The potential difference (V_A - V_B) between | vedclass

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(B) To find the potential difference $(V_A - V_B)$,we apply Kirchhoff's voltage law along the path from $A$ to $B$.Starting from point $A$ and moving

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$+9$

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(B) To find the potential difference $(V_A - V_B)$,we apply Kirchhoff's voltage law along the path from $A$ to $B$.Starting from point $A$ and moving towards $B$ in the direction of the current $I = 2 \, A$:$V_A - I R_1 - E - I R_2 = V_B$Where $R_1 = 2 \, \Omega$,$E = 3 \, V$,and $R_2 = 1 \, \Omega$.Substituting the values:$V_A - (2 \, A 	imes 2 \, \Omega) - 3 \, V - (2 \, A 	imes 1 \, \Omega) = V_B$$V_A - 4 \, V - 3 \, V - 2 \, V = V_B$$V_A - 9 \, V = V_B$Therefore,$V_A - V_B = 9 \, V$.

The potential difference $(V_A - V_B)$ between the points $A$ and $B$ in the given figure is ............ $V$.

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