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)

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