Q 12-14-171JEE MainJEE Main 2020 (3 Sep, Shift 1)Easy
When a diode is forward biased, it has a voltage drop of $0.5\ \text{V}$. The safe limit of current through the diode is $10\ \text{mA}$. If a battery of emf $1.5\ \text{V}$ is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is:
Answer: (C) $100\ \Omega$
$$R = \frac{1.5 - 0.5}{10\times10^{-3}} = 100\ \Omega$$
Solution by Sreeraj P, M.Sc Physics