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Q.

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

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a

40kΩ

b

4kΩ

c

200Ω

d

400Ω

answer is D.

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Detailed Solution

Potential Drop across R = E - Vled=6-2=4 VOhm's Law : R=P.Dcurrent=6−210×10−3=400Ω
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