In the given circuit, calculate the potential | vedclass

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(D) Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:$2 - I_{1}(2 + 3) = 0$$2 - 5I_{1} = 0$$I_{1} = 0.4 \, A$Applying Kirchhoff's Voltage Law

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$-5.2$

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(D) Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:$2 - I_{1}(2 + 3) = 0$$2 - 5I_{1} = 0$$I_{1} = 0.4 \, A$Applying Kirchhoff's Voltage Law $(KVL)$ to the right loop:$4 - I_{2}(3 + 5) = 0$$4 - 8I_{2} = 0$$I_{2} = 0.5 \, A$To find the potential difference between $A$ and $B$, we traverse the path from $A$ to $B$ through the middle branch:$V_{A} + I_{1}(3) - 4 = V_{B}$$V_{A} - V_{B} = 4 - 3I_{1}$$V_{A} - V_{B} = 4 - 3(0.4) = 4 - 1.2 = 2.8 \, V$Wait, re-evaluating the circuit diagram path from $A$ to $B$:Starting from $A$, moving towards $C$ (the node between the $2V$ and $4V$ batteries), the potential increases by $I_{1} 	imes 3 \, \Omega = 0.4 	imes 3 = 1.2 \, V$.Then, crossing the $4V$ battery from negative to positive terminal, the potential increases by $4 \, V$.Thus, $V_{B} - V_{A} = 1.2 + 4 = 5.2 \, V$.Therefore, $V_{A} - V_{B} = -5.2 \, V$.

In the given circuit, calculate the potential difference between $A$ and $B$ in $V$.

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