To Determine The Internal Resistance Of A Primary Cell Using Potentiometer

A potentiometer is a versatile and precise instrument used in electrical measurements to determine the internal resistance of a primary cell. By comparing the voltage drop across the cell with a known reference voltage, the internal resistance can be accurately calculated. This method is particularly advantageous as it does not draw significant current from the cell, preserving its true characteristics.

In real life, understanding the internal resistance of batteries is crucial for many applications, such as smartphones and electric vehicles. For instance, high internal resistance in a phone battery can lead to faster power drain and reduced battery life. Similarly, in electric vehicles, knowing the internal resistance helps optimize battery performance and longevity, ensure efficient energy use, and enhance the overall driving experience.

Aim

To determine the internal resistance of a given primary cell using a potentiometer.

Apparatus

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

Theory

The internal resistance of a cell is given by
$
r=R\left[\frac{l_1-l_2}{l_2}\right]
$

where $l_1$ and $l_2$ are the balancing lengths without a shunt and with a shunt, respectively, and R is the shunt resistance in parallel with the given cell.

Circuit Diagram

Procedure

1. Make the connections accordingly, as shown in the circuit diagram.
2. Clean the ends of the connecting wires with sandpaper and make tight connections according to the circuit diagram.
3. Tight the plugs of the resistance box.
4. Check the em.f. of the battery and cell and see that em.f. of the battery is more than that of the given cell, otherwise null or balance point will not be obtained (E' > E).
5. Take maximum current from the battery, making rheostat small.

6. To test the correctness of the connections. (Insert the plug in the key $\mathrm{K}_1$ and note the ammeter reading. Take out $2000 \Omega$ resistance plug from the resistance box. Place the jockey first at the end P of the wire and then at the end Q. If the galvanometer shows deflection in opposite directions in the two cases, the connections are correct).
7. Without inserting the plug in the key $\mathrm{K}_2$ adjust the rheostat so that a null point is obtained on the fourth wire of the potentiometer.

8. Insert the 2000 ohm plug back in its position in the resistance box and by slightly adjusting the jockey near the previously obtained position of null point, obtain the null point position accurately, using a set square.

9. Measure the balancing length $l_1$ between this point and the end P of the wire.
10. Take out the 2000 ohms plug again from the resistance box R.B. Introduce the plugs in key $\mathrm{K}_1$, as well as in key $\mathrm{K}_2$. Take out a small resistance $(1-5 \Omega)$ from the resistance box R connected in parallel with the cell.
11. Slide the jockey along the potentiometer wire and obtain the null point.

12. Insert 2000 ohms plug back in its position in R.B. and if necessary make a further
adjustment for sharp null point.
13. Measure the balancing length $l_1$ from end P
14. Remove the plug keys at $\mathrm{K}_1$ and $\mathrm{K}_2$. Wait for some time and for the same value of
current (as shown by the ammeter), repeat the steps 7 to 13.
15. Repeat the observations for different values of R repeating each observation twice.
16. Record your observations and on the basis of observations calculate the internal resistance.

Observations

Range of voltmeter $=\ldots \ldots$.
Least count of voltmeter $=$
E.M.F. of battery (or battery eliminator) $=\ldots$.
E.M.F. of cell $=\ldots$....

Calculations

1. For each set of observations find the mean and $l_2$
2. Calculate the value of r for each set.
3. Take the mean of values of r recorded.

Result

The internal resistance of the given cell is .......

Solved Examples Based on Determine the Internal Resistance of a Primary Cell Using Potentiometer

Example 1: A cell, shunted by a $8 \Omega$ resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2m when the cell is shunted by $4 \Omega$ resistance. The value of the internal resistance of the cell will be _______$\Omega$..

1) 8

2) 9

3) 10

4) 11

Solution:

$
\frac{\varepsilon}{\mathrm{R}+\mathrm{r}} \times \mathrm{R}=\mathrm{K} l
$
In the first case

$
\frac{\varepsilon}{8+r} \times 8=K \times 3
$
In the second case

$
\frac{\varepsilon}{4+r} \times 4=K \times 2
$
Solve (1) and (2)

$
\mathrm{r}=8 \Omega
$

Solve (1) and (2)
$
r=8 \Omega
$
Hence the answer is $8 \Omega$

Example 2: A potentiometer experiment is performed to determine the value of an unknown resistor. A known resistor of 10.0 Ω is connected to the potentiometer, and the balancing length is found to be 400 cm. When the unknown resistor is connected, the balancing length becomes 300 cm. Calculate the value of the unknown resistor.

1) 7.5 Ω

2) 8.3 Ω

3) 80.3 Ω

4) 10.5 Ω

Solution

Let 'R _known' be the known resistor, R_unknown be the unknown resistor, L₁ be the balancing length without the unknown resistor, and L₂ be the balancing length with the unknown resistor.

According to the principle of a potentiometer:
$
\frac{R_{\text {known }}}{R_{\text {unknown }}}=\frac{L_1}{L_2}
$
Given $R_{\text {known }}=10.0 \Omega, L 1=400 \mathrm{~cm}$, and $L 2=300 \mathrm{~cm}$, we can solve for $R_{\text {unknown }}$ :

$
R_{\text {unknown }=} \frac{R_{\text {known }} . L_2}{L_1}=\frac{10.0 .300}{400}=7.5 \Omega
$
Therefore, the value of the unknown resistor is 7.5 Ω.
Hence the answer is option (1).

Example 3: A potentiometer experiment is conducted to determine the internal resistance of a cell. A standard cell of known EMF 1.2V is connected to the potentiometer. The null point is obtained at a length of 60 cm. When the cell being tested is connected instead of the standard cell, the null point is obtained at a length of 45 cm. Calculate the internal resistance of the cell being tested.

1) 4.8 Ω

2) 79.5 Ω

3) 59.5 Ω

4) 3.0 Ω

Solution:

The internal resistance (r) of a cell can be determined using the formula:
r = Length of Potentiometer Wire × Known EMF/ Change in Potentiometer Length

Substitute the given values:
Length of Potentiometer Wire = 60 cm
Known EMF = 1.2 V

Change in Potentiometer Length = 60 cm − 45 cm = 15 cm
Calculate the internal resistance:
$r=\frac{60}{15} = 4.8\Omega$

Hence, the internal resistance of the cell being tested is 4.8 Ω.

Hence the answer is option (1).

Example 4: A potentiometer experiment is conducted to determine the internal resistance of a cell. A standard cell of known EMF 1.2V is connected to the potentiometer. The null point is obtained at a length of 60 cm. When the cell being tested is connected instead of the standard cell, the null point is obtained at a length of 48 cm. Calculate the internal resistance of the cell being tested.

1) 8.8Ω

2) 5.0Ω

3) 4.8 Ω

4) 5.10 Ω

Solution:

The internal resistance (r) of a cell can be calculated using the formula:

r = Length of Potentiometer Wire × Known EMF/ Change in Potentiometer Length

Substitute the given values:

Length of Potentiometer Wire = 60 cm
Known EMF = 1.2 V

Change in Potentiometer Length = 60 cm − 48 cm = 12 cm
Calculate the internal resistance:
$r=\frac{60 \times 1.2}{12}=6.0 \Omega$
Hence, the internal resistance of the tested cell is 6.0 Ω.

Hence the answer is option (3).

Example 5: An experiment is conducted to determine the internal resistance (r) of a cell using a potentiometer. A known emf of E = 1.5 V is connected to the potentiometer. The balancing length for the cell is L = 60 cm, and the total length of the potentiometer wire is L_total = 100 cm. If the internal resistance of the potentiometer is negligible, calculate the internal resistance of the cell.

1) 0.6 volt

2) 0.5 volt

3) 1.0 volt

4) 0.7 volt

Solution:

Given values:

Emf of the cell (E) = 1.5 V
Balancing length (L) = 60 cm
The total length of the potentiometer wire (L_total) = 100 cm

The potential difference across the length L of the potentiometer wire is given by:

$V=\frac{L}{L_{\text {total }}} E$

The potential difference V is also equal to the product of the current (I) and the total resistance (R) in the circuit

V = I.R

The total resistance R can be written as the sum of the internal resistance (r) of the cell and the resistance (R_pot) of the potentiometer wire

R = r + R_pot

Substitute the expressions for V and R and solve for r

$r=\frac{L}{L_{\text {total }}} E-R_{\text {pot }}$

Step 1: Calculate the resistance of the potentiometer wire:

$R_{\text {pot }}=\frac{L_{\text {total }}-L}{L} E$

Step 2: Calculate the internal resistance r of the cell:

$r=E-R_{p o t}$

Step 3: Substitute the given values and calculate r:

$r=1.5 \mathrm{~V}-\frac{40 \mathrm{~cm}}{60 \mathrm{~cm}} 1.5 \mathrm{~V}=0.5 \mathrm{~V}$

The internal resistance of the cell is 0.5 V.

Hence the answer is option (2).

Summary

A primary cell’s internal resistance can be determined using a potentiometer by taking readings of the cell’s potential difference in an open circuit (where there is no current flow) and also under a known external resistor, joining it. Thanks to its high precision, a potentiometer is able to accurately measure these potential differences. The internal resistance can then be obtained through the comparison of the emf (electromotive force) and the terminal voltage of the cell.