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(D) In the given circuit, the top diode is forward-biased, while the middle diode is reverse-biased.The reverse-biased diode acts as an open circuit (

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$0.04 A$

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(D) In the given circuit, the top diode is forward-biased, while the middle diode is reverse-biased.The reverse-biased diode acts as an open circuit (infinite resistance), so no current flows through the middle branch.The circuit simplifies to a series combination of the $10 	ext{ V}$ battery, the $150 \Omega$ resistor, the top diode (with $50 \Omega$ forward resistance), and the $50 \Omega$ resistor in that branch.The total resistance of the circuit is $R_{total} = R_{diode} + R_{top} + R_{series} = 50 \Omega + 50 \Omega + 150 \Omega = 250 \Omega$.The current $I$ flowing through the $150 \Omega$ resistor is given by Ohm's law: $I = \frac{V}{R_{total}} = \frac{10 	ext{ V}}{250 \Omega} = 0.04 	ext{ A}$.

Assume that each diode as shown in the figure has a forward bias resistance of $50 \Omega$ and an infinite reverse bias resistance. The current through the $150 \Omega$ resistance is

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