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Physics NCERT Exemplar Solutions Class 12th Chapter Eight Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor 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 $Ω$ is………….mA

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Payal Gupta | Contributor-Level 10

3 months ago

D₁ → forward biased

D₂ → Reversed biased

So, current will flow through only D₁

$l = \frac{(1 - 0.6) v}{(60 + 40) Ω} = \frac{0.4}{100}$

= 4 × 10^-³ A

= 4mA

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

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