Null point in the galvanometer is obtained when a cell of emf E and internal resistance r is connected across the length of 22 cm wire of the potentiometer- Now a resistance of 10 - is connected across the terminal of the cell by closing the key K and null point is obtained against the length of 20 cm- Then the internal resistance r of the cell is

Taking case I when the key is open and there is no current in the secondary circuit. E∝ l_1 here l_1=22 cm nbsp; nbsp; .....(i) Now taking the case II whe ...

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Standard XII

Physics

Series and Parallel Combination of Resistors

Null point in...

Question

Null point in the galvanometer is obtained when a cell of emf $E$ and internal resistance $r$ is connected across the length of 22 $c m$ wire of the potentiometer. Now a resistance of $10 Ω$ is connected across the terminal of the cell (by closing the key K) and null point is obtained against the length of 20 cm. Then the internal resistance $r$ of the cell is

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Solution

Taking case I when the key is open and there is no current in the secondary circuit.
$E \propto l_{1}$
here $l_{1} = 22 c m$ .....(i)
Now taking the case II when key is closed
$E - I r \propto l_{2}$ ...(ii)
here $l_{2} = 20 c m$
$I = \frac{E}{r + R}$
Dividing eq(i) by (ii)
$r = R (\frac{l_{1}}{l_{2}} - 1) = 10 (\frac{22}{20} - 1) = 1 Ω$

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Taking case I when the key is open and there is no current in the secondary circuit. E∝l1 here l1=22 cm .....(i) Now taking the case II when key is closed E−Ir∝l2 ...(ii) here l2=20 cm I=Er+R Dividing eq(i) by (ii) r=R(l1l2−1)=10(2220−1)=1Ω