A series parallel combination battery consisting of a large number `N = 300` of identical cells, each will an inernal resistance `r = 0.3 Omega`, is loaded with an external resistances `R = 10 Omega`. Find the number `n` of parallel groups consisting of an equal number of cells connected in series, at which the external resistance generates the highest thermal power.

Options

A. A) 2
B. B) 3
C. C) 4
D. D) 6

Answer

To solve the problem, we need to determine the number of parallel groups \( n \) that maximizes the thermal power generated in the external resistance \( R \) when connected to a series-parallel combination of identical cells.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have \( N = 300 \) identical cells, each with an internal resistance \( r = 0.3 \, \Omega \).
- The external resistance is \( R = 10 \, \Omega \).
- Let \( x \) be the number of cells in series in each parallel group. Therefore, the total number of parallel groups is \( n \).

2. **Relating Cells in Series and Parallel**:
- Since the total number of cells is \( N \), we can express this relationship as:
\[ n \cdot x = N \quad \text{or} \quad x = \frac{N}{n} \]

3. **Finding the Current**:
- For one loop consisting of \( x \) cells in series and the external resistance \( R \), using Kirchhoff's loop rule, we can write:
\[ I \cdot R + I \cdot (x \cdot r) = x \cdot \epsilon \]
where \( \epsilon \) is the emf of each cell. Rearranging gives:
\[ I \left(R + x \cdot r\right) = x \cdot \epsilon \]
Thus, the current \( I \) can be expressed as:
\[ I = \frac{x \cdot \epsilon}{R + x \cdot r} \]

4. **Power in the External Resistance**:
- The power \( P \) dissipated in the external resistance \( R \) is given by:
\[ P = I^2 \cdot R \]
- Substituting for \( I \):
\[ P = \left(\frac{x \cdot \epsilon}{R + x \cdot r}\right)^2 \cdot R \]

5. **Substituting for \( x \)**:
- Substitute \( x = \frac{N}{n} \):
\[ P = \left(\frac{\frac{N}{n} \cdot \epsilon}{R + \frac{N}{n} \cdot r}\right)^2 \cdot R \]
- Simplifying gives:
\[ P = \left(\frac{N \cdot \epsilon}{n \cdot R + N \cdot r}\right)^2 \cdot R \]

6. **Maximizing Power**:
- To find the value of \( n \) that maximizes \( P \), we can take the derivative \( \frac{dP}{dn} \) and set it to zero. This is a more complex calculus step, but we can use the result derived from the theory of maximum power transfer:
\[ n = \sqrt{\frac{N \cdot R}{r}} \]

7. **Calculating \( n \)**:
- Plugging in the values:
\[ n = \sqrt{\frac{300 \cdot 10}{0.3}} = \sqrt{10000} = 100 \]

8. **Final Result**:
- The number \( n \) of parallel groups that maximizes the thermal power in the external resistance is:
\[ n = 10 \]