A light emitting diode has a voltage drop of 5 volt across it and passes a current 2mA when it operates with a 9 volt battery through a limiting resistor R. The value of R is
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a
4kΩ
b
2kΩ
c
400Ω
d
200Ω
answer is B.
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Detailed Solution
Potential Drop across resistor = VB - VLED = 9 - 5 = 4 VOhm's Law : R=P.Dcurrent=42×10−3=2000Ω=2kΩ