Physics Class 12 Chapter 3 Current Electricity CBSE Class 12 Physics Chapter3 Current Electricity | Page 12

If R be the total resistance of potentiometer wire L its total length , then potential gradient , i . e ., fall in potential per unit length along the potentiometer will be
K = V / L = IR / L
= Eo R / ( Ro + R ) L where , Eo = emf of battery and Ro = resistance inserted by means of rheostat Rh .
Determination of emf of a Cell using Potentiometer
If with a cell of emf E on sliding the contact point we obtain zero deflection in galvanometer G when contact point is at J at a length I from the end where positive terminal of cell have been joined . then fall in potential along length i is just balancing the emf of cell . Thus , we have
E = Kl or E1 / E2 = l1 / l2
Determination of Internal Resistance of a Cell using Potentiometer
The arrangement is shown in figure . If the cell E is in open circuit and balancing length l1 , then
E = Kl1
But if by inserting key K2 circuit of cell is closed , then ooten difference V is balanced by a length l2 of potential where
V = Kl2 Internal resistance of cell
r = E – V / V , R = l1 – l2 / l2 * R
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