In the given figure,the potential difference | vedclass

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(D) The circuit consists of a $30 \, V$ battery connected to two $10 \, k\Omega$ resistors in series,with a diode connected to the midpoint. Assuming

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$15$

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(D) The circuit consists of a $30 \, V$ battery connected to two $10 \, k\Omega$ resistors in series,with a diode connected to the midpoint. Assuming the diode is ideal and forward-biased,the circuit can be analyzed as follows:$1$. The total resistance of the main branch is $R_{eq} = 10 \, k\Omega + 10 \, k\Omega = 20 \, k\Omega$.$2$. The current in the main branch is $i = \frac{30 \, V}{20 \, k\Omega} = 1.5 	imes 10^{-3} \, A$.$3$. The potential at the midpoint (before the diode) is $V_{mid} = 30 \, V - (1.5 	imes 10^{-3} \, A 	imes 10 	imes 10^3 \, \Omega) = 30 - 15 = 15 \, V$.$4$. Since the diode is forward-biased,the potential at $A$ will be $15 \, V$. The current flowing through the $10 \, k\Omega$ resistor connected to $A$ is $i_A = \frac{15 \, V}{10 \, k\Omega} = 1.5 	imes 10^{-3} \, A$.$5$. The potential difference between $A$ and $B$ is the voltage across the $10 \, k\Omega$ resistor connected to $A$,which is $15 \, V$.

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

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