A light emitting diode has a voltage drop of 2 volt across it and passes a current of 10 ma when it operates with 6 volt battery through a limiting resistor r. The value of r is :
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The value of the limiting resistor is 400Ω
Explanation :
Light emitting diode is a forward biased P-N junction which emits light.
The voltage across it = 2 v
Battery voltage = 6 v
Hence total voltage in the circuit = V = 6 - 2 = 4 v
since a LED has 0 resistance in the forward biased region, total resistance in the circuit
= 0 + r = r
current I = 10 mA = 10/1000 A = 0.01 A
we know that,
R = V/I
=> r = 4/0.01 = 400Ω
Hence the value of the limiting resistor is 400Ω
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