Question
In the circuit shown, the current through the ideal diode is:
Image

Answer

(c) 25 mA
Explanation: The given diode is ideal and forward-biased so there is short circuit. No current flow through the $20 \Omega$ and whole current flow through the short circuit. The current flow through the circuit is, $I =\frac{2 V}{80 \Omega}=25 mA$

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