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(C) In the given circuit,diode $D_1$ is reverse biased because its p-terminal is at a lower potential than its n-terminal,so it acts as an open circui

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$2.5$

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(C) In the given circuit,diode $D_1$ is reverse biased because its p-terminal is at a lower potential than its n-terminal,so it acts as an open circuit and blocks the current.Diode $D_2$ is forward biased because its p-terminal is at a higher potential than its n-terminal,so it acts as a closed switch (ideal diode).Hence,the equivalent circuit consists of the $10 \ V$ battery,resistance $R_1 = 2 \ \Omega$,and resistance $R_3 = 2 \ \Omega$ in series.The total resistance of the circuit is $R_{eq} = R_1 + R_3 = 2 \ \Omega + 2 \ \Omega = 4 \ \Omega$.The current flowing through the resistance $R_1$ is given by $I = \frac{V}{R_{eq}} = \frac{10 \ V}{4 \ \Omega} = 2.5 \ A$.

The given circuit has two ideal diodes connected as shown in the figure. The current flowing through the resistance $R_1$ will be.....$ A$

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