In the given figure,the potential difference | vedclass

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(B) The diode is in forward bias,so it acts as a short circuit (assuming an ideal diode).The circuit consists of a $30 \, V$ battery in series with a

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$5$

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(B) The diode is in forward bias,so it acts as a short circuit (assuming an ideal diode).The circuit consists of a $30 \, V$ battery in series with a $10 \, k\Omega$ resistor. This combination is connected to a parallel combination of a $10 \, k\Omega$ resistor and a series branch containing a diode and another $10 \, k\Omega$ resistor.Equivalent resistance of the parallel part:$R_p = \frac{10 \, k\Omega 	imes 10 \, k\Omega}{10 \, k\Omega + 10 \, k\Omega} = 5 \, k\Omega$Total resistance of the circuit:$R_{eq} = 10 \, k\Omega + 5 \, k\Omega = 15 \, k\Omega$Total current from the battery:$I = \frac{30 \, V}{15 \, k\Omega} = 2 \, mA$The voltage across the parallel combination (which is the voltage across the branch containing $A$ and $B$):$V_p = I 	imes R_p = 2 \, mA 	imes 5 \, k\Omega = 10 \, V$Since the diode is in series with the $10 \, k\Omega$ resistor in the branch $AB$,the current through this branch is:$I_{AB} = \frac{V_p}{10 \, k\Omega + 10 \, k\Omega} = \frac{10 \, V}{20 \, k\Omega} = 0.5 \, mA$The potential difference between $A$ and $B$ is the voltage across the $10 \, k\Omega$ resistor in that branch:$V_{AB} = I_{AB} 	imes 10 \, k\Omega = 0.5 \, mA 	imes 10 \, k\Omega = 5 \, V$

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

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