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

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 register R. The value of R is ……

A

`40kOmega`

B

`4kOmega`

C

`200Omega`

D

`400Omega`

Text Solution

Verified by Experts

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