For a cell, the terminal potential difference is `2.2 V`, when circuit is open and reduces to `1.8 V `. When cell is connected to a resistance `R=5Omega`, the internal resistance of cell `(R)` is

Options

A. `10/(9Omega)`
B. `9/(10Omega)`
C. `11/(9Omega)`
D. `5/(9Omega)`

Answer

To find the internal resistance of the cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Open circuit terminal potential difference (V_open) = 2.2 V
- Terminal potential difference when connected to the resistance (V_load) = 1.8 V
- Resistance (R) = 5 Ω

2. **Calculate the Current (I)**:
- When the cell is connected to the resistance, we can use Ohm's law to find the current flowing through the circuit.
- According to Ohm's law:
\[ V = I \times R \]
- Here, V is the potential difference across the resistor, which is 1.8 V. Thus:
\[ 1.8 = I \times 5 \]
- Rearranging gives:
\[ I = \frac{1.8}{5} = 0.36 \text{ A} \]

3. **Calculate the Potential Drop Across Internal Resistance (r)**:
- The potential drop across the internal resistance can be calculated as:
\[ V_{internal} = V_{open} - V_{load} \]
- Substituting the values:
\[ V_{internal} = 2.2 - 1.8 = 0.4 \text{ V} \]

4. **Use Ohm's Law to Find Internal Resistance (r)**:
- The potential drop across the internal resistance is given by:
\[ V_{internal} = I \times r \]
- Substituting the known values:
\[ 0.4 = 0.36 \times r \]
- Rearranging gives:
\[ r = \frac{0.4}{0.36} \approx 1.11 \text{ Ω} \]

5. **Conclusion**:
- The internal resistance of the cell is approximately **1.11 Ω**.