24 cells, each of emf 1.5V and internal resistance is `2Omega` connected to `12Omega` series external resistance. Then,

Options

A. in order to send the maximum current through a `12Omega` resistor, connect the two rows of 12 cells in series
B. the current is each row is `0.375A`
C. the current in each cell is 0.75A
D. potential difference across `12Omega` resistor is 4.5V

Answer

To solve the problem step by step, we will analyze the given data and apply the relevant formulas.

### Step 1: Identify Given Data
- Number of cells (N) = 24
- EMF of each cell (E) = 1.5 V
- Internal resistance of each cell (r) = 2 Ω
- External resistance (R) = 12 Ω

### Step 2: Determine Configuration of Cells
We need to connect the cells in such a way that we maximize the current through the external resistor. We can assume that the cells are arranged in `m` rows and `n` columns.

From the problem, we know:
- Total number of cells = m × n = 24 (Equation 1)
- For maximum current, the condition is R = n × r/m, where R is the external resistance, r is the internal resistance of each cell, and n is the number of cells in series in each row.

### Step 3: Set Up the Equation for Maximum Current
From the condition for maximum current:
\[ R = \frac{n \cdot r}{m} \]
Substituting the values:
\[ 12 = \frac{n \cdot 2}{m} \]
This simplifies to:
\[ n = 6m \] (Equation 2)

### Step 4: Solve the Equations
Now we have two equations:
1. \( m \cdot n = 24 \) (Equation 1)
2. \( n = 6m \) (Equation 2)

Substituting Equation 2 into Equation 1:
\[ m \cdot (6m) = 24 \]
\[ 6m^2 = 24 \]
\[ m^2 = 4 \]
\[ m = 2 \]

Now substituting back to find n:
\[ n = 6 \cdot 2 = 12 \]

### Step 5: Configuration of Cells
Thus, we have:
- m = 2 (rows)
- n = 12 (columns)

This means we have 2 rows of 12 cells connected in parallel.

### Step 6: Calculate Maximum Current
The maximum current through the external resistance can be calculated using the formula:
\[ I_{max} = \frac{m \cdot E}{R + (n \cdot r)} \]
Substituting the values:
\[ I_{max} = \frac{2 \cdot 1.5}{12 + (12 \cdot 2)} \]
\[ I_{max} = \frac{3}{12 + 24} \]
\[ I_{max} = \frac{3}{36} \]
\[ I_{max} = 0.0833 \, \text{A} \]

### Step 7: Calculate Current in Each Row
Since the rows are in parallel, the current in each row can be calculated as:
\[ I_{row} = \frac{I_{max}}{m} = \frac{0.0833}{2} = 0.04165 \, \text{A} \]

### Step 8: Calculate Potential Difference Across External Resistance
The potential difference across the external resistance can be calculated as:
\[ V = I_{max} \cdot R = 0.0833 \cdot 12 = 1 \, \text{V} \]

### Summary of Results
1. The configuration for maximum current is 2 rows of 12 cells each.
2. The current in each row is approximately 0.04165 A.
3. The potential difference across the 12 Ω resistor is 1 V.