Sensitivity of Potentiometer

Sensitivity of potentiometer : A potentiometer is said to be more sensitive. If it measure a small potential difference more accurately.

(i) The sensitivity of potentiometer is assessed by its potential gradient. The sensitivity is inversely proportional to the potential gradient.

(ii) In order to increase the sensitivity of potentiometer

(iii)The resistance in primary circuit will have to be decreased.

(iv) The length of potentiometer wire will have to be increased so that the length may be measured more accurately.

Difference between voltmeter and potentiometer :

Voltmeter Potentiometer
It’s resistance is high but finite Its resistance is infinite
It draws some current from source of emf It does not draw any current from the source of unknown emf
The potential difference measured by it is lesser than the actual potential difference The potential difference measured by it is equal to actual potential difference
Its sensitivity is low Its sensitivity is high
It is a versatile instrument It measures only emf or potential difference
It is based on deflection method It is based on zero deflection method

Example In a potentiometer arrangement, a cell of emf 1.25 gives a balance point at 35.0 cm length of the wire, If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?

Sol. Here,        E1 = 1.25 V  , l1 = 35.0 cm , E2 = ?  , l2 = 63.0 cm

$ \displaystyle \frac{E_2}{E_1} = \frac{l_2}{l_1} $

$ \displaystyle E_2 = \frac{l_2}{l_1}E_1 $

$ \displaystyle E_2 = \frac{63}{35}\times 1.25 $

= 2.25 V

Example : A potentiometer wire of length 100 cm has a total resistance of 10Ω. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A cell of emf 10 mV is balanced against a length of 40cm of potentiometer wire. What is the value of the external resistance R ?

Solution : As shown in the figure, if R is the unknown resistance,
the current in the circuit

$ \displaystyle I = \frac{V}{R + r} $

$ \displaystyle I = \frac{2}{10 + R} $

Now as the 100 cm wire has a resistance of 10 Ω,

The resistance of 40 cm of wire will be $ \displaystyle 40 \times \frac{10}{100}$ = 4 ohm.

Potential drop across 40 cm wire will be V = I x 4

but here V = 10 mv (given)

10 × 10-3 = I × 4

$ \displaystyle I = \frac{1}{400} $

and $ \displaystyle I = \frac{2}{10+R} = \frac{1}{400}$

i.e. R = 790 Ω

Also Read :

→ Electric Current , Drift velocity & Deduction of Ohm’s Law
→ Combination of Resistors
→ Grouping of Cells
→Concept of EMF & Terminal P.D
→Kirchhoff ‘ s Laws & Solved Examples
→Wheatstone Bridge Principle & Metre Bridge
→Ammeter and Voltmeter
→Potentiometer & its Applications
→Charging of capacitors
→Discharging of Capacitors
→ Heating effect of currents

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