A light emitting diode 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 :
 

As LED is connected to a battery through a resistance in series
The current flowing, 10mA is the same
The voltage drop across LED = 2V
As the battery has voltage of  6V, the potential difference across  R = 4V
$\therefore V=iR$

$\Rightarrow R=\frac{4}{i}=\frac{4}{10\times {10}^{-3}}\text{\hspace{0.17em}A}$

$\therefore R=400\text{\hspace{0.17em}Ω}$