What is the current through an ideal P-N junc | vedclass

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(C) In the given circuit,the $P-N$ junction diode is connected in forward bias because the positive terminal of the $3V$ battery is connected to the $

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$20$

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(C) In the given circuit,the $P-N$ junction diode is connected in forward bias because the positive terminal of the $3V$ battery is connected to the $P$-side (through the resistor) and the positive terminal of the $1V$ battery is connected to the $N$-side.For an ideal diode in forward bias,the resistance is zero.The net potential difference across the circuit is $V_{net} = 3V - 1V = 2V$.The resistance in the circuit is $R = 100\,\Omega$.Using Ohm's law,the current $i$ is given by $i = \frac{V_{net}}{R} = \frac{2V}{100\,\Omega} = 0.02\,A$.Converting to milliamperes,$i = 0.02 	imes 1000\,mA = 20\,mA$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

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

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