Two ideal diodes are connected to a battery a | vedclass

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(D) In the given circuit,the upper diode $D_{1}$ is connected in forward bias because its p-side is connected to the positive terminal of the battery.

Vedclass · Accepted Answer

$0.5$

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(D) In the given circuit,the upper diode $D_{1}$ is connected in forward bias because its p-side is connected to the positive terminal of the battery.Therefore,$D_{1}$ acts as a closed switch (ideal diode has zero resistance in forward bias).The lower diode $D_{2}$ is connected in reverse bias because its p-side is connected to the negative terminal of the battery.Therefore,$D_{2}$ acts as an open switch (ideal diode has infinite resistance in reverse bias).The circuit effectively consists of only the upper branch with a $5 \ V$ battery and a $10 \ \Omega$ resistor.The current supplied by the battery is $I = \frac{V}{R} = \frac{5 \ V}{10 \ \Omega} = 0.5 \ A$.

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

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