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(A) Let the potential at node $A$ be $V_A$. Applying Kirchhoff's Current Law $(KCL)$ at node $A$,the sum of charges on the plates connected to node $A

Vedclass · Accepted Answer

$6$

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(A) Let the potential at node $A$ be $V_A$. Applying Kirchhoff's Current Law $(KCL)$ at node $A$,the sum of charges on the plates connected to node $A$ must be zero (assuming the plates were initially uncharged):$C_1(V_A - 6) + C_2(V_A - 6) + C_3(V_A - 0) = 0$Given $C_1 = 2 \, \mu F$,$C_2 = 2 \, \mu F$,and $C_3 = 4 \, \mu F$:$2(V_A - 6) + 2(V_A - 6) + 4(V_A) = 0$$2V_A - 12 + 2V_A - 12 + 4V_A = 0$$8V_A = 24$$V_A = 3 \, V$Now,the charge on capacitor $C_1$ is $Q_1 = C_1(6 - V_A) = 2 \, \mu F 	imes (6 - 3) \, V = 2 \, \mu F 	imes 3 \, V = 6 \, \mu C$.

Three capacitors are connected as shown in the figure. Then the charge on capacitor $C_1$ is.....$\mu C$.

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