Question

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

(a) 40 KΩ 

(b) 4 KΩ 

(c) 200Ω 

(d) 400Ω

Answer 1

Answer is (d) 400Ω

Voltage drop of LED = 2V 

Voltage across resistor, 

(V) = 6 - 2 = 4V

So, value of R = \(\frac{V}{I}\) = \(\frac{4}{10\times10^{-3}}\) = 400Ω