A 300V battery is connected across capacitors of 3μF and 6μF in parallel. Calculate the energy stored in each capacitor.

Given,
Potential $V = 300\;V$
Capacitor $C_1 = 3\mu F$
Capacitor $C_2 = 6\mu F$
Energy stored in each capacitor $E = ?$

When the capacitors are connected in parallel, then the equivalent capacitance

$C = C_1 + C_2 = (3 + 6)\;\mu F = 9 \; \mu F$

We have,
Energy stored in a capacitor ($C_1$);

$E_1 = \frac{1}{2}C\;V^2 = \frac{1}{2} * 3 * 10^{-6} * 300^2 = 0.135\;J$

Energy stored in a capacitor ($C_2$);

$E_2 = \frac{1}{2}C\;V^2 = \frac{1}{2} * 6 * 10^{-6} * 300^2 = 0.27\;J$

Energy stored in the equivalence capacitance ($C$);

$E = \frac{1}{2}C\;V^2 = \frac{1}{2} * 9 * 10^{-6} * 300^2 = 0.405\;J$ 

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