There are a large number of cells available, each marked (6 V, 0.5 `Omega` ) to be used to supply current to a device of resistance `0.75 Omega` requiring 24 A current . How should the cells be arranged , so that maximum power is transmitted to the load using minimum number of cells ?

Options

A. Six rows , each containig four cells
B. Four rows, each containing six cells
C. Four rows, each containing four cells
D. Six rows , each contining six cells .

Answer

To solve the problem of arranging the cells to supply a current of 24 A to a load resistance of 0.75 Ω while maximizing power transfer, we can follow these steps:

### Step 1: Understand the Requirements
We need to supply a current of 24 A to a load resistance of 0.75 Ω. The cells have an EMF of 6 V and an internal resistance of 0.5 Ω each.

### Step 2: Set Up the Circuit Configuration
Assume we connect M cells in series and N such series connections in parallel. The total EMF of the series combination will be \( E_{total} = 6M \) volts, and the total internal resistance of the series combination will be \( R_{internal} = 0.5M \) ohms.

### Step 3: Calculate the Total Resistance
The total resistance in the circuit when the series combinations are connected in parallel can be expressed as:
\[ R_{total} = R_{load} + R_{internal} = 0.75 + \frac{0.5M}{N} \]
Where \( R_{load} = 0.75 \, \Omega \).

### Step 4: Apply Ohm's Law
Using Ohm's Law, we can express the voltage across the load:
\[ V = I \cdot R_{total} = 24 \cdot (0.75 + \frac{0.5M}{N}) \]
This voltage must also equal the total EMF:
\[ 6M = 24 \cdot (0.75 + \frac{0.5M}{N}) \]

### Step 5: Simplify the Equation
Expanding the equation gives:
\[ 6M = 18 + \frac{12M}{N} \]
Rearranging leads to:
\[ 6M - \frac{12M}{N} = 18 \]
Multiplying through by N to eliminate the fraction:
\[ 6MN - 12M = 18N \]
Thus:
\[ 6MN - 18N = 12M \]

### Step 6: Rearranging for M and N
Rearranging gives:
\[ M(6N - 12) = 18N \]
From this, we can express:
\[ M = \frac{18N}{6N - 12} \]

### Step 7: Maximum Power Transfer Condition
For maximum power transfer, the internal resistance should equal the load resistance:
\[ R_{internal} = R_{load} \]
Thus:
\[ \frac{0.5M}{N} = 0.75 \]
This leads to:
\[ 0.5M = 0.75N \implies M = \frac{1.5N}{0.5} \implies M = 3N \]

### Step 8: Solve the Two Equations
Now we have two equations:
1. \( M = \frac{18N}{6N - 12} \)
2. \( M = 3N \)

Substituting \( M = 3N \) into the first equation:
\[ 3N = \frac{18N}{6N - 12} \]
Cross-multiplying gives:
\[ 3N(6N - 12) = 18N \]
Expanding:
\[ 18N^2 - 36N = 18N \]
This simplifies to:
\[ 18N^2 - 54N = 0 \]
Factoring out \( 18N \):
\[ 18N(N - 3) = 0 \]
Thus, \( N = 0 \) or \( N = 3 \). Since \( N = 0 \) is not practical, we take \( N = 3 \).

### Step 9: Calculate M
Using \( N = 3 \) in \( M = 3N \):
\[ M = 3 \times 3 = 9 \]

### Conclusion
Thus, the cells should be arranged with 9 cells in series (M = 9) and 3 parallel lines (N = 3).

### Final Arrangement
The final arrangement is **3 rows, each containing 9 cells**.

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