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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 li

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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
A. `40kOmega`
B. `4kOmega`
C. `200Omega`
D. `400Omega`
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1 Answer

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Correct Answer - D
`R=(6-2)/(10xx10^(-3))=400Omega`
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