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JEE Main 202227 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2022 · 27 June · Evening Shift · Physics · Semiconductors

The cut-off voltage of the diodes (shown in figure) in forward bias...

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6V0 . 6V. The current through the resister of 40Ω40Ω is _____ mAmA.

The cut-off voltage of the diodes (shown in figure) in forward bias is 0 . 6V . The current through the resister of 40Ω

Correct answer

0

Step-by-step solution

Here diode D1{D}_{1} is forward biased and diode D2{D}_{2} is reverse biased.

So, no current will flow through diode D2{D}_{2} resistor.

Current through the resistor of 40Ω40Ω is i=1−0.660+40=0.4100=4mAi = \dfrac{1 - 0 . 6}{60 + 40} = \dfrac{0 . 4}{100} = 4mA

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