Question

Determine the internal resistance of the given primary cell using potentiometer.

Hint:

Crucially, the EMF of the driver cell must be greater than the EMF of the experimental cell; otherwise, a null point will not be obtained. Always connect the positive terminals of both the driver cell and the experimental cell to the same end (the 'zero' end) of the potentiometer wire.

Answer 1

Step 1: Understanding the Concept:
A potentiometer can be used to compare EMFs of two cells and also to find the internal resistance of a cell. The principle is that the potential drop across any portion of the potentiometer wire is directly proportional to the length of that portion, provided the wire is of uniform cross-section and a constant current flows through it. By balancing the EMF (\(\epsilon\)) and the terminal potential difference (V) of the cell against the potential drop on the wire, we can find the internal resistance.
Step 2: Key Formula and Apparatus:
Apparatus Required:
A potentiometer, a primary cell (whose internal resistance is to be determined), a driver cell (battery), a resistance box, a galvanometer, a rheostat, two one-way keys, a jockey, and connecting wires.
Key Formula:
The internal resistance (r) of a cell is given by: \[ r = \left( \frac{\epsilon}{V} - 1 \right) R \] Using the potentiometer principle, \( \epsilon \propto l_1 \) and \( V \propto l_2 \), where \(l_1\) is the balancing length for the EMF (open circuit) and \(l_2\) is the balancing length for the terminal voltage (closed circuit with external resistance R).
The formula becomes: \[ r = \left( \frac{l_1}{l_2} - 1 \right) R \] Step 3: Detailed Procedure:
1. Circuit Setup:
- Set up the primary circuit by connecting the driver cell, key \(K_1\), and rheostat in series with the potentiometer wire.
- Set up the secondary circuit by connecting the positive terminal of the experimental cell to the same end of the potentiometer wire as the positive terminal of the driver cell. Connect its negative terminal to the jockey through a galvanometer.
- Connect a resistance box (R) and a key (\(K_2\)) in parallel with the experimental cell.
2. Taking Measurements:
- Close key \(K_1\) and keep key \(K_2\) open.
- Slide the jockey along the potentiometer wire to find the null point where the galvanometer shows zero deflection. Let this balancing length be \(l_1\). This corresponds to the EMF (\(\epsilon\)) of the cell.
- Now, take out a small resistance (e.g., 10 \(\Omega\)) from the resistance box (R) and close the key \(K_2\). This draws current from the cell.
- Again, find the null point by sliding the jockey. Let this new balancing length be \(l_2\). This corresponds to the terminal potential difference (V) across the cell. Note that \(l_2\) will be less than \(l_1\).
3. Repeating for Accuracy:
- Repeat the experiment for 3-4 different values of R from the resistance box. For each R, find the corresponding \(l_2\) (while \(l_1\) remains constant).
Step 4: Calculation and Final Answer:
For each set of R and \(l_2\), calculate the internal resistance r using the formula: \[ r = R \left( \frac{l_1}{l_2} - 1 \right) \] Calculate the mean of all the values of r.
\[ r_{mean} = \frac{r_1 + r_2 + r_3 + ...}{n} \] The result is stated as: "The internal resistance of the given primary cell is \(r_{mean} \, \Omega\)."