The figure shows a network of four resistance | vedclass

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

Vedclass · Accepted Answer

$123$

Vedclass · Answer

(B) Let the potential at node $F$ be $0 \ V$. Let the potential at node $C$ be $V_C$. Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{V_C - 30}{4 + 2} + \frac{V_C - 6}{0} + \frac{V_C - 15}{1 + 2} = 0$ (Note: The middle branch has only a battery,so $V_C = 6 \ V$).If $V_C = 6 \ V$,the current in the left loop $(I_1)$ is: $I_1 = \frac{30 - 6}{4 + 2} = \frac{24}{6} = 4 \ A$.The current in the right loop $(I_2)$ is: $I_2 = \frac{15 - 6}{1 + 2} = \frac{9}{3} = 3 \ A$.The power dissipated in the resistances is $P = I_1^2 R_{left} + I_2^2 R_{right}$.$P = (4)^2 	imes (4 + 2) + (3)^2 	imes (1 + 2) = 16 	imes 6 + 9 	imes 3 = 96 + 27 = 123 \ W$.

The figure shows a network of four resistances and three batteries. The electrical power dissipated as heat in the circuit is .............. $W$.

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