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

Answer 1

Answer: (D)

The term LED is abbreviated as Light Emitting Diode. It is forward-biased p-n junction which emits spontaneous radiation. Current in the circuit $ =10\,mA=10\times {{10}^{-3}}A $ $ \text{Voltage}=6-2=4V $ From Ohms law $ V=IR $ $ R=\frac{V}{I}=\frac{4}{10\times {{10}^{-3}}}=400\,\Omega $