The measuring instrument called a **potentiometer** is essentially a voltage divider **used** for measuring electric potential (voltage); the component is an implementation of the same principle, hence its name. **Potentiometers** are commonly **used** to control electrical devices such as volume controls on audio equipment. We will study here about Comparison of emf of two cells using Potentiometer, **Determination of internal resistance of cell**, **Calibration of voltmeter**, **Calibration of ammeter, Measurement of small thermo emf.**

**The Potentiometer is mainly used for,**

- To compare the emfs of two primary cells
- To determine the internal resistance of a primary cell
- To determine the value of a high resistance
- To determine the emf of a cell.
- Calibration of Ammeter
- Calibration of Voltmeter
- measurement of small thermo emf

**PRECAUTIONS IN THE USE OF POTENTIOMETER**

**Determination of unknown EMF or Potential Difference **

[3] If unknown potential difference V is balanced for length $\ell$ than $V=x \ell=\left(\frac{E_{0}}{\ell_{0}}\right) l$

[4] If the length of potentiometer wire is changed from L to L’ then the new balancing length is $\ell^{\prime}=\left(\frac{L^{\prime}}{L}\right) \ell$ If length is increased L’ > L so $l^{\prime}>l$ and if length is decreased L’ < L so $l^{\prime}<l$ So change in balancing length $\Delta \ell=\left(\ell^{\prime}-\ell\right)$

[ 5] If the current flowing through resistance $\mathrm{R}$ is I then

$$ \mathrm{V}=\mathrm{IR}=\mathrm{x} \ell=\left(\frac{\mathrm{E}_{0}}{\ell_{0}}\right) \ell \quad \text { so } \quad \mathrm{I}=\frac{\mathrm{x} \ell}{\mathrm{R}}=\left(\frac{\mathrm{E}_{0}}{\ell_{0}}\right) \frac{\ell}{\mathrm{R}} $$

For determination of current we use a coil of standard resistance.

**Ex.** The current flowing through the primary circuit is 2A and resistance per unit length is

0.2 $\Omega / m$. If the potential difference across 10 ohm coil is balanced at 2.5 m then find current flowing through the coil.

**Sol.** The potential gradient $x=\frac{I R}{L}=2 \times 0.2=0.4 V / m$

Unknown potential $V=x \ell=I^{\prime} R$ so $I^{\prime}=\frac{x \ell}{R}=\frac{0.4 \times 2.5}{10}=0.1 A$

**Comparison of emf of two cells using Potentiometer**

[1] No standardisation is necessary.

[2] Let $E_{1}$ emf be balanced at length $\ell_{1}$ and $E_{2}$ emf be balanced at length $\ell_{2}$ then

$E_{1}=x \ell_{1}$ and $E_{2}=x \ell_{2}$ so $\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

[3] If two cells joined in series support each other then the balancing length is $\ell_{1}$ so $E_{1}+E_{2}=x \ell_{1}$

If two cells joined in series oppose each other then the balancing length is $\ell_{2}$ so $E_{1}-E_{2}=x \ell_{2}$

$\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{\ell_{1}}{\ell_{2}}$ or $\frac{E_{1}}{E_{2}}=\frac{\ell_{1}+\ell_{2}}{\ell_{1}-\ell_{2}}$

**Ex.** A uniform potential gradient is established across a potentiometer wire. Two cells of emf $E_{1}$ and $E_{2}$ connected to support and oppose each other are balanced over $\ell_{1}$ = 6m and $\ell_{2}$ = 2m. Find E1/E2.

**Sol. **$E _{1}+ E _{2}= x \ell_{1}=6 x$ and $E _{1}- E _{2}=2 x$

$\frac{E_{1}+E_{2}}{E_{1}-E_{2}}=\frac{6}{2}$ or $\frac{E_{1}}{E_{2}}=\frac{2}{1}$

**Determination of Internal Resistance of Cell**

[1] No standardization is necessary.

[2] Keeping $\mathrm{K}_{1}$ open the balancing length $\ell_{1}$ gives emf of cell so E = $x \ell_{1}$

[3] Keeping $K_{1}$ closed the balancing length $\ell_{2}$ for some resistance R gives potential difference so V = $x \ell_{2}$

Internal resistance $r=\frac{E-V}{V} \quad R=\frac{x \ell_{1}-x \ell_{2}}{x \ell_{2}}$

$R=\left(\frac{\ell_{1}-\ell_{2}}{\ell_{2}}\right) R$

**Calibration of Voltmeter **

[1] The voltmeters do not given accurate measurements because they do not have infinite resistance.

[2] The error in measurement is found by comparision with readings of potentiometer.

[3] Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$

[4] The unknown potential difference V’ is balanced at length $\ell_{1}$ then $V^{\prime}=x \ell_{1}=\frac{E_{0}}{l_{0}} \ell_{1}$

[5] If reading of voltmeter is V then error is V – V’ which can be positive, negative or zero.

[6] The calibration curve is obtained plotting voltmeter reading V on x axis and error on y axis.

**Calibration of Ammeter**

[1] V = IR and if R = 1$\Omega$ then V = I so potential difference across 1$\Omega$ resistance is equal to current through the resistance.

[2] Potential gradient is found by using a standard cell $x=E_{0} / \ell_{0}$

[3] The current through R or 1$\Omega$ coil is measured by

ammeter () and calculated by potentiometer as $V^{\prime}=I^{\prime} \times 1 \Omega=x \ell_{1}=\left(\frac{E_{0}}{\ell_{0}}\right) \ell_{1}$

[4] The error $I-I^{\prime}$can be found and calibration curve is obtained by plotting ammeter reading on x axis and error on y axis.

**Measurement of Small Thermo EMF**

[1] A high resistance box is used in primary circuit to reduce primary current.

[2] If galvanometer shows no deflection for some $R_{1}$ then potential difference across $R_{1}$ is $V=E_{0}=I R_{1}$ or $|=E_{0} / R_{1}$

[3] If balancing length for small thermo emf E is $\ell$ then $E=x \ell=\frac{E_{0}}{R_{1}} \cdot \frac{R}{L} \ell$

**Precautions in use of Potentiometer**

[1] All high potential terminals of primary and secondary circuit should be connected at same point of Potentiometer.

[2] The known emf of primary circuit must be greater than the unknown potential difference to be measured in secondary circuit.

[3] The balancing length is always measured from the point of higher potential.

[4] In state of zero deflection no current flows in galvanometer circuit but it flows in primary circuit.

**Potentiometer Working Principle****Difference Between Potentiometer and Voltmeter****Sensitivity of Potentiometer****Advantages of Potentiometer****Potentiometer Important Questions**

Watch out the Video: **Applications of Potentiometer & its Construction by Saransh Sir.**

**Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)**