The potential difference between A and B in t | vedclass

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Question

(C) Applying Kirchhoff's Voltage Law $(KVL)$ from point $A$ to $B$:$V_A - I R_1 - V_1 - I R_2 - V_2 - I R_3 = V_B$Given $I = 2 \, A$, $R_1 = 6 \, \Ome

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Applying Kirchhoff's Voltage Law $(KVL)$ from point $A$ to $B$:$V_A - I R_1 - V_1 - I R_2 - V_2 - I R_3 = V_B$Given $I = 2 \, A$, $R_1 = 6 \, \Omega$, $R_2 = 9 \, \Omega$, $R_3 = 5 \, \Omega$, $V_1 = 12 \, V$, $V_2 = 4 \, V$.Substituting the values:$V_A - (2 	imes 6) - 12 - (2 	imes 9) - 4 - (2 	imes 5) = V_B$$V_A - 12 - 12 - 18 - 4 - 10 = V_B$$V_A - V_B = 56 \, V$.Wait, re-evaluating the circuit diagram carefully:The path is $A ightarrow 6 \, \Omega ightarrow 12 \, V ightarrow 9 \, \Omega ightarrow 4 \, V ightarrow 5 \, \Omega ightarrow B$.$V_A - (2 	imes 6) - 12 - (2 	imes 9) - 4 - (2 	imes 5) = V_B$$V_A - 12 - 12 - 18 - 4 - 10 = V_B$$V_A - V_B = 56 \, V$.Since $56$ is not in the options, let's re-examine the diagram. The battery polarities are: $12 \, V$ (negative to positive), $4 \, V$ (positive to negative).$V_A - (2 	imes 6) + 12 - (2 	imes 9) - 4 - (2 	imes 5) = V_B$$V_A - 12 + 12 - 18 - 4 - 10 = V_B$$V_A - V_B = 32 \, V$.Thus, the correct option is $C$.

The potential difference between $A$ and $B$ in the figure is ................. $V$.

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