- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
If the potential of a capacitor having capacity of $6\,\mu F$ is increased from $10\, V$ to $20\, V$, then increase in its energy will be
A
$12 \times 10^{-6} \,J$
B
$9 \times 10^{-4} \,J$
C
$4 \times 10^{-6} \,J$
D
$4 \times 10^{-9} \,J$
Solution
(b) The increase in energy of the capacitor $\Delta U = \frac{1}{2}C(V_2^2 – V_1^2) = \frac{1}{2}(6 \times {10^{ – 6}})\,({20^2} – {10^2})$
$ = 3 \times {10^{ – 6}} \times 300 = 9 \times {10^{ – 4}}\,J$
Standard 12
Physics
