The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be

3.13 A

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2.5 A

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10.0 A

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1.43 A

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Solution

The correct option is A $2.5 A$
From figure, ideal diode $D_{1}$ is reversed biased whereas ideal diode $D_{2}$ is forward biased. Thus $D_{1}$ acts as an open switch while $D_{2}$ as a closed switch as shown in the equivalent circuit.
Thus current flowing through $R_{1}$ , $I = \frac{V}{R_{1} + R_{3}} = \frac{10}{2 + 2} = 2.5 A$

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