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Q.

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

200Ω

d

400Ω

answer is B.

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Detailed Solution

Potential Drop across resistor = VB - VLED = 9 - 5 = 4 V

Ohm's Law : R=P.Dcurrent=42×10−3=2000Ω=2kΩ

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