To Determine the Internal Resistance of a Given Primary Cell Using Potentiometer

A uniform wire AB with a low temperature coefficient of resistance is stretched over a uniform wooden scale to make a potentiometer. The potentiometer's wire is organised into two or three rows, summing to about $400\,cm$. Typically, a wooden board with a $100\,cm$ scale is fastened next to it. 

The cell produces EMF, i.e. Electromotive Force. This is the energy it provides to the electrons of the conductor by setting up an electric potential. Due to this, the electrons are able to drift and thus are able to produce an electric current. However, other than the resistance provided by the constituent atoms of the conductor, the cell also provides some resistance. This is known as the internal resistance of the cell.

Table of Contents

1. Aim

2. Apparatus Required

3. Theory 

4. Procedure

5. Observations 

6. Result

7. Precautions

8. Lab Manual Questions 

9. Viva Questions 

10. Practical Based Questions 

Aim

To set up a potentiometer and use it to determine the internal resistance of a given primary cell using potentiometer

Apparatus Required

1. Potentiometer

2. Battery (or battery eliminator)

3. Two one-way keys

4. Rheostat of low resistance

5. Galvanometer

6. High resistance box

7. Fractional resistance box

8. Ammeter

9. Voltmeter

10. Cell

11. Jockey

12. Set square

13. Connecting wires 

14. Sand paper

Theory

The device works on the principle that when two points are at the same potential, no current flows through the wire connecting those points. To use the device, a potential is set across the wire of the potentiometer. Each unit length of the potentiometer has a certain potential, known as the potential gradient. It can mathematically be expressed as $\text{Potential gradient} = \dfrac{\text{Potential across potentiometer wire}}{\text{Total length of potentiometer wire}}$.

The potentiometer wire is an external resistance in the circuit. In the absence of an external resistance, a closed circuit is short-circuited. This can be calculated from the equation $V = E  - Ir$, ……..(1)

Where $V$ is potential difference, $E$ is EMF of the cell, $I$ is the current through circuit and $r$ is the internal resistance. By putting the value of $I = \dfrac{E}{r}$ in (1), we get $V=0$

Similarly, if the terminals of cell are not connected, it is an open circuit and the potential difference across this cell is equal to its EMF as $I=0$ (from 1)

But, if an external resistance $R$ is connected in the circuit, then equation (1) becomes $V = E - \left ( \dfrac{E}{R+r} \right )r$ ….(2)

Rearranging (2), we get $\dfrac{E}{V}= 1 + \dfrac{r}{R}$ …(3)

When the jockey connected to one end of the cell of unknown internal resistance is slid over the potentiometer wire, it tries to balance the EMF of the cell with the potential of the wire. As soon as the length of the potentiometer to suffice the EMF of the cell is reached, the galvanometer shows no deflection. Thus the null point is said to be obtained. This way $V$ as well as $E$ can be obtained by arranging the circuit accordingly.

Mathematically, EMF of cell $(E)$ or potential across cell $V$ and length of potentiometer wire $(L)$ can be related as $E = k \times L_0$ or $V = k \times L$

Where, $k$ is the potential gradient of the potentiometer wire.

Now, we can write the ratio of $E$ and $V$ in terms of the length of the null point as $\dfrac{E}{V}=\dfrac{L_0}{L}$.

Therefore, equation (3) becomes $\dfrac{L_0}{L} = 1 + \dfrac{r}{R}$

Or, $\dfrac {L_0 – L}{L} = \dfrac{r}{R}$

Therefore, internal resistance of cell can be calculated by $r = \left ( \dfrac{L_0-L}{L} \right )R$

Procedure

1. Connect the circuit as shown in the diagram. In the diagram, $E_1$ is the cell of unknown internal resistance $(r)$ and $E$ is the cell providing potential difference to the potentiometer circuit. 

Circuit diagram

2. After checking the connection, insert a plug in key $K_1$ to connect the potentiometer circuit.

3. Keep the keys $K_2$ and $K_3$ open so that $R_{BOX-1}$ is not in the circuit whereas $R_{BOX-2}$ is. Take out a protective resistance from $R_{BOX-2}$.

4. Now, insert a plug in key $K_3$ to short circuit $R_{BOX-2}$ and find the balance length $L_0$ using a jockey.

5. Take $R=10 \Omega$ from $R_{BOX-1}$ and close the key $K_2$ to include $R_{BOX-1}$ into the circuit. With this set up, quickly measure balance length $L$ and then pull the plug so that $K_2$ is open again.

6. Throughout the above observations, keep the ammeter reading constant.

7. Reduce the value of resistance $R$ from $R_{BOX-1}$ in equal increments of $1 \, \Omega$ and get the balance length $L$ for each value of $R$.

8. Find the value of $L_0$ by opening $K_2$ and repeating step 4.

Observations

Observation Table

Calculate the value of internal resistance $(r)$ using the formula $ r \ = \left ( \dfrac{L_0-L}{L} \right )R$ for each set of observations. Find the mean of the internal resistance $(r)$.

Calculate the value of $r$ using a graph as well-

• The equation $r= \left ( \dfrac{L_0-L}{L} \right )R$ can be rearranged to $\dfrac{1}{R}=\dfrac{L_0}{r}\left ( \dfrac{1}{L} \right )- \dfrac{1}{r}$ . 

• Plot a graph of $\dfrac{1}{R}$ versus $\dfrac{1}{L}$. The graph will come out to be a straight line graph having slope $\dfrac{L_0}{r}$. 

• Extend the slope to find the intercept. This intercept will give a value of $\dfrac{1}{r}$, which can be reciprocated to obtain the value of internal resistance $(r)$.

Result

• The internal resistance $(r)$ of the cell according to mean value is …………….

• The internal resistance $(r)$ of the cell according to the graph is …………….

Precautions

1. It is important to avoid disturbing the primary cell of unknown internal resistance, since doing so might alter the accuracy of its internal resistance.

2. The primary cell, $E_1$, should have a higher EMF than battery $E$.

3. Positive terminals of both cells $E$ and $E_1$ both should be connected to the same point on the potentiometer.

4. Always measure distance up to the balancing point starting from end A on the potentiometer, which is the location where the positive terminals of the battery are attached.

Lab Manual Questions

1. The potential difference is measured using a voltmeter connected in series with a battery's terminals. What distinguishes it from emf?

Ans: A voltmeter can only measure the difference in voltage between a cell's two terminals. However, it adds some resistance to the circuit. Therefore, it takes up some current and shows potential less than the actual EMF.

2. What conclusions would you draw if the galvanometer needle in your test setup kept swaying?

Ans: This may be due to error in connections. All connections should be checked and tightened.

3. Find the internal resistance of a freshly prepared Leclanche cell. Does its internal resistance change with $R$ ?

Ans: The internal resistance of a cell is independent of the external resistance $R$ connected. It only depends on the concentration of electrolyte, the temperature and the distance between electrodes.

4. State the factors on which the internal resistance of a cell depends.

Ans: The concentration of electrolyte, the temperature and the distance between electrodes of the cell affects the internal resistance of a cell.

Viva Questions

1. Explain what is the internal resistance of a cell?

Ans: It is the resistance provided by the cell, against the flow of current.

2. What is a potentiometer and state its uses?

Ans: Potentiometer is a device having a long wire attached, across which a potential gradient is set up. It is used to measure the EMF and internal resistance of a cell.

3. What is the difference between the Voltage across a cell when it is connected to an external resistance and when it’s not?

Ans: The potential difference $V$, the EMF of cell $E$, the internal resistance $r$ and the current $I$ can be related by the equation $V = E - Ir$. When there is no external resistance connected, by Ohm’s law the current $I = \dfrac{E}{r}$. Adding this value in the first equation, $V = E - \dfrac{E}{r}r$. Therefore, $V= 0$. However, in the presence of external resistance, $I = \dfrac{E}{R+r}$. Therefore, $V = E - \dfrac{E}{R+r}r$. Therefore, $V = \dfrac{ER}{R+r}$ 

4. What is the preferred substance for making a potentiometer wire?

Ans: The wire used in potentiometers is typically constructed of an alloy, such as manganin or nichrome. This is due to the high resistivity and low temperature coefficient of resistance of such alloys.

5. How does the internal resistance of a cell depend on current drawn from it?

Ans: A cell's internal resistance is independent of the current that is being drawn from it. Internal resistance depends on the type of the electrolyte and is independent of current because it is just the amount of resistance the cell's electrolyte offers to the flow of electric current. However, it increases with increase in temperature and temperature increases due to heat generated by the flowing current.

6. What factors affect the internal resistance of a cell?

Ans: The concentration of electrolyte, the temperature and the distance between electrodes of the cell affects the internal resistance of a cell.

7. State the possible sources of error in the experiment.

Ans: The possible sources of error are as follows:

• The cross-section of a potentiometer wire need not be uniform.

• The resistance of the brass strips at their ends might be limited.

• The auxiliary battery's emf, which causes the potential drop down the wire, could vary over the duration of the experiment.

• Some mistake may be introduced if the potentiometer wire is heated by current.

8. What is a rheostat?

Ans: A rheostat is an adjustable resistor used in situations where varying resistance or adjusting current in an electric circuit is necessary.

9. Why does a galvanometer show no deflection at the null point?

Ans: The potential of both the terminals to which the galvanometer is connected is the same and thus there is no potential difference at the null point. Thus, no current flows and the galvanometer needle does not deflect.

10. What are the voltmeter and ammeter used for?

Ans: The voltmeter is connected in parallel to the resistance, the voltage or potential difference across which is to be measured. The ammeter, on the other hand, is connected in series with the resistance the current flowing through which is to be measured.

Practical Based Questions

1. The internal resistance of which among the following cannot be calculated

1. Leclanche cell

2. Daniel cell

3. Nickel metal Hydride

4. Dry cell

Ans: (c)

2. The current $(I)$ flowing through a cell of EMF $E$ and internal resistance $r$ when it is connected to an external resistance $R$ is:

1. $\dfrac{E}{R+r}$ Coulomb

2. $0$ Coulomb

3. $\dfrac{R+r}{E}$ Coulomb

4. Infinite

Ans: (a)

3. The potential across a cell of EMF $E$ and internal resistance $r$ is:

1. Infinite

2. $E$

3. $\dfrac{E}{r}$

4. Zero

Ans: (d)

4. The current $(I)$ flowing through a cell of EMF $E$ and internal resistance $r$ is:

1. $0$ Coulomb

2. $\dfrac{E}{r}$ Coulomb

3. $\dfrac{r}{E}$ Coulomb

4. Infinite

Ans: (b)

5. Potential gradient of potentiometer wire depends on

1. Current passing through the potentiometer wire

2. Resistivity of material of the potentiometer wire

3. Area of cross-section of the potentiometer wire

4. All of the above

Ans: (d)

6. The sensitivity of the potentiometer can be increased by:

1. Decreasing the thickness of the potentiometer wire

2. Increasing the length of the potentiometer wire

3. Increasing the temperature of the surrounding

4. All of the above

Ans: (b)

7. Which instrument is used to measure deflection due to current:

1. Voltmeter

2. Rheostat

3. Galvanometer

4. Keys

Ans: (c)

8. Which value cannot be measured by a potentiometer? 

1. Capacity

2. Resistance

3. Electric current

4. Potential

Ans: (a)

9. If the radius of the potentiometer wire is doubled, the length of the null point will

1. Remain same

2. Increase

3. Decrease

4. Increase and then decrease

Ans: (c)

10. If the length of the potentiometer wire is increased but same current is flowed through it, its sensitivity will:

1. Increase

2. Decrease

3. Remain constant

4. Cannot be predicted

Ans: (a)

Summary

A potentiometer works on the principle that when two points are at the same potential, no current flows through the wire connecting those two points. Since it works on the principle of no deflection, it is the most accurate method. A jockey is slid on the potentiometer wire to obtain a null point since the length of the null point is proportional to the EMF of cell $E_1$.

This method is repeated in the presence of external resistance to find the new potential across the cell. The null point is the length which, multiplied by the potential gradient of the wire (potential per unit length), gives the EMF $E$ of and potential $V$ across the cell. After the ratio of the length of the null point of $E$ and $V$ is found, it can be equated with their potentials. Therefore, $\dfrac{E}{V}=\dfrac{L_0}{L}$. This can then be used to find internal resistance of cell from formula $ r = \left ( \dfrac{L_0-L}{L} \right )R$