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(B) For the given circuit, the diode is forward-biased because the potential at the anode $(3 	ext{ V})$ is higher than the potential at the cathode

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$20$

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(B) For the given circuit, the diode is forward-biased because the potential at the anode $(3 	ext{ V})$ is higher than the potential at the cathode $(1 	ext{ V})$.Since the diode is ideal, its resistance in the forward-biased state is $0 \Omega$.The potential difference across the resistor is $\Delta V = 3 	ext{ V} - 1 	ext{ V} = 2 	ext{ V}$.Using Ohm's law, the current $i$ flowing through the circuit is:$i = \frac{\Delta V}{R} = \frac{2 	ext{ V}}{100 \Omega} = 0.02 	ext{ A}$.Converting this to milliamperes:$i = 0.02 	imes 1000 	ext{ mA} = 20 	ext{ mA}$.

Assuming that the junction diode is ideal, find the current in the arrangement shown in the figure. (in $\text{mA}$)

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