The potential difference (V_A - V_B) between | vedclass

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(D) 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 $A$,the curren

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$23$ $V$

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(D) 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 $A$,the current $I = 3 	ext{ A}$ flows through the $4 \ \Omega$ resistor,causing a potential drop of $I 	imes R = 3 	imes 4 = 12 	ext{ V}$. Then,we encounter the $5 	ext{ V}$ battery from the positive terminal to the negative terminal,which represents a potential drop of $5 	ext{ V}$. Finally,the current flows through the $2 \ \Omega$ resistor,causing a potential drop of $I 	imes R = 3 	imes 2 = 6 	ext{ V}$. Thus,the equation is: $V_A - (3 	imes 4) - 5 - (3 	imes 2) = V_B$ $V_A - 12 - 5 - 6 = V_B$ $V_A - V_B = 12 + 5 + 6$ $V_A - V_B = 23 	ext{ V}$

The potential difference $(V_A - V_B)$ between the points $A$ and $B$ in the given part of the circuit is:

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