Q 12-03-245JEE MainJEE Main 2019 (10 Jan, Shift 1)Easy
A $2\ \text{W}$ carbon resistor is colour coded with green, black, red and silver respectively. The maximum current which can be passed through this resistor is:
Answer: (C) $20\ \text{mA}$
Green $= 5$, black $= 0$, red $=\times10^2$: $R = 50\times10^2 = 5000\ \Omega$ (silver $= \pm10\%$).
$$I_{max} = \sqrt{\frac{P}{R}} = \sqrt{\frac{2}{5000}} = 0.02\ \text{A} = 20\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics