 \times {10^{ - 4}}\,J$
D
$9 \times {10^{ - 6}}\,J$
Solution
(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$
Standard 12
Physics
