The figure shows a network of five resistance | vedclass

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(C) Let the potential at node $F$ be $0 \ V$. Then the potential at node $C$ is $V_C$. Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{V_C

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(C) Let the potential at node $F$ be $0 \ V$. Then the potential at node $C$ is $V_C$. Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{V_C - 30}{4 + 2} + \frac{V_C - 0}{3} + \frac{V_C - 15}{1 + 2} = 0$$\frac{V_C - 30}{6} + \frac{V_C}{3} + \frac{V_C - 15}{3} = 0$Multiplying by $6$:$(V_C - 30) + 2V_C + 2(V_C - 15) = 0$$V_C - 30 + 2V_C + 2V_C - 30 = 0$$5V_C = 60 \implies V_C = 12 \ V$.The current in the left branch is $I_1 = \frac{30 - 12}{6} = 3 \ A$.The current in the middle branch is $I_2 = \frac{12 - 0}{3} = 4 \ A$.The current in the right branch is $I_3 = \frac{12 - 15}{3} = -1 \ A$ (meaning current flows from $D$ to $C$).The power consumed by the circuit is the sum of power dissipated in resistors: $P = I_1^2 R_1 + I_2^2 R_2 + I_3^2 R_3$$P = (3^2 	imes 6) + (4^2 	imes 3) + ((-1)^2 	imes 3) = 54 + 48 + 3 = 105 \ W$.

The figure shows a network of five resistances and two batteries. The total electrical power consumed by the circuit is ............... $W$.

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