The current through the ideal diode as shown in the figure is:

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0.02A

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0.04A

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0.06A

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Solution

The correct option is B
0.02A

6V - 3V = 150I diode is forward biased

$I = \frac{3}{150}$

$I = 0.02 A$

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Current through the ideal diode as shown in the figure is

R_{1}

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