Q: A light emitting diode (LED) has a voltage drop of 2 volt across it and passes a current of 10 mA when it operates with a 6 volt battery through a limiting resistor R. The value of R is
(a) 40 k Ω
(b) 4 kΩ
(c) 200 Ω
(d) 400 Ω
Click to See Answer :
Ans: (d)
Sol: $\large R = \frac{6-2}{10 \times 10^{-3}}$
R = 400 Ω