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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2019 · 11 January · Evening Shift · Physics · Semiconductors

The circuit shown below contains two ideal diodes, each with a...

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: The circuit shown below contains two ideal diodes, each with a forward resistance of 50 . If the battery voltage is 6

Options

  1. A0.036
  2. B0.02
  3. C0.027
  4. D0.03

Correct answer

B. 0.02

Step-by-step solution

As D2D_{2} is reversed biased, so no current through 75Ω75 \Omega resistor. now Req=150+50+100=300ΩR_{e q}=150+50+100 =300 \Omega So, required current I= BatteryVoltage 300I=\frac{\text{ BatteryVoltage }}{300}\mathrm{I}=6300=0.02=\frac{6}{300}=0.02
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