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(B) In the given circuit, the top diode is in forward bias because its p-side is connected to the positive terminal of the battery. The bottom diode i

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$0.2$

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(B) In the given circuit, the top diode is in forward bias because its p-side is connected to the positive terminal of the battery. The bottom diode is in reverse bias because its n-side is connected to the positive terminal of the battery.For the diode in forward bias, the resistance is $0 \ \Omega$. Thus, the current flows only through the top branch containing the $10 \ \Omega$ resistor.For the diode in reverse bias, the resistance is $\infty$, so no current flows through the bottom branch.Using Ohm's law, the current $I$ supplied by the cell is:$I = \frac{V}{R} = \frac{2 \ V}{10 \ \Omega} = 0.2 \ A$.

$A$ $2 \ V$ cell is connected across points $A$ and $B$ as shown in the figure. Assume that the resistance of each diode is zero in forward bias and infinity in reverse bias. The current supplied by the cell is (in $A$)

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