Question

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

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

Answer 1

The correct option is D $400k\Omega$

Maximum current in $LED=10mA$
$=10\times {10}^{-3}A$
Resistance of $LED=\frac{2}{10\times {10}^{-3}}=2\times {10}^2\Omega$
Now, applying ohm's law
$\frac{6}{200+R}=10\times {10}^{-3}={10}^{-2}=\frac{1}{100}$
$200+R=600\Rightarrow R=400\Omega$