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(D) The diode is ideal and forward-biased because the potential at the anode $(+6 	ext{ V})$ is greater than the potential at the cathode $(+5 	ext{

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$10$ mA

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(D) The diode is ideal and forward-biased because the potential at the anode $(+6 	ext{ V})$ is greater than the potential at the cathode $(+5 	ext{ V})$.In forward bias, an ideal diode acts as a short circuit (zero resistance).Thus, the current $I$ flowing through the circuit is determined by Ohm's law:$I = \frac{V_{	ext{applied}}}{R} = \frac{6 	ext{ V} - 5 	ext{ V}}{100 	ext{ } \Omega} = \frac{1 	ext{ V}}{100 	ext{ } \Omega} = 0.01 	ext{ A} = 10 	ext{ mA}$.Therefore, the correct option is $(D)$.

What is the current flowing through the given circuit? $A$ given diode is an ideal diode.

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