Two ideal junction diodes $D_{1}, D_{2}$ are connected as shown in the figure. A $3 V$ battery is connected between A and B. The current supplied by the battery, if its positive terminal is connected to B, is

0.15 A

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0.3 A

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3 A

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1 A

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Solution

The correct option is A $0.15 A$
Given:
$V = 3 V$
$R_{1} = 10 Ω$
$R_{2} = 20 Ω$
Let, $D_{1}$ is the diode connected in series with resistance $R_{1}$ and $D_{2}$ is the diode connected in series with resistance $R_{2}$
Here, both the diodes are ideal hence, when the positive terminal of the battery is connected to B diode $D_{1}$ becomes reverse biased and offers infinite resistance to the current. Also, diode $D_{2}$ becomes forward biased and offers zero resistance to the current.
Therefore current will flow only through the resistance $R_{2}$ .
$I = \frac{V}{R_{2}}$
$I = \frac{3}{20}$
$I = 0.15 A$

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