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(B) The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$.Let the potential at $A$ be $0 \ V$ and at $C$ be $10 \ V$. L

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$6$

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(B) The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$.Let the potential at $A$ be $0 \ V$ and at $C$ be $10 \ V$. Let the potential at node $B$ be $V_B$.Using Kirchhoff's current law at node $B$,the sum of charges on the plates connected to $B$ must be zero (assuming initial charge is zero): $10(V_B - 0) + 5(V_B - 0) + 4(V_B - 10) + 6(V_B - 10) = 0$.$15V_B + 10V_B - 40 - 60 = 0 \implies 25V_B = 100 \implies V_B = 4 \ V$.Now,calculate the energy $U = \frac{1}{2} C(V_{diff})^2$ for each capacitor:For $10 \ \mu F$: $U = \frac{1}{2} 	imes 10 	imes (4 - 0)^2 = 5 	imes 16 = 80 \ \mu J$.For $5 \ \mu F$: $U = \frac{1}{2} 	imes 5 	imes (4 - 0)^2 = 2.5 	imes 16 = 40 \ \mu J$.For $4 \ \mu F$: $U = \frac{1}{2} 	imes 4 	imes (4 - 10)^2 = 2 	imes 36 = 72 \ \mu J$.For $6 \ \mu F$: $U = \frac{1}{2} 	imes 6 	imes (4 - 10)^2 = 3 	imes 36 = 108 \ \mu J$.The maximum energy is stored in the $6 \ \mu F$ capacitor.

In the circuit shown in the figure,four capacitors are connected to a battery. The maximum energy is stored in the capacitor of.....$\mu F$

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