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) `E = (1.02)/(67.3) xx 82.3 = 1.25 V`
(b) The high resistance `R` keeps the current from standard cell within permissible limit and prevent a large current to flow through the galvanometer when far away from the balance point.
( c) Al null point, no current flow through `R`, hence no effect no null point.
(d) The null point depends on terminal voltage of `E_0` and the emf of `E` only, hence no effect.
( e) If `p.d` across `AB` due to the driver cell `E_0` become less than the emf of cell `E_1`, method will fail, there would be no null point on the wire `AB`
(i) `p.d` across `AB` due to cell `E_0` of emf `2 V` will be less than `1 V`. Since emf of the standard cell is greater than this value, no null point on `AB`
(ii) No null point on `AB`.