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

Answer 1

Answer: (D)

$400\,\Omega$

Answer 2

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