JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
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 ........... \(mA\).
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
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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