A battery of emf ε is connected the end terminal A
and B of potentiometer wire with rheostat Rh, and key K in series.
This is called an auxiliary circuit. Cell of emf which to be measure is
connected as shown in the figure.
Now we find the no deflection position of galvanometer by
closing the key K. we get
ε=Kl1 …..1
Now we close the key K1, resistance R is also added
in circuit and we find the no deflection position of galvanometer at l2
therefore
V= Kl2 …..2
Divide 1by 2 we get
ε/ V= l1/l2 …..3
We know that the internal resistance r1 of a cell
of emf E, when resistance R connected in its circuit is given by
r1 = (ε-V) R/ V =
(ε/ V-1) R
…..4
Putting the value from 3 to
4, we get
r1 = (l1/l2
-1) R
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