Correct diagram for the determination of internal resistance of a primary cell by potentiometer

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Answer

To determine the internal resistance of a primary cell using a potentiometer, we can follow these steps to set up the correct circuit diagram and understand its functioning:

### Step-by-Step Solution

1. **Understand the Components**:
- Identify the components needed: a primary cell (whose internal resistance we want to measure), a potentiometer, a galvanometer, a known resistor (external resistance), and a battery.

2. **Connect the Circuit**:
- Connect the positive terminal of the primary cell to one end of the potentiometer wire.
- Connect the negative terminal of the primary cell to the negative terminal of the battery.
- Connect the positive terminal of the battery to the other end of the potentiometer wire.

3. **Insert the Galvanometer**:
- Connect the galvanometer in parallel with the potentiometer wire. This will help in detecting the null point where no current flows through the galvanometer.

4. **Add the External Resistance**:
- Include a known external resistance in series with the primary cell. This will allow us to measure the current flowing through the circuit.

5. **Adjust for Null Point**:
- Adjust the position of the sliding contact on the potentiometer wire until the galvanometer shows zero deflection (null point). This indicates that the potential difference across the galvanometer is zero.

6. **Measure the Voltage**:
- At the null point, measure the length of the potentiometer wire that corresponds to the potential difference across the known resistor. Use the formula \( V = k \cdot L \), where \( V \) is the voltage, \( k \) is the potential gradient of the potentiometer, and \( L \) is the length of the wire.

7. **Calculate Internal Resistance**:
- Use Ohm's law and the known values to calculate the internal resistance \( r \) of the primary cell using the formula:
\[ r = \frac{E - V}{I} \]
where \( E \) is the electromotive force (emf) of the cell, \( V \) is the voltage across the external resistor, and \( I \) is the current through the circuit.