The figure shows a potentiometer using a cell `E` of emf `2.0 V` and internal resistance `0.40 Omega` connected to a resistor wire `AB`. A standard cell of constant emf of `1.02 V` gives a balance point a t `67.3 cm` length of the wire. A very high resistance `R = 100 k Omega` is put in with the standard cell. This resistance is shorted by inserting switch `S` when close to the balance point. The standard cell is then replaced by a cell of unknown emf `E` and the null point turns out to be `82.3 cm` length of the wire.
(a) What is the value of `E` ?
(b) What is the purpose of using the high resistance `R` ?
( c) Is the null point affected by this high resistance ?
(d) Is the null point affected by the internal resistance of the cell `E` ?
( e) Would this method work if
(i) the internal resistance of cell `E` were higher than the resistance of wire `AB` and
(ii) the emf of cell `E` were `1.0 V` instead of `2.0 V` ?

Explanation

(a) Here, `E_(1) =1.02 V , =67.3cm , E_(2) =E =? , l_(2) = 82.3cm`
Since, `E_(2)/E_(1) = l_(2)/l_(1) :. E =l_(2)/l_(1) xxE_(1) = 82.3/67.3xx1.02 = 1.247 V`
(b) The purpose of using high resistance of `600k Omega` is to allow very small current through the galvanometer when the movable contact is far from the balance point.
(c) No, the nalance point is not affected by the presence of this resistance.
(d) No, the balance point is not affected by the internal resistance of the driver cell.
(e) No, the method will not work as the balance point will not be obtained on teh potentiometer wire is the e.m.f. of the driver cell is less than the e.m.f. of the other cell.
(f) The circuit will not work for measuring extremely small e.m.f. because in that case, the balance point will be just close to the end `A`. To modify the circuit, we have to use a suitable high resistance in series with the cell of `2.0V`. This would decrease the current in the potentiometer wire. Therefore, potential difference/cm of wire wil decrease. Then only extremely small e.m.f. can be measured.