3.22 Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω 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 up to 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value of 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 1.0 V instead of 2.0 V?

(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?


The resistance of each resistor connected in the given circuit, R=1Ω
Equivalent resistance of the given circuit=R'
The network is infinite.
Hence, equivalent resistance is given by the relation,
R'=2+R'(R'+1)
(R)2-2R'-2=0
R'=2±4+82
=2±122=1±3
Negative value of R' cannot be accepted.
Hence, equivalent resistance,
R'=(1+3)=1+1.73=2.73 Ω
Internal resistance of the circuit, r=0.5Ω
Hence, total resistance of the given circuit=2.73+0.5=3.23 Ω
Supply voltage, V=12 V
According to Ohm's Law, current drawn from the source is given by the ratio, 123.23=3.72 A