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(B) The diode is forward-biased as the potential at $A$ is higher than the potential at $B$. Assuming an ideal diode,it acts as a short circuit.Let th

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$10$

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(B) The diode is forward-biased as the potential at $A$ is higher than the potential at $B$. Assuming an ideal diode,it acts as a short circuit.Let the potential at $B$ be $0 \ V$. The circuit consists of a $30 \ V$ source connected to a $10 \ k\Omega$ resistor in series with a parallel combination of two $10 \ k\Omega$ resistors (one connected between $A$ and $B$,and the other in series with the diode).The equivalent resistance of the two parallel $10 \ k\Omega$ resistors is $R_p = \frac{10 	imes 10}{10 + 10} = 5 \ k\Omega$.The total resistance of the circuit is $R_{eq} = 10 \ k\Omega + 5 \ k\Omega = 15 \ k\Omega$.The current from the battery is $I = \frac{30 \ V}{15 \ k\Omega} = 2 \ mA$.The potential at $A$ relative to $B$ is the voltage drop across the parallel combination $R_p$:$V_{AB} = I 	imes R_p = 2 \ mA 	imes 5 \ k\Omega = 10 \ V$.

In the figure,the potential difference between $A$ and $B$ is......$V$.

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