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Question

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.


Solution

Measurement of internal resistance ofa cell using potentiometer is shown in figure. The cell of emf, E is connected across a resistance box R through key K2.

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When K2 is open, balance length is obtained at length AN1=l1
            E  l1
           E=ϕ l1 ........ (i)
When K2 is closed :
Let V be the terminal potential difference of cell and the balance is obtained at AN2=l2
        V  l2
         V=ϕ l2 ........ (ii)
From equations (i) and (ii), we get
         EV=l1l2 ..... (iii)
        E=I(r+R)
       V=IR
         EV=r+RR ....... (iv)
From equations (iii) and (iv), we get
         R+rR=l1l2
       r=R(l1l21)
We know l1,l2 and R, so we can calculate r.

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