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The emf of driver cell in a potentiomete...

The emf of driver cell in a potentiometer circuit is 22 V. The length of the potentiometer wire is 1 m and its resistance of `1 Omega` A cell of emf 2 V is balanced against a length `(1)/(10)`.

Text Solution

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We know that , `V = ((E )/( R+r)) (r )/(L) I`
where , V = emf of cell to be measured ,
E = emf of driving cell ,
r = resistance of potentiometer wire ,
L = length of potentiometer wire ,
l = the length of potentiometer wire at balance point
and R = resistance of driving cell .
So , `2 = ((22)/(R+10)) (10)/(1) xx (1)/(10)`
`rArr " " 2R + 20 = 22`
`rArr" " R = 1 Omega`
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