To solve the problem of arranging the cells to supply current to a device with a resistance of \(0.75 \, \Omega\) while requiring \(24 \, A\) of current, we need to consider the specifications of the cells and the principles of maximum power transfer.
### Step-by-Step Solution:
1. **Identify the Parameters**:
- Each cell has an EMF of \(6 \, V\) and an internal resistance of \(0.5 \, \Omega\).
- The load resistance \(R_L = 0.75 \, \Omega\).
- The required current \(I = 24 \, A\).
2. **Calculate the Total Voltage Required**:
- According to Ohm's Law, the voltage across the load can be calculated as:
\[
V_L = I \times R_L = 24 \, A \times 0.75 \, \Omega = 18 \, V
\]
3. **Determine the Number of Cells Needed in Series**:
- The total EMF from \(n\) cells in series is:
\[
V_{total} = n \times 6 \, V
\]
- To find \(n\), set \(V_{total}\) equal to \(V_L\):
\[
n \times 6 = 18 \implies n = \frac{18}{6} = 3
\]
4. **Calculate the Total Internal Resistance**:
- The total internal resistance \(R_{internal}\) of \(n\) cells in series is:
\[
R_{internal} = n \times 0.5 \, \Omega = 3 \times 0.5 = 1.5 \, \Omega
\]
5. **Determine the Total Resistance for Maximum Power Transfer**:
- For maximum power transfer, the total resistance \(R_{total}\) should equal the load resistance \(R_L\):
\[
R_{total} = R_{internal} + R_L
\]
- Rearranging gives:
\[
R_{internal} = R_L \implies 1.5 \, \Omega = 0.75 \, \Omega
\]
- This indicates that we need to arrange cells in parallel to reduce the internal resistance.
6. **Calculate the Number of Parallel Rows**:
- Let \(m\) be the number of parallel rows. The total internal resistance of \(m\) rows of \(n\) cells in series is given by:
\[
R_{total} = \frac{R_{internal}}{m} = \frac{1.5}{m}
\]
- Setting this equal to \(R_L\):
\[
\frac{1.5}{m} = 0.75 \implies m = \frac{1.5}{0.75} = 2
\]
7. **Final Arrangement**:
- We need \(n = 3\) cells in series and \(m = 2\) rows in parallel.
- Therefore, the total number of cells used is:
\[
\text{Total cells} = n \times m = 3 \times 2 = 6
\]
### Conclusion:
To achieve maximum power transfer to the load while using the minimum number of cells, arrange the cells in **2 parallel rows of 3 cells each**.
