8

To determine the internal resistance of a cell by a potentiometer.

Theory: The potentiometer primarily measures potential difference. The potentiometer circuit is shown in the figure 1. AB is a length of uniform resistive wire. Terminals X and Y are connected across the potential difference (V1) being measured in the same way as those of a voltmeter would be.

If the positive terminal of the driving cell is connected to X (as shown in the diagram), then X must be connected to the positive side of the potential difference (V1) being measured.

The potentiometer is said to be balanced when the jockey (sliding contact) is at such a position on AB that there is no current through the galvanometer.

At balance, IG = 0, and therefore

I = I1;(by applying kirchhoffs 1st law at C)

And so

I2 = 0 ;(by applying kirchhoffs 1st law at A)

I1I1GalvanometerXYABPotential difference being measured (V1)DrivingBatteryII2I1IGJockeyCRh Figure - 1

Therefore, Potential at X = Potential at A (since I2 = 0)

And, Potential at Y = Potential at C (since IG = 0)

Therefore, Potential difference between X and Y = Potential difference between A and C i.e. V1 = I (R1) ; where R1 is resistance of wire between points A and C.

Now, let ; where L1 is length of wire between points A and C.

Therefore we have .(1)If V1 is replaced by another potential difference V2 and a new balance length L2 is found, then

.(2)

Dividing equation (1) by equation (2) gives ..(3)

Thus, by measuring the respective balance lengths, two potential differences can be compared. At balances, IG =I2 = 0, and therefore no current is drawn from the potential differences being compared.

Cell or any other source which supplies a potential difference to the circuit to which it is connected, has within it some resistance called internal resistance. When there is no current in the cell i.e., at open circuit, the potential difference E between its terminal is maximum and is called its electromotive force (e.m.f.). When the cell is discharging i.e., at closed circuit, its terminal potential difference between its terminal is reduced to V because of internal drop of potential across its internal resistance r. R

K2

G E1 r A K1

DC Power Supply

C J Rh

B

Figure- 2 : Finding internal resistance of a cell by a potentiometer.

In the above figure the balance point for the cell E1 whose internal resistance r is to be determined , is found out as usual , at a distance L1 from the end A of the potentiometer with the key K2 open .

.(4)Then the resistance R is introduced in the resistance box and the key K2 is closed, now i current is flowing through the circuit ( through external resistances R and internal resistance r ). the potential difference between the terminals of the cell E1 falls because of internal drop of potential across its internal resistance r.

E1 = Potential difference between the terminals + lost voltage in the cell

As E1 and V are the potential differences between the terminals of the cell at the open and closed circuits.

A balance point is now found at a distance L2 from the end A of the potentiometer and V is the potential difference between the terminals of the cell at closed circuit. Therefore

.(5)from equation (4) and (5) we have

or

i.e. .(6)By knowing L1 , L2 and R we can determine internal resistance of a cell by equation (6)

Apparatus : Potentiometer, DC power supply, Cell E1 , Resistance box , Rheostat (Rh) , Galvanometer , Connecting wires.Connection : Connect the positive terminal of the DC power supply to the screw A (end A) of the potentiometer and the negative terminal of the DC power supply through a rheostat (Rh) to the screw B (end B) of the potentiometer with a key K1 (crocodile clip of the wire can be as key K1, when clip is connected K1 (circuit) is closed). Join the positive terminal of the cell E1 to the screw A of the potentiometer and negative terminal of the cell E1 through the galvanometer G to the jockey J. Also connect the resistance box (R) through a key K2 to the two terminals of the cell E1.

Procedure:1. Adjust a small resistance in the rheostat Rh and close the key K1 by connecting crocodile clip to the DC power supply. Keep the key K2 open(do not connect resistance R with battery at the beginning)

2. At first press the jockey near point the end A and then near the end B of the potentiometer wire. If the galvanometer deflections are in the same direction, then either the resistance in the Rh is too great or voltage of the DC power supply is too small. Decrease the resistance in Rh or increase the voltage in the DC power supply until the opposite deflections are obtained at the two ends A and B. Adjustment of Rh should be such that as to get a null point on the 9th or 10th wire.

3. Now find out the balance point accurately when K1 is closed and K2 is open . Open the key K1 and calculate the distance L1 of the balance point from the end A of the potentiometer wire. Determine L1 three times and calculate the mean value of L1.

4. Close the key K2 (connect resistance R with battery) by connecting crocodile clip without changing Rh and take out a resistance 10 ohms from the resistance box and determine the balance point from end A of the potentiometer and find the length L2 of the balance point from the end A. Then remove 20, 30, 40, 50 ohms from resistance box and determine the value of L2 in each case.

5. Calculate the value of r using the relation (6) for each value of R and then calculate the mean value of r.

Discussions: 1) The internal resistance of a cell depends on the strength of the current. It decreases as the current increases. It is , therefore, better to change the external resistance over a range of 40 ohms and then to calculate the mean values of r.2) After every reading, the key K1 should be opened to allow the wire to cool.3) Care should be taken to see that K2 is open when determining l2.4) The internal resistance of a cell can be determined with voltmeter and ammeter also but this method is more accurate.Data sheet:

ResistanceIn

R

ohmsValue of

L1

cmMean of

L1

cmResistanceIn

R

ohmsValueOf

L2

cmInternal resistance

rof the cell

ohms

Mean

ohms

Key

K1

Closed

and

K2

open

infinityBothKey

K1

and

K2

closed10

20

30

40

50

Calculation:

= ..

Error calculation:

=

Results :

Internal resistance of the cell, = .ohmsSample oral Questions and AnswersExperiment No: 02

1. What do you understand by the internal resistance of a cell?Ans: When the external circuit is complete, within the cell (or battery) a current flow from the plate at a lower potential to the plate at higher potential and the medium between the plates offers a resistance to the flow of current. This resistance is known as the internal resistance of the cell.

2. On what factors does the internal resistance of cell depend?Ans: It depends on (a) the conductivity of the medium between the plates. (b) The distance between the plates (c) the area of those portions of the plates that are immersed in the electrolytes.

3. What do you mean by e.m.f. of a cell?Ans : The e.m.f. of a cell is the driving force which drives electric current through the circuit and is caused by the chemical changes taking place in the cell and has a definite direction.

4. What is the difference between e.m.f and potential difference?Ans : Electromotive force is the sum of the potential drops in both external and internal circuits and is the total driving force of the current in the circuit. It has a definite direction. The potential difference between two points in the circuit is the product of the current and the resistance between these two points. Hence the magnitude and direction of potential difference depends on those of the current where as the e.m.f. of the circuit is constant both in magnitude and direction. E.M.F. is present both in the open and closed circuit while potential difference is present in the closed circuit.

5. If even with correct connections, deflections on the first and last wires are in same direction for the cell, what would you do?Ans : The current sent by the driver cell is such that the total potential drop across the wire is greater than the e.m.f. of one cell but less than the e.m.f. of the other cell. The potentiometer current is to be increased by diminishing the rheostat resistance.

6. Does the resistance of the galvanometer interfere with the position of the null point?Ans : No, it saves the galvanometer from damage.

7. How can you increase the sensitiveness of the potentiometer?Ans : By increasing the balancing length of the potentiometer as long as possible.

8. What kind of wire would you use as the potentiometer wire and why?Ans : The material should be so chosen that it may have high resistance ( such as eureka , manganin , etc.). Otherwise the fall of potential will not be sufficient even for the whole length of the wire. They should have also low temperature coefficient of resistance.

9. If the wire be heated by the passage of current, would the null point change?Ans : Yes , since the resistance of the wire changes with the change of temperature. Hence the potentiometer circuit should be closed only for the time during which the null point is determined.

Potential at X = Potential at A ; (since I2 = 0)

Potential at Y = Potential at C ; (since IG = 0)

Therefore,

Potential difference between X and Y = Potential difference between A and C i.e. V1 = I (R1) ;Where R1 is resistance of wire between points A and C.

Now, let ; Where L1 is length of wire between points A and C.Therefore we have

.(1)I1I1GalvanometerXYABPotential difference being measured (V1)DrivingBatteryII2I1IGJockeyCRh

1. If negative terminal of the DC power supply (Driving cell) in the diagram below is connected to point A, then will the null point occur? Why?Ans: No.

2. After getting one null point, the circuit should be left open for 2-3 minutes before the determination of the next null point is taken up. Why?Ans: Resistance changes.

3. If resistance of the rheostat (Rh) is increased and others are kept constant, then where does the null point occur; closer or further from the point A? Why?

Ans: Further from the point A. We know . When Rh is increased current I is decreased. As V1 is constant therefore L1 is increased.

4. If DC power supply voltage (Driving battery) is smaller than that of cell voltage ( E1) . Will the null point occur? Why? Ans: No5. If voltage of the DC power supply voltage(Driving cell) is increased and others are kept constant, then where does the null point occur; closer or further from the point A? Why?

Ans: Closer to the point A. We know . When DC power supply voltage(Driving cell) is increased current I is increased. As V1 is constant therefore L1 is decreased.6. If resistance R is connected to the cell E1 then where does the null point occur; closer or further from the point A? Why?

Ans: Closer to the point A. We know . When V1 is decreased and current I is constant therefore L1 is decreased.Pre-Lab Exercise:

1. What would you expect the voltmeterinthe adjacent circuit to do when the potentiometer wiper is moved to the right?........................................................................................

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2. Which way would you have to move thepotentiometerwiper, to the left or to the right,inorder to increase current through resistor R1 in the adjacent circuit?

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3. What is the difference between e.m.f and potential difference?

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4. Why does internal resistance of a battery occur?

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5. If resistance of the rheostat (Rh) is increased and others are kept constant in the adjacent circuit, then where does the null point occur; closer or further from the point A? And why?

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I1I1GalvanometerXYABPotential difference being measured (V1)DrivingBatteryII2I1IGJockeyCRh

6. If voltage of the DC power supply voltage(Driving cell) in the above circuit is increased and others are kept constant, then where does the null point occur; closer or further from the point A? And why?...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................