A light emitting diode `(LED)` has a voltage drop of `2V` across it and passes a current of `10 mA`. When it operates with a `6V` battery through a limiting resistor `R`. The value of `R` is

Options

A. `40k Omega`
B. `4k Omega`
C. `200 Omega`
D. `400 Omega`

Answer

d

Explanation

The term `LED` is abbreviated as 'Light' Emitting Diode, It is forward-biased `p-n` junction which emits spontaneous radiation. Crrrent in the circuit `=10 m A =10xx10^(-3) A` and voltage in the circuit `=6-2= 4V`
From Ohm's law.
`V=IR`
`:. R=(V)/(I)= (4)/(10xx10^(-3))= 400 Omega`