Question

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internalresistance 0.40 W maintaining a potential drop across the resistorwire 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 fromthe standard cell, a very high resistance of 600 kW is put in serieswith it, which is shorted close to the balance point. The standardcell is then replaced by a cell of unknown emf e and the balancepoint found similarly, turns out to be at 82.3 cm length of the wire. (a) What is the value e ? (b) What purpose does the high resistance of 600 kW have? (c) Is the balance point affected by this high resistance? (d) Would the method work in the above situation if the driver cellof the potentiometer had an emf of 1.0V instead of 2.0V?(e) 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?

Answer 1

Given: The constant emf of the standard cell is 1.02 V, the balance point on the wire is 67.3 cm, the new balance point on the wire 82.3 cm, the resistance is 600 kΩ, the internal resistance is 0.40 Ω, the cell voltage is 2.0 V.

a)

Consider the given figure below.

The relation between EMF and balance point is given as,

E 1 l 1 = ε l ε= l l 1 × E 1

Where, the constant EMF of the standard cell is E 1 , the balance point on the wire is l 1 , the new balance point on the wire is l and the unknown emf is ε.

By substituting the given values in the above expression, we get

ε= 82.3 67.3 ×1.02 =1.247 V

Thus, the unknown value of emf is 1.247 V.

b)

The high resistance 600 kΩ is used to reduce the flow of current through the galvanometer when the movable contact is far from the balance point.

c)

The presence of high value resistance does not affect the balance point.

d)

The internal resistance of the driver cell does not affect the balance point.

e)

If the emf of the driver cell of the potentiometer is less than the emf of the other cell, then we would not get any balance point on the wire.

Thus, this method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V.

f)

There would be large percentage of error for determining extremely small emf because circuit would be unstable and the balance point would be close to point A. Thus, the circuit would not work well for determining extremely small emf.

The percentage error can be reduced by connecting a series resistance AB so that the potential drop across AB slightly greater than the measured emf.

Thus, the circuit can be modified by connecting a series resistance AB.