The figure shows a network of four resistance | vedclass

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(A) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. Applying Kirchhoff's voltage law to the left loop $(ABCF)$:

Vedclass · Accepted Answer

The potential difference $V_C - V_F = 6\,V.$

Vedclass · Answer

(A) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. Applying Kirchhoff's voltage law to the left loop $(ABCF)$: $30 - 4I_1 - 6 - 2I_1 = 0 \implies 6I_1 = 24 \implies I_1 = 4\,A.$ Applying Kirchhoff's voltage law to the right loop $(CDEF)$: $15 - 1I_2 - 6 - 2I_2 = 0 \implies 3I_2 = 9 \implies I_2 = 3\,A.$ The current flowing through the branch $CF$ is $I = I_1 + I_2 = 4 + 3 = 7\,A.$ Since the current $I$ flows from $C$ to $F$,the potential difference $V_C - V_F$ across the $6\,V$ battery is given by $V_C - V_F = E + Ir = 6 + 7(0) = 6\,V$ (assuming internal resistance is zero). Thus,the potential difference $V_C - V_F = 6\,V$ and the current flows from $C$ to $F$. Therefore,none of the given options $(A), (B), (C)$ are strictly correct based on the standard interpretation of the circuit,but $(A)$ is a correct statement about the potential difference.

The figure shows a network of four resistances and three batteries. Choose the correct alternative.

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