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

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(B) First,calculate the equivalent capacitance of $C_1$ and $C_2$ in series:$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 	imes 6}{2 + 6} = \frac{12}

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

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(B) First,calculate the equivalent capacitance of $C_1$ and $C_2$ in series:$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 	imes 6}{2 + 6} = \frac{12}{8} = 1.5 \,\mu F$.Next,calculate the total equivalent capacitance $C_{eq}$ of the circuit,where $C_{12}$ is in parallel with $C_3$:$C_{eq} = C_{12} + C_3 = 1.5 \,\mu F + 4 \,\mu F = 5.5 \,\mu F = 5.5 	imes 10^{-6} \,F$.The energy supplied by the battery is given by the formula $E = \frac{1}{2} C_{eq} V^2$,where $V = 2 \,V$:$E = \frac{1}{2} 	imes (5.5 	imes 10^{-6}) 	imes (2)^2$$E = \frac{1}{2} 	imes 5.5 	imes 10^{-6} 	imes 4$$E = 11 	imes 10^{-6} \,J$.

Two capacitors $C_1 = 2 \,\mu F$ and $C_2 = 6 \,\mu F$ are connected in series. This combination is then connected in parallel with a third capacitor $C_3 = 4 \,\mu F$. This entire arrangement is connected to a battery of $2 \,V$. What is the energy supplied by the battery to charge the capacitors?

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