Question

Two capacitors of capacitances 4⋅0 µF and 6⋅0 µF are connected in series with a battery of 20 V. Find the energy supplied by the battery.

Answer 1

Given:
$$\begin{gathered} C_1=4\mathrm{\mu F} \\ {C}_2=6\mathrm{\mu F} \\ V=20\mathrm{V} \end{gathered}$$

Now,
The equivalent capacitance is given by
$C_{\mathrm{eq}}=\frac{C_1{C}_2}{C_1+C_2}=\frac{4\times 6}{4+6}=2.4\mathrm{\mu F}$
The energy supplied by the battery is given by
$$\begin{gathered} E=C_{\mathrm{eq}}{V}^2 \\ =\left(2.4\mathrm{\mu F}\right)\times {\left(20\right)}^2 \\ =\left(2.4\mathrm{\mu F}\right)\times 400=960\mathrm{\mu J} \end{gathered}$$