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