State the working principle of potentiometer.Explain with the help of circuit diagram .How the potentiometer is used to determine the internal resistance of the given primary cell.
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Answers
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Answer:
Working principle : Potential difference across length of potentiometer wire is proportional to the length of the wire.
E=kL where, E is the emf and k is the potential gradient.
In the figure:
Cell emf e whose internal resistance r is to be determined is connected via resistance box R.B. through key K
2
. This key when open, balance is obtained at l
1
(AN
1
) .
So, e=kl
1
If v is terminal potential difference, balance is obtained at length l
2
(AN
2
).
So, v=kl
2
Now, e=I(R+r) where I is the current.
and v=IR
Dividing the equations.
v
e
=
l
2
l
1
=
r
R+r
Therefore, r=R(
l
2
l
1
−1)
Emf of the cell, E
1
=1.25 V
Balance point of the potentiometer, l
1
=35 cm
Cell is replaced by another cell of emf, E
2
Balance point of the potentiometer, l
2
=63 cm
E
2
E
1
=
l
2
l
1
E
2
=1.25×
35
63
=2.25 V