Potentiometer 

A potentiometer is an electrical device used to measure and compare voltages accurately without drawing current from the circuit. It consists of a resistive wire and a sliding contact (wiper) that moves along the wire to adjust the voltage. Potentiometers are mainly used to determine the internal resistance of a cell, compare EMFs of two cells, and measure potential differences in circuits. Its key advantage is that it provides precise and reliable measurements, making it an important tool in physics experiments.

1.0Definition of Potentiometer

A potentiometer is an electrical device used to measure and compare potential differences (voltages) accurately. It consists of a long uniform resistive wire and a sliding contact, which helps obtain a balance point where no current flows through the galvanometer. This makes the measurement highly precise.

2.0Necessity of potentiometer

Practically voltmeter has finite resistance. (ideally it should be ) in other words it draws some current from the circuit. To overcome this problem potentiometer is used because at the instant of measurement, it does not draw current from the circuit.A potentiometer is a linear conductor of uniform cross-section with a steady current set up in it. This maintains a uniform potential gradient along the length of the wire. Any potential difference which is less than the potential difference maintained across the potentiometer wire can be measured using this. 

The wire should have high resistivity and low expansion coefficient.For example: Manganin or Constantine wire etc.

3.0Circuit of Potentiometer

• Primary circuit contains constant source of voltage rheostat or Resistance Box.
•  The secondary circuit contains a galvanometer and unknown cell.
• Potential Gradient (x) V/m

The fall of potential per unit length of potentiometer wire is called potential gradient.

$x=\frac{V}{L}=\frac{currentresistanceofpotentiometerwire}{lengthofpotentiometerwire}=I(\frac{R_p}{L})$

4.0Application of Potentiometer

(A) To measure unknown emf of cell

The unknown voltage V is connected across the potentiometer wire as shown in figure. The positive terminal of the unknown voltage is kept on the same side as of the source of the top most battery. When the reading of the galvanometer is zero then we say that the meter is balanced. In that condition V = xl.

(B) To find emf of unknown cell and compare emf of two cells

In case 1 , In figure, (2) is joint to (1) then balance length = $l_1$

${\epsilon }_1=x{l}_1\dots \dots (1)$

$\text{In case 2 , In figure, (3) is joint to (2)}$

$\text{Then balance length }=l_2$

${\epsilon }_2=x{l}_2\dots \dots (2)$

$\frac{{\epsilon }_1}{{\epsilon }_2}=\frac{l_1}{l_2}$

(C) To find current if resistance is known

$V_A-V_C=x{l}_1$

$I{R}_1=x{l}_1;I=\frac{x{l}_1}{R_1}$

Similarly, we can find the value of R2also. Potentiometer is ideal voltmeter because it does not draw any current from circuit,at the balance point.

(D) To find the internal resistance of cell

By First arrangement, ${\epsilon }^{\prime }=x{l}_1$

By second Arrangement, $IR=x{l}_2$

$I=\frac{x{l}_2}{R}$

Also $I=\frac{{\epsilon }^{\prime }}{r^{\prime }+R}$

$$\begin{gathered} \frac{{\epsilon }^{\prime }}{r^{\prime }+R}=\frac{x{l}_2}{R} \\ \frac{x{l}_1}{r^{\prime }+R}=\frac{x{l}_2}{R} \\ {r}^{\prime }=\left[\frac{l_1-l_2}{l_2}\right]R \end{gathered}$$

Note:

• If the same polarity of 1 cell and 2 cell are not connected with each other then the balancing condition (null deflection) cannot be obtained.
• If potential drop across wire is less than potential drop of secondary circuit then balancing condition cannot be obtained.

5.0Difference Between Potentiometer And Voltmeter

6.0Sensitivity of Potentiometer

It is the ability of potentiometers to read very small values of potential difference easily. So, sensitivity is inversely proportional to potential gradient of potentiometer wire.

$Sensitivity\propto \frac{1}{P.G}\propto \frac{l}{V}\propto \frac{Lengthofwire}{Potentialdropacrosswire}$

7.0Half Deflection Method

To find the resistance of a galvanometer by half deflection method and find its figure of merit.

Theory: The connections for finding the resistance of a galvanometer by the half deflection method are shown in Fig. When the key $K_1$ is closed, keeping the key $K_2$ open, the current Ig through the galvanometer is given by

$I_g=\frac{E}{R+G}$, where E=E.M.F. of the cell

R=Resistance from the resistance  box R.B.

G=Galvanometer Resistance

If $\theta$ is the deflection produced, then

$\frac{E}{R+G}=k\theta$ …..(1)

If now the key $K_2$ is closed and the value of the shunt resistance S is adjusted so that the deflection is reduced to half of the first value, then current flowing through the galvanometer Ig'is given by

$I_g^{\prime }=\frac{E}{R+\frac{GS}{G+S}}\left(\frac{S}{G+S}\right)=\frac{k\theta }{2}$

$I_g^{\prime }=\frac{ES}{R(G+S)+GS}=\frac{k\theta }{2}\dots \dots (2)$

Comparing(1) and (2)

$(R+G)2S=R(G+S)+GS$

$(R-S)G=RS\text{or}G=\frac{RS}{R-S}$

If the value of R is very large as compared to S, then $\frac{R}{R-S}$ is nearly equal to unity. Hence GS

Figure of Merit:

The merit of a galvanometer is that much current is sent through the galvanometer in order to produce a deflection of one division on the scale.

If k is the figure of merit of the galvanometer and '' be the number of divisions on the scale, then current Ig through the galvanometer is given by

$Ig=k\theta$

Precautions:

1. The value of ‘R’ should be large

2. To decrease the deflection, the shunt resistance should be decreased and vice-versa.

3. In this method it is assumed that the deflection is proportional to the current. This is possible only in a Weston type moving coil galvanometer.

4. The connections must be tight and the ends of connecting wires should be cleaned.

Illustration-1.A potentiometer uses a wire 10 meters long with a resistance of 10 Ω. It is connected in series with a 2 V supply and a rheostat. If the potential gradient along the wire is adjusted to 1 μV per mm, what is the required resistance of the rheostat?

Solution:

$l=10m,R=10\Omega ,E=2Volts,\frac{dV}{dl}=1\mu V/mm$

$\frac{dV}{dl}=\frac{1\times 10^{-6}}{1\times 10^{-3}}V/m=1\times 10^{-3}V/m$

Across wire potential drop, $\frac{dV}{dl}\times l=1\times 10^{-3}\times 10=0.01$ Volts

$I=\frac{0.01}{10}=0.001=\frac{E}{R+R^{\prime }}(R^{\prime }=ResistanceofRheostat)$

$R^{\prime }=\frac{E}{0.001}-R=\frac{2}{0.001}-10=2000-10=1990\Omega$

Illustration-2:In an experiment to determine the emf of an unknown cell, its emf is compared with a standard cell of known emf ${\epsilon }_1=1.12V$. The balance point is obtained at 56 cm with a standard cell and 80 cm with the unknown cell. Determine the emf of the unknown cell.

Solution:

Here ${\epsilon }_1=1.12V;l_1=56cm;l_2=80cm$

Using equation 

${\epsilon }_1=x{l}_1\dots \dots (1)$

${\epsilon }_2=x{l}_2\dots \dots (2)$

$\frac{{\epsilon }_1}{{\epsilon }_2}=\frac{l_1}{l_2}\Rightarrow {\epsilon }_2={\epsilon }_1\left(\frac{l_2}{l_1}\right)$

${\epsilon }_2=1.12\left(\frac{80}{56}\right)=1.6V$