Verified Solution • 2025–26 Session moderate

QJEEM22Phb414: JEE Main 2022 Physics - Class 12 JEEM22Phb414 Solution: The cut-off voltage of the diodes (shown in figure) in forward…

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 Ω\Omega is __________ mA.

JEE Main 2022 (Online) 27th June Evening Shift Physics - Semiconductor Question 80 English

Step-by-Step Verified Solution

D1 : conducting

D2 : open circuit

i=1−0.660+40Ai = {{1 - 0.6} \over {60 + 40}}A

=0.4100A= {{0.4} \over {100}}A

⇒i=4\Rightarrow i = 4 mA

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring