The `n` rows each containinig `m` cells in series are joined parallel. Maximum current is taken from this combination across jan external resistance of `3 Omega` resistance. If the total number of cells used are 24 and internal resistance of each cell is `0.5 Omega` then

Options

A. m = 8, n = 3
B. m = 6, n = 4
C. m = 12, n = 2
D. m = 2, n = 12

Answer

To solve the problem, we need to determine the values of `m` (the number of cells in each row) and `n` (the number of rows) given the total number of cells and the internal resistance of each cell. We know that the total number of cells is 24 and the internal resistance of each cell is 0.5 ohms. The equivalent resistance of the entire combination is given as 3 ohms.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have `n` rows, each containing `m` cells in series.
- The total number of cells is given by the equation:
\[ n \times m = 24 \]

2. **Calculating Internal Resistance of Each Row**:
- Each cell has an internal resistance of \( r = 0.5 \, \Omega \).
- The total internal resistance of one row containing `m` cells in series is:
\[ R_1 = m \times r = m \times 0.5 \]

3. **Finding the Equivalent Resistance of the Parallel Combination**:
- Since there are `n` such rows connected in parallel, the equivalent resistance \( R_{eq} \) of the parallel combination can be expressed as:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_1} + \ldots + \frac{1}{R_1} \quad (n \text{ times}) \]
- This simplifies to:
\[ \frac{1}{R_{eq}} = \frac{n}{R_1} \]
- Therefore, we can express \( R_{eq} \) as:
\[ R_{eq} = \frac{R_1}{n} \]

4. **Substituting for \( R_1 \)**:
- Substituting \( R_1 \) into the equation gives:
\[ R_{eq} = \frac{m \times 0.5}{n} \]

5. **Setting Up the Equation**:
- We know from the problem that \( R_{eq} = 3 \, \Omega \). Thus, we can set up the equation:
\[ \frac{m \times 0.5}{n} = 3 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ m \times 0.5 = 3n \]
- Simplifying further:
\[ m = \frac{3n}{0.5} = 6n \]

7. **Substituting into the Total Cells Equation**:
- We already have \( n \times m = 24 \). Substituting \( m = 6n \) into this equation gives:
\[ n \times (6n) = 24 \]
- This simplifies to:
\[ 6n^2 = 24 \]
- Dividing both sides by 6:
\[ n^2 = 4 \]
- Taking the square root:
\[ n = 2 \]

8. **Finding \( m \)**:
- Now substituting \( n = 2 \) back into the equation for \( m \):
\[ m = 6n = 6 \times 2 = 12 \]

### Final Values:
- \( n = 2 \)
- \( m = 12 \)

### Conclusion:
The values of \( m \) and \( n \) that satisfy the given conditions are \( m = 12 \) and \( n = 2 \).