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Q.

The circuit shown below contains two ideal diodes, each with a forward resistance of  50  Ω. If the battery voltage is 6V, the current through the 100  Ω resistance (in Amperes) is :

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a

0.036

b

0.027

c

0.030

d

0.020

answer is B.

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Detailed Solution

As D2  is reversed biased, so no current through  75 Ω resistor.

Now Req = 150 + 50 + 100 = 300   Ω

So, required current  I = Battery  Voltage300

I = 6300 = 0.02A

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