Potentiometer Problem
Student observation in a potentiometer
A student sets up a circuit as shown in the figure given below to measure the emf of a test cell.
- (i) Explain why he is unable to find a balance point and state the change he must make in order to achieve the balance.
- (ii) State how he would recognize the balance point.
- (iii) He obtained the balance point for distance 37.5 cm using standard cell of emf 1.50 V. And for the test cell, the balance distance AB was 25.0 cm. Calculate the emf of the test cell.
- (iv) He could have used an ordinary voltmeter to measure the emf of the test cell directly. The student, however, argues that the above instrument is more precise than an ordinary voltmeter. Justify his logic.
(i)Problem in circuit
The positive terminals of both the driving cell and the test cell must be connected to the same terminal A. If their polarities oppose each other at the joint (A), the potentials will add up (a case of series combination) making a null deflection impossible.
(ii) Balance point
The Voltmeter (V) will show null deflection once the balance point is achieved indicating the same potential difference between the potentiometer wire (AB section) and the secondary circuit.
(iii) Emf of test cell
Using the principle of potentiometer, $$V \propto l$$
We get,$$\frac{1.5}{E_2}=\frac{37.5}{25}$$
$$E2=1 V$$
Emf of a test cell = 1 V
(iv) Preciseness of a potentiometer
The potentiometer works based on the 'null deflection' and hence draws no current from the cell thereby measuring the open circuit potential difference which is the emf of a cell.
An ordinary voltmeter has however a finite internal resistance. So, it draws a small amount of current from the cell and hence accurate reading is not measured. It is because it alters the p.d. of the circuit by drawing the current.
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