Question

A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R, the value of R is

• A. 200 $\Omega$
• B. 400 $\Omega$
• C. 40 $k\Omega$
• D. 4 $k\Omega$

Answer 1

Answer: (B)

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