The potential difference (V_A - V_B) between | vedclass

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(C) To find the potential difference $(V_A - V_B)$,we apply Kirchhoff's Voltage Law $(KVL)$ starting from point $A$ to point $B$.Moving from $A$ to $B

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$9$

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(C) To find the potential difference $(V_A - V_B)$,we apply Kirchhoff's Voltage Law $(KVL)$ starting from point $A$ to point $B$.Moving from $A$ to $B$ in the direction of the current $I = 2 \ A$:$V_A - I \cdot R_1 - E - I \cdot R_2 = V_B$Here,$R_1 = 2 \ \Omega$,$R_2 = 1 \ \Omega$,and $E = 3 \ V$.The current flows from $A$ to $B$,so we encounter the $2 \ \Omega$ resistor,then the battery (entering the positive terminal,so we subtract $3 \ V$),and finally the $1 \ \Omega$ resistor.$V_A - (2 \ A \cdot 2 \ \Omega) - 3 \ V - (2 \ A \cdot 1 \ \Omega) = V_B$$V_A - 4 \ V - 3 \ V - 2 \ V = V_B$$V_A - 9 \ V = V_B$$V_A - V_B = 9 \ V$Thus,the potential difference is $9 \ V$.

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

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