Two capacitors $C_1 = 2 \mu F$ and $C_2 = 6 \mu F$ in series, are connected in parallel to a third capacitor $C_3 = 4 \mu F$ . This arrangement is then connected to a battery of e.m.f. = 2V, as shown in the figure. How much energy is lost by the battery in charging the capacitors
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$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} + C_3 = \frac{2 \times 6}{2 + 6} + 4 = 5.5 \, \mu F$
Energy supplied $(E) = QV = CV^{2} = 22 \times 10^{-6} \text{ J}$
P.E. stored $(U) = \frac{1}{2} C_{eq} V^{2} = \frac{1}{2} \times 5.5 \times (2)^{2} = 11 \times 10^{-6} \mathrm{J}$
⇒ ⇒ Energy lost $= E - U = 11 \times 10^{-6} \text{ J}$
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