A light emitting diode has a voltage drop of 2 v across it and passes a current of `10muA`. When it operates with a 6 v battery through a limiting resistor R, the value of R is

Options

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

Answer

D

Explanation

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