Two capacitors, C_1 = 2 text{ mF} and C_2 = 8 | vedclass

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(A) When capacitors are connected in series, the equivalent capacitance $C_{net}$ is given by:$C_{net} = \frac{C_1 	imes C_2}{C_1 + C_2} = \frac{2

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The charge on each capacitor is $480 	imes 10^{-3} 	ext{ C}$

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(A) When capacitors are connected in series, the equivalent capacitance $C_{net}$ is given by:$C_{net} = \frac{C_1 	imes C_2}{C_1 + C_2} = \frac{2 	imes 8}{2 + 8} = 1.6 	ext{ mF} = 1.6 	imes 10^{-3} 	ext{ F}$.The total charge $Q$ supplied by the source is:$Q = C_{net} 	imes V = 1.6 	imes 10^{-3} 	imes 300 = 0.48 	ext{ C} = 480 	imes 10^{-3} 	ext{ C}$.In a series circuit, the charge on each capacitor is the same, so $Q_1 = Q_2 = 480 	imes 10^{-3} 	ext{ C}$.The potential difference across $C_1$ is $V_1 = \frac{Q}{C_1} = \frac{480 	imes 10^{-3}}{2 	imes 10^{-3}} = 240 	ext{ V}$.The potential difference across $C_2$ is $V_2 = \frac{Q}{C_2} = \frac{480 	imes 10^{-3}}{8 	imes 10^{-3}} = 60 	ext{ V}$.The total energy stored is $U = \frac{1}{2} C_{net} V^2 = \frac{1}{2} 	imes 1.6 	imes 10^{-3} 	imes (300)^2 = 0.8 	imes 10^{-3} 	imes 90000 = 72 	ext{ J}$.

Two capacitors, $C_1 = 2 \text{ mF}$ and $C_2 = 8 \text{ mF}$, are connected in series across a $300 \text{ V}$ source. Then:

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