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(D) Let the potential of the negative terminal of the battery be $0 \ V$. Then the potential of the positive terminal is $10 \ V$.Let the potential at

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$0$

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(D) Let the potential of the negative terminal of the battery be $0 \ V$. Then the potential of the positive terminal is $10 \ V$.Let the potential at point $B$ be $V_B$ and at point $A$ be $V_A$.For the upper branch,the resistors $8 \ \Omega$ and $6 \ \Omega$ are in series. The potential at $B$ is determined by the voltage divider rule: $V_B = 10 	imes \frac{6}{8+6} = 10 	imes \frac{6}{14} = \frac{60}{14} = \frac{30}{7} \ V$.For the lower branch,the resistors $4 \ \Omega$ and $3 \ \Omega$ are in series. The potential at $A$ is determined by the voltage divider rule: $V_A = 10 	imes \frac{3}{4+3} = 10 	imes \frac{3}{7} = \frac{30}{7} \ V$.The potential difference between $A$ and $B$ is $|V_A - V_B| = |\frac{30}{7} - \frac{30}{7}| = 0 \ V$.

The potential difference between point $A$ and $B$ is ............. $V$.

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