To determine the internal resistance of a dry cell using an ammeter and a voltmeter.

Notice that when no current is being drawn from the cell ,there is no potential drop across the internal resistance r .Hence the potential difference between the terminals is equal to the e.m.f. However ,if the cell is connected across an external resistor as shown amount of energy per unit charge is wasted in getting through the cell.

V = ξ – Ir

Therefore ,V is less than ξ,then not all the energy per unit charge supplied by the cell is transformed in the external circuit into other forms of energy . It implies that a certain amount of energy per unit charge is wasted in getting through the cell.

The internal resistor of a cell depends on several factors and is seldom constant . However ,it is sometimes useful to know its rough value .The internal resistance of a cell can be estimated as:

Procedure :

1. The circuit was connected as shown below.

2. With the switch S open ,the e.m.f. of the cell was measured by using the voltmeter.

3. The switch S was closed and the rheostat was adjusted so that the cell supplied a small current through the circuit.

4. The ammeter reading and the voltmeter reading were recorded .

5. The rheostat was adjusted and different pairs of ammeter reading and voltmeter reading were recorder(note :over the widest possible range).

6. The experiment was repeated by replacing the cell X by cell Y.

Results :

(i) Experiment with the cell X

Measured e.m.f. = 1.40 V

The graph of the potential difference across the cell V against the current I :

(ii)Experiment with the cell Y

Measured e.m.f. = 1.40 V

The graph of the potential difference across the cell V against the current I :

From the graphs show above ,find the internal resistance and the e.m.f. of each cell .

(i)

           slope of the graph = - r

(1.30-1.34) / (0.0450-0.0250) = - r

            2.00 ohm = r

         The internal resistance of cell X is 2.00 ohm.

V = ξ – Ir

                    1.34 =ξ- 0.025 X 2.00

1.39 V =ξ

The e.m.f. of cell X is 1.39V.

(ii)

slope of the graph = - r

(0.9-0.6) / (0.085-0.13) = - r

6.00 ohm= r

         The internal resistance of cell Y is 6.00 ohm.  

V = ξ – Ir

                   0.9=ξ- 0.085 X 6.00

1.41 V =ξ

          The e.m.f. of cell Y is 1. 41 V.

Question for Discussion :

1. What do you notice about the values of the internal resistance of the cells X and Y?

Internal resistance of X(2 ohm) is smaller than that of Y(6 ohm) by 4 ohm

2. Explain why a high resistance voltmeter is necessary to give an accurate value of the e.m.f of the cell.

The voltmeter reading shows the terminal voltage of the cell.

Voltmeter should draw most of the voltage which is given by the cell in order to give the accurate value of the terminal voltage of the cell ,which is then used to find out the e.m.f. of the cell .

The higher is the resistance ,the more accurate is the e.m.f. of the cell .

3. State the main precautions in this experiment .

1. Take the readings only when the values shown on the voltmeter and ammeter are steady.

2. Don’t touch the electric wires which are not wrapped by plastic in order to avoid electric shock.

   

3. The equipments should be connected correctly.

4. Discuss the sources of error in this experiment .

1. The internal resistance is seldom constant .However ,sometimes we assume it to be constant for simplicity.

2. The resistance of equipments such as wires ,ammeter and voltmeter are assume to be negligible .In fact ,some of the electric energy is dissipated by Joule heating inside the battery.

3. The readings may not reach the final value.