Figure 6.36 shows a potentiometer with a cell of emf `2.0 V` and internal resistance `0.4 Omega` maintaining a potential drop across the resistor wire `AB`. A standard cell that maintains a constant emf of `1.02 V` (for very moderate current up to emf `mu A`) gives a balance point at `67.3 cm` length of the wire. To ensure very low current is drawn the standard cell, a very high resistance of `600 k Omega` 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 `epsilon` and the balance point found, similary, turns out to be at `82.3 cm` length of the wire.
a. What is the value of `epsilon`?
b.What purpose does the high resistance of `600 k Omega` have?
c. Is the balance point affected by this high resistance?
d. Is the balance point affected by 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 `1.0 V` instead of `2.0 V`?
f. Would the circuit work well for determining an extermely small emf, say of the order of a few `mV` (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ?

Answer

(a) `1.2474 V` (b) To reduced current (c ) No (d) Yes (e) No (f) No, suitable resistance is put in series with wire `AB`.

Explanation

(a) By formula, `(epsilon)/(epsilon_(standard)) = (l)/(l_(standard))`
or `epsilon= (l xx epsilon_(standard))/(l_(standard)) = (82.3 xx 1.02)/(67.3) = 1.2474 V`
(b) To reduce the current through the galvanometer when the movable constant is far from the balance point.
(c ) No.
(d) Yes.
(e) No, if `epsilon` as it is, would be unsuited, because the balance point (for `e` of the order of a few `mV`) will be very close to the end `A` and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor `r` in series with the wire `AB` so that potential drop across `AB` is only slightly greater than the emf to be measured. Then the balance point will be at larger length of the wire and the percentage error will be much smaller.