The cut-off voltage of the diodes (shown in figure) in forward bias… | JEE Main 2022 (27 Jun, Shift 2)

Q66JEE Main 2022 · 27 JUN · SHIFT 2Medium

P-N Junction Diode

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resistor of 40 $\Omega$ is __________ mA.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$D_1$ and $D_2$ face opposite ways, so for either battery polarity exactly one branch conducts: one $60\ \Omega$ resistor plus one diode.
That diode needs its cut-in voltage $0.6$ V, leaving $1 - 0.6 = 0.4$ V for the resistors.
Series resistance $= 60 + 40 = 100\ \Omega$ .
$I = \frac{0.4}{100} = 4 \times 10^{-3}$ A $= 4$ mA.

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