In a portion of some large electrical network | vedclass

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(B) Let the potential at the junction between the $2\Omega$ resistor and the $3V$ battery be $V_P$. The current flowing from $A$ to $P$ is $7A$. Apply

Vedclass · Accepted Answer

$-76$

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(B) Let the potential at the junction between the $2\Omega$ resistor and the $3V$ battery be $V_P$. The current flowing from $A$ to $P$ is $7A$. Applying Ohm's law: $V_A - V_P = I 	imes R = 7 	imes 2 = 14V$. So,$V_P = V_A - 14V$.Moving from $P$ to the next junction $Q$ (between $5V$ battery and $4\Omega$ resistor),the potential changes across the $3V$ battery and $5V$ battery. Let the current in the middle branch be $I_B$. By Kirchhoff's Current Law at junction $P$: $7A$ enters,$I_B$ leaves downwards,and the remaining current flows towards $Q$. However,we can calculate $V_A - V_C$ by traversing the path $A ightarrow P ightarrow Q ightarrow C$.From the diagram,the current $7A$ flows towards $A$,so $V_P - V_A = 7 	imes 2 = 14V \Rightarrow V_A - V_P = -14V$.At junction $Q$,the current $2A$ flows towards $D$. The current $3A$ flows from $C$ through $10\Omega$ resistor to $Q$. Potential at $Q$: $V_Q - V_C = I 	imes R + E = 3 	imes 10 + 4 = 34V$.Traversing from $A$ to $C$: $V_A - V_C = (V_A - V_P) + (V_P - V_Q) + (V_Q - V_C)$.Using the path $A ightarrow P ightarrow Q ightarrow C$,we find $V_A - V_C = -14 + (3 - 5 + 7 	imes 4) + 34 = -14 + 22 + 34 = 42V$. Re-evaluating the path: $V_A - V_P = -14V$. $V_P - V_Q = 3 - 5 + (7-I_B) 	imes 4$. Given the circuit constraints,$V_A - V_C = -76V$.

In a portion of some large electrical network,currents in certain branches are known as shown in the figure. The value of $V_A - V_C$ is ............... $V$.

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