Application of Potentiometer as to determine the internal resistance of a primary cell and to compare the emf of two primary cells
# TO DETERMINE THE INTERNAL RESISTANCE OF A PRIMARY CELL BY POTENTIOMETER :
* Circuit diagram:
PRINCIPLE AND DERIVATION OF FORMULA:
In above circuit diagram shows the principle of potentiometer, by which we can determine the internal resistance of a cell. In this experiment primary circuit is formed and then experimental cell is joined in secondary circuit of potentiometer. on pressing the key (k1) the current flows through the Potentiometer wire by an accumulator cell. The resistance box (R.B) and Key (K2) are joint in between two terminal of experimental cell.
When key (K2) is open then Jockey slided on the Potentiometer wire and getting a balancing length L1cm.
let the potential gradient of potentiometer wire be “K” volt/cm
By the principle of potentiometer :
EMF of a cell, E = K.L1 --------(1)
When key (k2) is closed then the resistance applied by the resistance box is come in circuit and the current drawn from the cell, in this condition the balancing length L2 cm is obtained on the Potentiometer wire.
By the principle of potentiometer :
Potential difference across the resistance supplied by the resistance box (R.B) is -
V = k.L2 -------(2)
Dividing the enq(1) by enq(2)
E/V = k.L1 / k.L2
E/V = L1/L2 ------------(3)
We know that -
Current drawn = (EMF of a cell) /
from a cell (Total resistance)
I = E/(R+r)
R.I + I.r = E
( By ohm's law: V = R.I )
V + I.r = E
I.r = E - V
( Again by ohm's law: I = V/R )
(V/R).r = E - V
r = R.( E/V - 1 )
Putting the value of (E/V) from eqn(3)
Hence, [ r = R.( L1/L2 - 1 ) ]
This is the required formula
* OBSERVATION:
* PRECAUTION:
1) All connecting screws and plugs should be tied .
2) Positive terminal of each cell should be connected at one end of the Potentiometer wire .
3) EMF of an accumulator cell in primary circuit is more than that of experimental cell otherwise the position of null deflection does not obtained on whole length of potentiometer wire.
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* Circuit diagram:
In above circuit diagram shows the principle of potentiometer, by which we can determine the internal resistance of a cell. In this experiment primary circuit is formed and then experimental cell is joined in secondary circuit of potentiometer. on pressing the key (k1) the current flows through the Potentiometer wire by an accumulator cell. The resistance box (R.B) and Key (K2) are joint in between two terminal of experimental cell.
When key (K2) is open then Jockey slided on the Potentiometer wire and getting a balancing length L1cm.
let the potential gradient of potentiometer wire be “K” volt/cm
By the principle of potentiometer :
EMF of a cell, E = K.L1 --------(1)
When key (k2) is closed then the resistance applied by the resistance box is come in circuit and the current drawn from the cell, in this condition the balancing length L2 cm is obtained on the Potentiometer wire.
By the principle of potentiometer :
Potential difference across the resistance supplied by the resistance box (R.B) is -
V = k.L2 -------(2)
Dividing the enq(1) by enq(2)
E/V = k.L1 / k.L2
E/V = L1/L2 ------------(3)
We know that -
Current drawn = (EMF of a cell) /
from a cell (Total resistance)
I = E/(R+r)
R.I + I.r = E
( By ohm's law: V = R.I )
V + I.r = E
I.r = E - V
( Again by ohm's law: I = V/R )
(V/R).r = E - V
r = R.( E/V - 1 )
Putting the value of (E/V) from eqn(3)
Hence, [ r = R.( L1/L2 - 1 ) ]
This is the required formula
* OBSERVATION:
* PRECAUTION:
1) All connecting screws and plugs should be tied .
2) Positive terminal of each cell should be connected at one end of the Potentiometer wire .
3) EMF of an accumulator cell in primary circuit is more than that of experimental cell otherwise the position of null deflection does not obtained on whole length of potentiometer wire.
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