Q. 224.0( 21 Votes )

Figure 3.33 shows

Answer :

(a) Constant emf of the given standard cell, E1 = 1.02 V


Balance point on the wire, l1 = 67.3 cm


A cell of unknown emf, ε, replaced the standard cell. Therefore, new balance point on the wire,


l = 82.3 cm


The relation connecting emf and balance point is,


E1/l1 = E/l


E = l × E1/l1


E = 82.3 cm × 1.02 V/67.3 cm


E = 1.247 Volt.


The value of unknown emf is 1.247 V.


(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.


(c) The balance point is not affected by the presence of high resistance.


(d) The point is not affected by the internal resistance of the driver cell.


(e) The method would not work if the driver cell of the potentiometer has an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then we cannot obtain balance point on the wire.


(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a high percentage of error. The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.


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