EMF of an electric cell is 2.1 V and its internal resistance is 0.1 ohm. When the two poles of the cell are connected with an external resistance, a potential difference of 2.08V is obtained. determine the value of the external resistance.
PLZ HELP ME FRNDZZZ. ...NO SPAM.....PLZZZZ.....
Answers & Comments
2.1-i(R+0.1)=0
or,2.1-iR-i x0.1=0
or,2.1-2.08-i x0.1=0. (iR=2.08V because according to question potential difference bw the poles is 2.08 V when external resistance is applied.)
or,i=0.02/0.1
or,i=0.2A
and since iR=2.08
or,R=2.08/i
or,R=2.08/0.2
or,R=10.4 ohm
Hope it helps.