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. `40kOmega`
B. `4 kOmega`
C. `200Omega`
D. `400Omega`

Answer

D

Explanation

As LED is connected to a battery through a resistance in series.
The currect flowing, 10 mA is the same.
The volate drop across LED= 2V.
`:.` As the battery has 6,v the potential difference across `R=4V` ltbr. `:. iR= 4 V rArr R= (4V)/(10xx10^(-3)A)=400 Omega`