To solve the problem, we need to find the values of \( m \) (the number of cells in series in each row) and \( n \) (the number of parallel rows) that will maximize the current through a resistance of \( 3 \, \Omega \). We are given that the total number of cells is \( 24 \) and the internal resistance of each cell is \( 0.5 \, \Omega \).
### Step 1: Understanding the Configuration
We have \( n \) parallel rows of \( m \) cells in series. The total number of cells is given by:
\[
n \cdot m = 24
\]
### Step 2: Finding the Equivalent Resistance
The equivalent resistance \( R_{eq} \) of the configuration can be expressed as:
\[
R_{eq} = R + r_{eq}
\]
where \( R \) is the external resistance (given as \( 3 \, \Omega \)) and \( r_{eq} \) is the internal resistance of the cells.
The internal resistance for \( m \) cells in series is:
\[
r_{series} = m \cdot r = m \cdot 0.5
\]
The equivalent resistance for \( n \) rows in parallel is:
\[
r_{eq} = \frac{r_{series}}{n} = \frac{m \cdot 0.5}{n}
\]
### Step 3: Total Resistance
Thus, the total resistance \( R_{total} \) is:
\[
R_{total} = R + r_{eq} = 3 + \frac{m \cdot 0.5}{n}
\]
### Step 4: Current Maximization
To maximize the current \( I \), we can use Ohm's law:
\[
I = \frac{V}{R_{total}}
\]
where \( V \) is the total voltage of the cells. Assuming each cell has a voltage \( V_c \), the total voltage \( V \) is given by:
\[
V = m \cdot n \cdot V_c = 24 \cdot V_c
\]
Thus, we can express the current as:
\[
I = \frac{24 \cdot V_c}{3 + \frac{m \cdot 0.5}{n}}
\]
### Step 5: Substitute \( n \) in Terms of \( m \)
From \( n \cdot m = 24 \), we can express \( n \) as:
\[
n = \frac{24}{m}
\]
Substituting this into the equation for total resistance gives:
\[
R_{total} = 3 + \frac{m \cdot 0.5}{\frac{24}{m}} = 3 + \frac{m^2 \cdot 0.5}{24}
\]
### Step 6: Finding Maximum Current
We can differentiate the current \( I \) with respect to \( m \) and set the derivative to zero to find the maximum current. However, for simplicity, we can analyze the values of \( m \) and \( n \) directly.
### Step 7: Testing Values
Let’s test integer values of \( m \) that satisfy \( n \cdot m = 24 \):
- If \( m = 2 \), then \( n = 12 \)
- If \( m = 3 \), then \( n = 8 \)
- If \( m = 4 \), then \( n = 6 \)
- If \( m = 6 \), then \( n = 4 \)
- If \( m = 8 \), then \( n = 3 \)
- If \( m = 12 \), then \( n = 2 \)
### Step 8: Calculate for Maximum Current
We can calculate the total resistance for these combinations and find the maximum current. After testing, we find that:
For \( m = 12 \) and \( n = 2 \):
\[
R_{total} = 3 + \frac{12 \cdot 0.5}{2} = 3 + 3 = 6 \, \Omega
\]
Thus, the maximum current occurs at:
\[
I = \frac{24 \cdot V_c}{6}
\]
### Conclusion
After testing various combinations, we find that the best configuration for maximum current is:
- \( m = 12 \)
- \( n = 2 \)
### Final Answer
Thus, the values of \( m \) and \( n \) are:
- \( m = 12 \)
- \( n = 2 \)
