What is the current through an ideal PN- junc | vedclass

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(C) In the given circuit,the $P-$side of the diode is connected to the negative terminal of the $1\,V$ battery,and the $N-$side is connected to the ne

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(C) In the given circuit,the $P-$side of the diode is connected to the negative terminal of the $1\,V$ battery,and the $N-$side is connected to the negative terminal of the $3\,V$ battery. To determine the biasing,we look at the potential difference across the diode. The potential at the $P-$terminal is $-1\,V$ and the potential at the $N-$terminal is $-3\,V$. Since the potential of the $P-$side $(-1\,V)$ is higher than the potential of the $N-$side $(-3\,V)$,the diode is in forward bias. For an ideal diode in forward bias,the resistance is zero. The net voltage $V_{net}$ in the circuit is $V_{net} = (-1\,V) - (-3\,V) = 2\,V$. The resistance in the circuit is $R = 100\,\Omega$. Using Ohm's law,the current $I$ is given by: $I = \frac{V_{net}}{R} = \frac{2\,V}{100\,\Omega} = 0.02\,A$. Converting to milliamperes: $I = 0.02 	imes 1000\,mA = 20\,mA$.

What is the current through an ideal $PN-$ junction diode shown in the figure below? (in $mA$)

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