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Home » 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 upto 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.
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 upto 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.
CBSE | Class 12 | Current Electricity | NCERT | Physics
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 upto 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 ε? 

(b) What purpose does the high resistance of 600 kΩ have?

(c) Is the balance point affected by this high resistance 

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

(e) 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?

Answer –

(a) We are given that,

Constant emf of the standard cell is E1 = 1.02 V

Balance point on the wire is l1 = 67.3 cm

The standard cell is then changed and a cell of unknown emf ε is connected and the balance point changes to l =  82.3 cm.

The relation between Emf and the balancing point can be written as –

(E1/l1) = (ε/l)

ε = (l x E1/l1) = (82.3  x 1.02)/67.3 

ε  = 1.247 V

(b) When the movable contact is far from the balance point, a high resistance of 600 k is used to reduce current via the galvanometer.

(c) No.

(d) No. There will be no balance point on the wire AB if is bigger than the emf of the potentiometer’s driver cell.

(e) The circuit will not work because the balance point (of the order of a few mV) will be very close to the end A, resulting in a substantial percentage of measurement error. By connecting a suitable resistor R to the wire AB, the circuit can be tweaked so that the potential drop across AB is only slightly bigger than the emf to be measured. The balance point will thus be at a longer length of wire, resulting in a significantly smaller percentage mistake.

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