Q 12-14-121JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
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.
Numerical value type. Enter your answer.
Answer: 4
The diodes face opposite ways, so whichever way the cell drives current, exactly one branch conducts (the other diode is reverse biased).
$$I = \frac{1 - 0.6}{60 + 40} = \frac{0.4}{100} = 4\times10^{-3}\ \text{A} = 4\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics