Question

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 £2 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 at 67.3 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 g 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 the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

Answer 1

(a) Here,

I₁ = 67.3 cm

I₂ = 82.3 cm

E₁ = 1.02 V

(b) The purpose of using high resistance is to allow only a very small current to flow through the galvanometer when the balance point has not been obtained.
(c) No, the balance point is not affected by the internal resistance of the driver cell.
(d) No, the arrangement will not work. If the emf of the driver cell is less than that of the cell whose emf is to be found, the balance point will not be obtained.
(e) The circuit is not suitable for measuring extremely small emf. It is because in such a case, the balance point will be just close to end A.