Potentiometer

The presentation explain principal, working and construction and application of Potentiometer it is useful for senior secondary students of Indian school

1. Potentiometer Presented by Vasudev Shrivastava P.G.T.(Physics) Jawahar Navodaya Vidyalaya Nowgong District Chhatarpur (M.P.)

2. Potentiometer is a device used to measure the internal resistance of a cell, to compare the e.m.f . of two cells and potential difference across a resistor. It consists of a long wire of uniform cross sectional area and of 10 m in length. The material of wire should have a high resistivity and low temperature coefficient.  The wires are stretched parallel to each other on a wooden board. The wires are joined in series by using thick copper strips. A metre scale is also attached on the wooden board. It works on the principle that when a constant current flows through a wire of uniform cross sectional area, potential difference between its two points is directly proportional to the length of the wire between the two points. Potentiometer

3. Potentiometer

4. A potentiometer is a null type resistance network device used for measuring potential differences and comparison of emfs of potential difference sources.

5. Principle of potentiometer A potentiometer works on the principle of balancing one voltage against another in parallel with it . Construction and working of potentiometer It consists of a long resistance wire AB, sometimes 6 m in length, and of uniform diameter. This resistance wire R, a working battery and a variable resistance R x all connected in series. R x enables us to develop a convenient output voltage V between A and B, maintaining a constant current I in the circuit. Now, according to Ohm’s law I = V/R

6. A source of emf ε which is to be measured is connected along with a Galvanometer through the sliding contact C. See figure.  As C slides over AB, a variable potential difference is available between A and C. If r represents the resistance between A and C, the potential drop between A and C is given by

7. Now if the PD between A and B is greater than ε  , then there must be some point C between A and B, at which the potential drop across AC is equal to ε. This point is found by sliding C over AB until the current in the galvanometer is zero. The current in galvanometer becomes zero because under such condition, called null condition, both the ends of the galvanometer are at the same potential. So under null conditions

8. Here r is the value of the wire resistance of length AC at null condition . Let the wire is of uniform cross-section A and AB = L and AC = l , then We know that resistance is directly proportional to the length and inversely proportional to the cross-sectional area of the wire .

9. Divide one equation by the other

10. Where L is the total length of the wire and l is the length AC. This equation gives the unknown emf in terms of the ratio of two known lengths and voltage applied. Ordinarily, the potential drop along the wire is calibrated and the balanced point gives directly the unknown emf ε

11. Comparison of emfs  Potentiometer can also be used for the comparison of two emf sources. The balancing length of emf ε 1 of a cell is found and compared with the balancing length of a standard cell which is also found separately. If ε and ε 1 are the two emf sources nulled at lengths l and l 1 respectively, then, Dividing one equation by other,

12. Dividing one equation by other ,

13. Internal resistance of primary cell Internal Resistance is the resistance which is present within the battery that resists the current flow when connected to a circuit. Thus it causes a voltage drop when current flows through it. It is the resistance provided by the electrolyte and electrodes which is present in a cell .

14. Relation between e.m.f (E)., potential difference(V) , and internal resistance(r) of a cell. If a cell of emf E and internal resistance r, connected to an external resistance R, then the circuit has the total resistance ( R+r ). The current I in the circuit is given by, Hence,                              This means, V is less than E by an amount equal to the fall of potential inside the cell due to its internal resistance . From the above equation, 

15. Or;   The internal resistance of the cell,  

16. Determination of internal resistance of a primary by potentiometer Using a potentiometer, we can adjust the rheostat to obtain the balancing lengths l 1 and l 2 of the potentiometer for open and closed circuits respectively.  Then, E= k l 1   and V = k l 2  ; where k is the potential gradient along the wire.  Now we can modify the equation for getting the internal resistance of the given cell, by using the above relations as;

17. Problems related to potentiometer

18. Home assignment Following Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V(for very moderate currents upto a few mA ) gives a balance point at67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire . (a) What is the value ε ? ( b) What purpose does the high resistance of 600 kΩ have ? ( c) Is the balance point affected by this high resistance ? (d) Is the balance point affected by the internal resistance of thedriver cell ? ( e) Would the method work in the above situation if the driver cellof the potentiometer had an emf of 1.0V instead of 2.0V ? ( f ) Would the circuit work well for determining an extremely smallemf , say of the order of a few mV (such as the typical emf of athermo -couple)? If not, how will you modify the circuit?