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 seriesThe current flowing, 10mA is the sameThe voltage drop across LED = 2VAs the battery has voltage of  6V, the potential difference across  R = 4V∴V=iR⇒R=4 i=410×10−3 A∴R=400 Ω