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(B) In the given circuit,the diode $D_2$ is connected in reverse bias because its $n$-terminal is connected to the positive terminal of the battery. T

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$0.25$ $A$

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(B) In the given circuit,the diode $D_2$ is connected in reverse bias because its $n$-terminal is connected to the positive terminal of the battery. Therefore,no current flows through the branch containing the $40 \Omega$ resistor. The circuit simplifies to a single loop with the $40 \Omega$ resistor and the diode $D_1$ in forward bias. Using Ohm's law,the current $I$ supplied by the battery is $I = \frac{V}{R} = \frac{10 	ext{ V}}{40 \Omega} = 0.25 	ext{ A}$.

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is

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