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(D) In the given circuit,the upper diode $D_1$ is forward-biased because its p-side is connected to the positive terminal of the battery.The lower dio

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$0.5$

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(D) In the given circuit,the upper diode $D_1$ is forward-biased because its p-side is connected to the positive terminal of the battery.The lower diode $D_2$ is reverse-biased because its n-side is connected to the positive terminal of the battery.An ideal diode in forward bias acts as a short circuit (zero resistance),and an ideal diode in reverse bias acts as an open circuit (infinite resistance).Therefore,no current flows through the branch containing $D_2$.The total current supplied by the battery flows only through the branch containing $D_1$ and the $10 \ \Omega$ resistor.Using Ohm's law,$I = \frac{V}{R} = \frac{5 \ V}{10 \ \Omega} = 0.5 \ A$.

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

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