Two identical cells connected in series send 1.0A current through a `5 Omega` resistor. When they are connected in parallel, they send 0.8 A current through the same resistor. What is the internal resistance of the cell?

Options

A. `0.5 Omega`
B. `1.0 Omega`
C. `1.5 Omega`
D. `2.5 Omega`

Explanation

To find the internal resistance of the cell, we can use the information provided about the current flowing through the resistor when the cells are connected in series and in parallel.

### Step-by-Step Solution:

1. **Understanding the Series Connection:**
- When two identical cells are connected in series, the total EMF (E_total) is the sum of the individual EMFs:
\[ E_{\text{total}} = 2E \]
- The total resistance (R_total) in the circuit is the sum of the internal resistances and the external resistance:
\[ R_{\text{total}} = 2r + 5 \, \Omega \]
- According to Ohm's Law, the current (I) flowing through the circuit can be expressed as:
\[ I = \frac{E_{\text{total}}}{R_{\text{total}}} = \frac{2E}{2r + 5} \]
- Given that the current (I) is 1.0 A when connected in series:
\[ 1 = \frac{2E}{2r + 5} \quad \text{(Equation 1)} \]

2. **Understanding the Parallel Connection:**
- When the cells are connected in parallel, the total EMF remains the same (E), and the total internal resistance (R_total) is halved:
\[ R_{\text{total}} = r + r + 5 = 2r + 5 \]
- The current flowing through the circuit can be expressed as:
\[ I = \frac{E}{R_{\text{total}}} = \frac{E}{2r + 5} \]
- Given that the current (I) is 0.8 A when connected in parallel:
\[ 0.8 = \frac{E}{5} \quad \text{(Equation 2)} \]

3. **Solving the Equations:**
- From Equation 2, we can solve for E:
\[ E = 0.8 \times 5 = 4 \, \text{V} \]
- Substitute E back into Equation 1:
\[ 1 = \frac{2 \times 4}{2r + 5} \]
\[ 1 = \frac{8}{2r + 5} \]
- Cross-multiplying gives:
\[ 2r + 5 = 8 \]
\[ 2r = 8 - 5 = 3 \]
\[ r = \frac{3}{2} = 1.5 \, \Omega \]

4. **Final Result:**
- The internal resistance of each cell is:
\[ r = 1.5 \, \Omega \]