Q. 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 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 $\Omega$
Explanation
(d) Current in the circuit = 10 mA = ${10 \times 10}^{- 3}$ A
and voltage in the circuit = 6$- 2 =$ 4 V
from Ohm’s law,
$V = IR$
$\therefore = \frac{V}{I} = \frac{4}{{10 \times 10}^{- 3}} = \ 400\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.