question_answer 22)

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40  maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant e.m.f. of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k 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 e.m.f.  and the balance point found simllarly, turns out to be at 82.3 cm length of the wire.                 (a) What is the value ?                 (h) What purpose does the high resistance of 600 k have?                 (c) Is the balance point affected by this high resistance?                 (d) Is the balance point affected by the internal resistance of the driver cell?                 (e) Would the method work in the above situation if the driver cell of the potentiometer had an e.m.f. of 1.0 V instead of 2.0 V?                 (f) Would the circuit work cell for determining an extremely small e.m.f., say of the order of a few mV (such as the typical e.m.f. of a thermo-couple?) If not, how will you modify the circuit?  

Answer:

(a) Using E  1, we get                                   or E2 =                   (b) Purpose of using high resistor is that it will allow very small current to flow through the galvanometer when the circuit is not balanced.                 (c) No, the balance point is not affected by the high resistance.                 (d) No, the balance point is not affected by the internal resistance of the driver cell.                 (e) No, the method will not work in this situation because the balance point is not obtained if driver cell of e.m.f. IV instead of 2V is used.                 (f) The circuit will not work well for determining extremely small e.m.f. because in such cases the balance point will he almost on end A.