Semiconductors

JEE Main level, single correct, medium

91 of 209

Question 91 of 209

The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω50 \Omega. If the battery voltage is 6 V,6 \mathrm{~V}, the current through the 100Ω100 \Omega resistance (in Amperes) is:

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Answer: (B) 0.02

As D2D_{2} is reversed biased, so no current through 75Ω75 \Omega resistor.

now Req=150+50+100R_{e q}=150+50+100

=300Ω=300 \Omega

So, required current I= BatteryVoltage 300I=\frac{\text { BatteryVoltage }}{300} I=6300=0.02\mathrm{I}=\frac{6}{300}=0.02

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