Question

Draw the circuit diagram of a potentiometer which can be used to determine the internal resistance r of a given cell of emf Explain briefly how the internal resistance of the cell is determined?​

Answer 1

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Determination of Internal Resistance of Potentiometer.

$\red{\fbox{\tt{Circuit:}}}$

A battery B1 a rhecostat (Rh) and a key K is connected across the ends A and B of the potentiometer wire such that positive terminal of battery is connected to point A.This completes the primary circuit.

Now the given cell C is connected such that its positive terminal is connected to A and negative terminal is jockey K through a galvanometer. A resistance (R) and a key K1 are connected across the cell.This completes the secondary circuit.

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Initially key K is closed and a potential difference is applied across the wire AB.Now rheostat Rh is so adjusted that on touching the jockey K at ends A and B of potentiometer wire, the deflection in the galvanometer is on both sides.Suppose that in this position the potential gradient on the wire is k.

Now Key k1 is kept open and the position of null deflection is obtained by sliding and pressing the jockey on the wire,Let this position be P1 and

$AP1=l1$

In this situation the cell is in open circuit, therefore the terminal potential difference will be equal to the emf of cell , i.e.,

$emfε=kl1$

___________________(i)

Now s suitable resistance R is taken in the resistance box and key K−1 is closed.Again, the position of null point is obtained on the wire by using jockey J.Let this position on wire be P2 and

$AP2=l2$

In this situation the cell is in closed circuit,therefore the terminal potential difference (V) of cell will be equal to the potential difference across external resistance R,i.e.,

$V=kl2$

____________________(ii)

Dividing (i) by (ii),we get

$\frac{ε}{V}= \frac{l1}{l2}$

∴ Internal resistance of cell,

$r = ( \frac{ε}{v} - 1)r \\ = (\frac{l1}{l2} - 1)r$

From this formula r may be calculated.

Answer 2

Answer:

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