The current in the following circuit is:

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(D) The diode is connected such that the p-side is at $+5 \ V$ and the n-side is at $+3 \ V$. Since the potential at the p-side is higher than the pot

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$10^{-2} \ A$

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(D) The diode is connected such that the p-side is at $+5 \ V$ and the n-side is at $+3 \ V$. Since the potential at the p-side is higher than the potential at the n-side,the diode is forward-biased. Assuming an ideal diode,the potential difference across the resistor is $V = 5 \ V - 3 \ V = 2 \ V$. The resistance $R$ is $200 \ \Omega$. Using Ohm's law,the current $I$ is given by $I = \frac{V}{R} = \frac{2 \ V}{200 \ \Omega} = 0.01 \ A = 10^{-2} \ A$.

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