1.

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

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


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