If V_{A}-V_{B}=7 text{ V},then find the store | vedclass

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(C) The equivalent capacitance $C_{eq}$ of the two capacitors $C_1 = 3 \mu 	ext{F}$ and $C_2 = 6 \mu 	ext{F}$ connected in series is given by:$\frac

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$\frac{8}{3} \mu 	ext{ J}$

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(C) The equivalent capacitance $C_{eq}$ of the two capacitors $C_1 = 3 \mu 	ext{F}$ and $C_2 = 6 \mu 	ext{F}$ connected in series is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \mu 	ext{F}^{-1}$So,$C_{eq} = 2 \mu 	ext{F}$.Applying Kirchhoff's Voltage Law $(KVL)$ in the loop:$V_A - \frac{q}{C_1} - 5 - \frac{q}{C_2} - V_B = 0$$V_A - V_B - 5 = q \left( \frac{1}{C_1} + \frac{1}{C_2} ight)$Given $V_A - V_B = 7 	ext{ V}$,we have:$7 - 5 = q \left( \frac{1}{3} + \frac{1}{6} ight)$$2 = q \left( \frac{1}{2} ight)$$q = 4 \mu 	ext{C}$.The energy stored in capacitor $C_1$ is given by:$U_{C_1} = \frac{q^2}{2 C_1} = \frac{(4 \mu 	ext{C})^2}{2 	imes 3 \mu 	ext{F}} = \frac{16}{6} \mu 	ext{J} = \frac{8}{3} \mu 	ext{J}$.

If $V_{A}-V_{B}=7 \text{ V}$,then find the stored energy in $C_1$.

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