SemiconductorMedium
Question
A light emitting diode (LED) has a voltage drop of 2 volt across it and passes a current of 10 mA.When it operates with a 6 volt battery through a limiting resistor R. the value of R is
Options
A.40 k
B.4 k
C.200 k
D.400 k
Solution
Maximum current in LED = 10 mA
= 10 × 10-3 A
Resistance of LED =
= 2 × 102 Ω
Now, applying ohm′s law
= 10 × 10-3 = 10-2 = 
200 + R = 600 ⇒ R400Ω
= 10 × 10-3 A
Resistance of LED =
= 2 × 102 ΩNow, applying ohm′s law
= 10 × 10-3 = 10-2 = 200 + R = 600 ⇒ R400Ω
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