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(A) To find the potential difference $(V_G - V_H)$,we apply Kirchhoff's Voltage Law along the path from $G$ to $H$.Starting from point $G$ and moving

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$12$

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(A) To find the potential difference $(V_G - V_H)$,we apply Kirchhoff's Voltage Law along the path from $G$ to $H$.Starting from point $G$ and moving towards $H$ through the $4 \, \Omega$ resistor,the $3 \, V$ battery,and the $2 \, \Omega$ resistor:$V_G - (I_1 	imes R_1) - E_1 - (I_1 	imes R_2) = V_{node}$Given $I_1 = 2 \, A$,$R_1 = 4 \, \Omega$,$E_1 = 3 \, V$,$R_2 = 2 \, \Omega$:$V_G - (2 	imes 4) - 3 - (2 	imes 2) = V_{node}$$V_G - 8 - 3 - 4 = V_{node}$$V_G - 15 = V_{node} \implies V_{node} = V_G - 15$  --- (Equation $1$)Now,from the node to $H$ through the $1 \, \Omega$ resistor:$V_{node} - (I_2 	imes R_3) = V_H$From the circuit,the current $I_2$ is flowing upwards from $H$ to the node with a value of $3 \, A$. Therefore,the potential at the node is higher than $V_H$ by the voltage drop across the $1 \, \Omega$ resistor:$V_{node} - (3 	imes 1) = V_H$$V_{node} - 3 = V_H \implies V_{node} = V_H + 3$ --- (Equation $2$)Equating Equation $1$ and Equation $2$:$V_G - 15 = V_H + 3$$V_G - V_H = 15 + 3 = 18 \, V$.Wait,re-evaluating the path: The current $3 \, A$ flows from $H$ to the node. Thus,$V_H - (3 	imes 1) = V_{node} \implies V_H - 3 = V_{node}$.Substituting $V_{node} = V_G - 15$ into $V_H - 3 = V_{node}$:$V_H - 3 = V_G - 15$$V_G - V_H = 15 - 3 = 12 \, V$.

In the part of a circuit shown in the figure,the potential difference $(V_G - V_H)$ between points $G$ and $H$ will be ............... $V$.

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