For an ideal diode,the current in the followi | vedclass

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(B) In the given circuit,the $p$-side of the diode is connected to $+2 V$ and the $n$-side is connected to $-2 V$ through a $400 \Omega$ resistor.Sinc

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$10 mA$

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(B) In the given circuit,the $p$-side of the diode is connected to $+2 V$ and the $n$-side is connected to $-2 V$ through a $400 \Omega$ resistor.Since the potential of the $p$-side is higher than the potential of the $n$-side,the diode is forward biased.For an ideal diode,the forward resistance is $0 \Omega$.Therefore,the total resistance in the circuit is $R = 400 \Omega$.The potential difference across the circuit is $V = 2 V - (-2 V) = 4 V$.Using Ohm's law,the current $I$ is given by $I = V / R$.$I = 4 V / 400 \Omega = 1 / 100 A = 0.01 A$.Converting to milliamperes,$I = 0.01 	imes 1000 mA = 10 mA$.

For an ideal diode,the current in the following arrangement is:

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