A battery of emf

**is connected the end terminal A and B of potentiometer wire with rheostat R***ε*_{h}, 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

**…..1**

*ε=Kl*_{1 }
Now we close the key K

_{1}, resistance R is also added in circuit and we find the no deflection position of galvanometer at l_{2}therefore

*V= Kl*_{2}_{ }…..2

Divide 1by 2 we get

*ε/ V= l*_{1}/l_{2}_{ }…..3

We know that the internal resistance

**of a cell of emf E, when resistance R connected in its circuit is given by***r*_{1}**…..4**

*r*_{1}= (ε-V) R/ V = (ε/ V-1) R
Putting the value from 3 to
4, we get

*r*_{1}= (l_{1}/l_{2 }-1) R
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