To Compare EMF Of Two Given Primary Cells Using Potentiometer

A potentiometer is a versatile instrument used in physics labs to measure the electromotive force (EMF) of cells with high precision. By comparing the EMF of two primary cells, we can determine their voltage without drawing any current from them, ensuring accurate results. This technique is crucial in understanding the efficiency and performance of various power sources. In real-life applications, such as in battery manufacturing and quality control, potentiometers ensure that batteries provide consistent and reliable power. For instance, they help in comparing the EMF of different battery brands to determine which one offers longer-lasting performance, thereby guiding consumer choices and industrial standards.

Aim

To compare the EMF of two given primary cells using the potentiometer.

Apparatus

A potentiometer, a Leclanche cell, a Daniel cell, an ammeter, a voltmeter, a galvanometer, a battery (or battery eliminator), a rheostat of low resistance, a resistance box, a one-way key, a two-way key, a jockey, a set
square, connecting wires and a piece of sandpaper.

Theory

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

where, $\mathrm{E}_1$ and $\mathrm{E}_2$ are the e.m.f. of two given cells and $\mathrm{l}_1$ and $\mathrm{l}_2$ are the corresponding balancing lengths on potentiometer wire.

Circuit Diagram

Procedure

1. Arrange the apparatus as shown in the circuit diagram figure.
2. Remove the insulation from the ends of the connecting copper wires with sandpaper.

3. Measure the e.m.f. (E) of the battery and the e.m.fs. (E₁, and E₂ ) of the cells. See that $E>E_1$ and also $E>E_2$
4. Connect the positive pole of the battery (a battery of constant e.m.f.) to the zero end (P) of the potentiometer and the negative pole through a one-way key, an ammeter, and a low resistance rheostat to the other end (Q) of the potentiometer.

5. Connect the positive poles of the cells E₁ and E₂ to the terminal at the zero end (P) and the negative poles to the terminals a and b of the two-way key.
6. Connect the common terminal c of the two-way key through a galvanometer (G) and a resistance box (R.B.) to the jockey J.

7. Take maximum current from the battery making rheostat resistance zero.

8. Insert the plug in the one-way key (K) in the circuit and also in between the terminals a and c of the two-way.
9. Take out a 2000 ohms plug from the resistance box (R.B).
10. Press the jockey at the zero end and note the direction of deflection in the galvanometer.

11. Press the jockey at the other end of the potentiometer wire. If the direction of deflection is opposite to that in the first case, the connections are correct. (If the deflection is in the same direction then either the connection is wrong or e.m.f. of the the auxiliary battery is less).
12. Slide the jockey gently over the potentiometer wires till you obtain a point where the galvanometer shows no deflection.

13. Put the 2000 ohms plug back in the resistance box and obtain the null point position accurately, using a set square.
14. Note the length $l_1$ of the wire for the cell $E_1$ Also note the current as indicated by the ammeter.
15. Disconnect the cell $E_1$ by removing the plug from the gap AC of the two-way key and connect the cell $E_2$ by inserting the plug into the gap be of the two-way key.

16. Take out a 2000 ohms plug from resistance box R.B. and slide the jockey along the potentiometer wire so as to obtain no deflection position.
17. Put the 2000 ohms plug back in the resistance box and obtain an accurate position of the null point for the second cell $E_2$

18. Note the length $l_2$ of wire in this position for the cell $E_2$ . However, make sure that the ammeter reading is the same as in step 14.

19. Repeat the observations alternately for each cell again for the same value of current.
20. Increase the current by adjusting the rheostat and obtain at least three sets of observations in a similar way.
21. Record your observations and on the basis of observations compare the emf of two given primary cells

Observations

1. Range of voltmeter $=\ldots$.

Least count of voltmeter $=\ldots .$.
E.M.F. of battery (or battery eliminator),$\quad E=\ldots \ldots$.
E.M.F. of Leclanche cell, $\quad E_1=$
E.M.F. of Daniel cell,
$E_2=$

Calculations

1. For each observation find the mean $l_1$ and mean $l_2$
2. Find $\frac{E_1}{E_2}$ for each set, by dividing mean $l_1$ by mean $l_2$
3. Find the mean $\frac{E_1}{E_2}$

Result

The ratio of E.M.Fs, $\frac{E_1}{E_2} \cong \ldots$

Solved Examples Based on Comparison EMF of Two Given Primary Cells Using Potentiometer

Example 1: In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potentiometer wire. The potential difference across the ends of resistance is found to balance against 220 cm of the 0.5wire. The corresponding reading of the voltmeter is 0.5 volt. The error in the reading of the voltmeter will be : (in volts)

1) -0.05

2) 0.15

3) 0.5

4) -0.15

Solution:

To compare the emf of two given primary cells using a potentiometer
$
\begin{aligned}
& \frac{E_1}{E_2}=\frac{K l_1}{K l_2}=\frac{l_1}{l_2} \\
& \mathrm{E}_1=\text { Emf of the first cell } \\
& \mathrm{E}_2=\text { Emf of the second cell } \\
& \text { wherein } \\
& \mathrm{K}=\text { Constant of proportionality } \\
& l_1, l_2=\text { balancing length's of both } \\
& V=K l \Rightarrow V \propto l \\
& E=1.1 \text { volt }, l_1=440 \mathrm{~cm}, V=0.5 \mathrm{Volt}, l_2=220 \mathrm{~cm}
\end{aligned}
$
Let Error in the reading of Voltmeter be $\Delta V$

$
\begin{aligned}
& \text { then, }(0.5 \Delta V)=220 K \\
& \frac{1.1}{440}=\frac{0.5-\Delta V}{220}, \quad \therefore V=-0.05 \mathrm{Volt}
\end{aligned}
$

Hence, the answer is (-0.05).

Example 2: In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3:2, the difference in the balancing length of the potentiometer wire in the above two cases will be ______ cm.

1) 25

2) 26

3) 27

4) 28

Solution:

For potentiometer
$
\begin{aligned}
& \text { E } \propto \mathrm{l} \\
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{l}_1}{\mathrm{l}_2} \\
& \frac{3}{2}=\frac{75 \mathrm{~cm}}{\mathrm{l}_2} \\
& \mathrm{l}_2=50 \mathrm{~cm}
\end{aligned}
$
The difference in the balancing Length of the potentiometer wire in the above two cases is $\Delta \mathrm{l}=\mathrm{l}_1-\mathrm{l}_2$ $\Rightarrow \Delta \mathrm{l}=25 \mathrm{~cm}$

Hence, the answer is (25).

Example 3: A potentiometer wire of length 10 m and resistance $20 \Omega$ is connected in series with a 25 V battery and an external resistance $30 \Omega$. A cell of emf E in the secondary circuit is balanced by a 250 cm-long potentiometer wire. The value of E in (volt) is $\frac{x}{10}$. The value of is $\qquad$

1) 2.5

2) 3.5

3) 4.5

4) 5.5

Solution:

$
\begin{array}{ll}
\mathrm{P} \rightarrow \text { Balancing point (null point) } \\
\mathrm{v}=25 \mathrm{v} & \\
R=30 \Omega & \phi=\text { potential gradient } \\
\mathrm{l}_{\mathrm{AB}}=10 \mathrm{~m} & \phi=\frac{V_{A B}}{l_{A B}}=\frac{I R_{A B}}{l_{A B}} \\
\mathrm{R}_{\mathrm{AB}}=20 \Omega &
\end{array}
$
For $p$ to be the null point

$
\begin{aligned}
\mathrm{E} & =\mathrm{V}_{\mathrm{AP}}=\mathrm{IR}_{\mathrm{AP}}=\phi \mathrm{l}_{\mathrm{AP}} \\
E & =\frac{I\left(R_{A B}\right)}{\left(l_{A B}\right)} \times\left(l_{A P}\right) \\
& =\left(\frac{V}{R+R_{A B}}\right) \times\left(\frac{R_{A B}}{l_{A B}}\right) \times l_{A P} \\
E & =2.5 \mathrm{~V}
\end{aligned}
$

Hence, the value of x is 2.5.

Example 4: In a potentiometer arrangement, a cell of emf 1.20V gives a balance point at 36 cm the length of the wire. This cell is now replaced by another cell of emf 1.80V. The difference in balancing length of the potentiometer wire in the above conditions will be__________ cm.

1) 18

2) 19

3) 20

4) 21

Solution:

For potentiometer,

$
\begin{aligned}
& \mathrm{E}=\phi \mathrm{l} \\
& \phi \rightarrow \text { potential gradient } \\
& \mathrm{l} \rightarrow \text { Balancing length } \\
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{l}_1}{\mathrm{l}_2} \\
& \frac{1.2}{1.8}=\frac{36}{\mathrm{l}_2} \\
& \mathrm{l}_2=\frac{36 \times 1.8}{1.2} \\
& \mathrm{l}_2=54 \mathrm{~cm}
\end{aligned}
$
The difference in the balancing length of the potentiometer wire in the above conditions will be $l_2-l_1=18 \mathrm{~cm}$

Hence, the answer is (18).

Example 5: A potentiometer experiment is performed to compare the electromotive forces (EMFs) of two primary cells, Cell X and Cell Y. The potentiometer wire has a total length of 120 cm. When Cell X is connected to the potentiometer, the null point is found at a length of 72 cm. When Cell Y is connected instead of Cell X, the null point is found at a length of 90 cm. Given that the EMF of Cell X is 1.5V, calculate the EMF of Cell Y.

1) 1.875V

2) 17.60 V

3) 2.50 V

4) 20.35 V

Solution:

The ratio of the EMFs of two cells is equal to the ratio of the lengths of wire on either side of the null point:
$
\frac{\text { EMFofcell' } X^{\prime}}{\text { EMFofcell' } Y^{\prime}}=\frac{\text { Lengthofcell' } X^{\prime}}{\text { Lengthofcell' }^{\prime} Y^{\prime}}
$
Substitute the given values:

$
\frac{1.5}{\text { EMFofcell' }^{\prime}}=\frac{72}{90}
$
Solving for the EMF of Cell Y:

$
\text { EMFofcell' }^{\prime} Y^{\prime}=\frac{1.5 \times 90}{72}=1.875 \mathrm{~V}
$
Hence, the electromotive force of Cell Y is approximately 1.875 V.

Hence, the answer is the option (1).

Summary

To compare the electromotive force (EMF) of two primary cells, a potentiometer is used to measure voltage differences between them without drawing any current from the cells. A potentiometer, which is known for its high degree of precision, finds the balance point where there is no current through the galvanometer. The ratio of EMFs can be determined by comparing the balance points of each cell. This method is extremely accurate because it eliminates errors caused by internal resistance thus making it possible to make a true comparison of the EMF of two cells.