Internal resistance of primary cell :
Internal resistance of primary cell
Close the key K1 . A constant current flows through the potentiometer wire. With key K2 kept open, move the jockey along AB till it balances the emf e of the cell. Let l1 be the balancing length of the wire. If k is the potential gradient, then emf of the cell will be : e = kl1
With the help of resistance box R.B., introduce a resistance R and close key K2 . Find the balance point for the terminal potential difference V of the cell. If l2 is the balancing length, then V = kl2
Divide the above to equation
Let r be the internal resistance of the cell. If current I flows through cell when it is shunted with resistance R, then from Ohm’* law we get ɛ= I (R + r) and V = IR
