Semiconductors

JEE Main level, numerical answer, medium

61 of 209

Question 61 of 209

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6V0.6V. The current through the resister of 40Ω40\Omega is _____ mAmA.

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Answer: 4

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 40Ω40\Omega is i=1−0.660+40=0.4100=4mAi=\dfrac{1-0.6}{60+40}=\dfrac{0.4}{100}=4mA

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