Determine the internal resistance of a cell using a potentiometer.

Experiment to measure the internal resistance of a cell using a potentiometer by comparing EMF and terminal voltage, applying the equilibrium and resistance relation.

Apparatus

• Potentiometer

• Battery

• Ammeter

• Resistance box

• Rheostat

• Two keys

• Galvanometer

• Given cell

• Shunt wire and connecting wires

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Theory

A potentiometer is a precise device used to measure or compare electrical potentials without drawing current from the circuit.

Principle

When a steady current passes through a wire of uniform cross-section, the potential difference across any length is proportional to that length. The internal resistance of a cell is the resistance offered by its electrolyte to current flow. It depends on the size and separation of the plates.

Consider a cell of EMF $E$ and internal resistance $r$ connected to an external resistance $R$ via a key $K$. A voltmeter can measure the potential difference across the cell’s terminals. When the external resistance is connected, current $I$ flows through the circuit. By Ohm’s law:

$$E=I(R+r)$$

Here, $IR$ is the voltage across the external resistance, which a voltmeter reads as $V$. The difference $E-V=Ir$ represents the internal potential drop, which cannot be measured directly but is determined from the balance lengths on the potentiometer.

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Procedure

1. Draw the circuit as shown in the diagram.

2. Clean the wire ends and arrange the apparatus: battery $E$, key K1K_1K1​, and rheostat in series with the potentiometer wire AB.

3. Connect the cell ExE_xEx​ to the resistance box RBR_BRB​ via key K2K_2K2​. The positive poles of both cells connect to end A. Connect the negative pole of the cell to a jockey through a galvanometer.

4. Introduce resistance via the rheostat. Close K1K_1K1​ and press the jockey near ends A and B to check the galvanometer deflection. Opposite directions indicate a correct connection.

5. Close K1K_1K1​ (keep K2K_2K2​ open) and find the balance point. Measure the distance l1l_1l1​ from end A.

6. Connect a suitable resistance $R$ through the resistance box. Close both keys. Find the new balance length l2l_2l2​. Repeat three times with slightly different resistances.

7. Calculate the internal resistance using:

r=l1−l2l2Rr = \frac{l_1 – l_2}{l_2} Rr=l2​l1​−l2​​R

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Observations & Calculations

• Balance length with K2K_2K2​ open: l1=_________cml_1 = \_\_\_\_\_\_\_\_\_ \text{cm}l1​=_________cm

• Balance lengths with resistances applied: l2=_________cml_2 = \_\_\_\_\_\_\_\_\_ \text{cm}l2​=_________cm

• Internal resistance: $r=\_\_\_\_\_\_\_\_\_\Omega$

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Precautions

• Ensure all connections are clean and tight.

• Maintain constant current in the ammeter during each observation.

• Open keys after each reading to prevent heating.

• Connect the positive poles of the battery and cell to the zero end of the wire.

• Maintain uniform diameter of potentiometer wire to ensure consistent resistance per unit length.

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Viva Voce

Q1. What is electric potential?
A: It indicates the work done in moving a unit positive charge from infinity to the point.

Q2. Define EMF.
A: EMF is the potential difference across the cell terminals due to chemical reactions.

Q3. What is the short-circuiting of a cell?
A: Connecting cell terminals with negligible resistance causes large currents and may damage the cell.

Q4. When does a battery give maximum current?
A: When the external resistance equals the internal resistance.

Q5. What is a potentiometer?
A: A device to measure or compare potentials accurately.

Q6. Working principle?
A: Potential difference across a wire is proportional to its length.

Q7. Can EMF be measured accurately?
A: Yes, without drawing current.

Q8. Uses of a potentiometer:

• Determine the EMF of a cell

• Compare the EMF of two cells

• Act as a continuous potential divider

• Measure the internal resistance of a cell

Q9. What is internal resistance?
A: Resistance offered by the electrolyte to current flow.

Q10. Is internal resistance constant?
A: No, it changes with current.

Q11. Difference between a cell and a battery?
A: A cell cannot be recharged; a battery can.

Q12. Internal resistance magnitude?
A: Usually very small (~0.1 Ω).

Q13. What is an accumulator?
A: A rechargeable battery.

Q14. Why stop battery current after null observation?
A: To prevent heating and resistance changes.

Q15. Why should the wire diameter be uniform?
A: To maintain consistent resistance per unit length.