Questions
A light emitting diode has a voltage drop of 2 volt across it and passes a current of 10mA, when it operates with a 6 volt battery through a limiting resistor R. The value of R is
detailed solution
Correct option is D
Potential Drop across R = E - Vled=6-2=4 VOhm's Law : R=P.Dcurrent=6−210×10−3=400ΩTalk to our academic expert!
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