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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 ........... \(mA\).

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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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