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(B) In the given circuit, the $p$-side of the junction diode is connected to a potential of $3 \, V$ and the $n$-side is connected to a potential of $

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$20$

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(B) In the given circuit, the $p$-side of the junction diode is connected to a potential of $3 \, V$ and the $n$-side is connected to a potential of $1 \, V$ through a resistor of $100 \, \Omega$.Since the potential at the $p$-side is higher than the potential at the $n$-side, the diode is forward-biased.For an ideal diode, the resistance in the forward-biased condition is zero.Therefore, the effective potential difference across the resistor is $V = 3 \, V - 1 \, V = 2 \, V$.Using Ohm's law, the current $I$ in the circuit is given by:$I = \frac{V}{R} = \frac{2 \, V}{100 \, \Omega} = 0.02 \, A$.Converting this to milliamperes, we get $I = 0.02 	imes 1000 \, mA = 20 \, mA$.

Assuming that the junction diode is ideal, the current in the arrangement shown in the circuit diagram is (in $ mA$)

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