Two capacitors C_1 = 2,mu F and C_2 = 6,mu F | vedclass

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(B) The equivalent capacitance $C_{eq}$ of the circuit is calculated as follows:First,$C_1$ and $C_2$ are in series,so their equivalent capacitance $C

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$11 	imes 10^{-6}\,J$

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(B) The equivalent capacitance $C_{eq}$ of the circuit is calculated as follows:First,$C_1$ and $C_2$ are in series,so their equivalent capacitance $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 	imes 6}{2 + 6} = \frac{12}{8} = 1.5\,\mu F$.Then,$C_{12}$ is in parallel with $C_3$,so $C_{eq} = C_{12} + C_3 = 1.5 + 4 = 5.5\,\mu F$.The total charge supplied by the battery is $Q = C_{eq} V = 5.5 	imes 10^{-6} 	imes 2 = 11 	imes 10^{-6}\,C$.The total energy supplied by the battery is $E = QV = (11 	imes 10^{-6}) 	imes 2 = 22 	imes 10^{-6}\,J$.The potential energy stored in the capacitors is $U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} 	imes 5.5 	imes 10^{-6} 	imes (2)^2 = 11 	imes 10^{-6}\,J$.The energy lost by the battery is the difference between the energy supplied and the energy stored in the capacitors: $\Delta E = E - U = 22 	imes 10^{-6} - 11 	imes 10^{-6} = 11 	imes 10^{-6}\,J$.

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 $e.m.f. = 2\,V$,as shown in the figure. How much energy is lost by the battery in charging the capacitors?

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