JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 28)
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
4
Explanation
D1 : conducting
D2 : open circuit
$$i = {{1 - 0.6} \over {60 + 40}}A$$
$$ = {{0.4} \over {100}}A$$
$$ \Rightarrow i = 4$$ mA
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