JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2022
- Medium
- Numerical
The cut-off voltage of the diodes (shown in figure) in forward bias is \(0.6 \,V\). The current through the resister of \(40 \,\Omega\) is ........... \(mA\).

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(1- I (60)-0.6- I (40)=0\)
\(\frac{0.4}{100}= I\)
\(I =4 \,mA\)
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