Question

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

• A. $40k\Omega$
• B. $4k\Omega$
• 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 $=10mA$
$=10\times 10^{-3}A$ and voltage in the circuit $=6-2=4V$ From Ohm's law,
$V=IR$
$\therefore R=\frac{V}{I}=\frac{4}{10\times 10^{-3}}=400\Omega$