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A light-emitting diode (LED) has a volta...

A light-emitting diode (LED) has a voltage drop of 2V across it and passes a current of `10 muA`, when it operates with a 6V battery with a limiting resistor R. what is the value of R ?

A

`40kOmega`

B

`4kOmega`

C

`200Omega`

D

`400Omega`

Text Solution

Verified by Experts

The correct Answer is:
D

potential difference across R = 6 - (P.D. across LED)
`=6-2=4VandI=10mAthereforeR=(4)/(10xx10^(-3))=400Omega`
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