N cells each of e.m.f. E & internal resistance r are grouped into sets of K cells connected in series. The (N/K) sets ae connected in parallel to a load of resistance R, then,

Options

A. a.Maximum power is delivered to the load is `K=sqrt((NR)/r)`
B. b.Maximum power is delivered to the load if `K=sqrt((r)/(NR))`
C. c.Maximum power delivered to the load is `E^(2)/(4Nr)`
D. d.Maximum power deliverd to the load is `(N(E)^(2))/(4r)`

Answer

To solve the problem, we need to analyze the arrangement of the cells and derive the expressions for the maximum power delivered to the load. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Configuration
We have N cells, each with an electromotive force (e.m.f.) E and internal resistance r. These cells are grouped into sets of K cells connected in series. Therefore, the total e.m.f. of one set is:

\[ E_{\text{set}} = K \cdot E \]

The total internal resistance of one set of K cells in series is:

\[ r_{\text{set}} = K \cdot r \]

### Step 2: Finding the Number of Sets
The total number of sets formed is given by:

\[ \text{Number of sets} = \frac{N}{K} \]

### Step 3: Equivalent Circuit for Sets in Parallel
The sets of K cells are connected in parallel to a load resistance R. The equivalent internal resistance of the sets in parallel can be calculated as follows:

The equivalent resistance \( R_{\text{eq}} \) of the sets in parallel is given by:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{r_{\text{set}}} \cdot \frac{N}{K} = \frac{1}{K \cdot r} \cdot \frac{N}{K} = \frac{N}{K^2 r} \]

Thus, the equivalent internal resistance is:

\[ R_{\text{eq}} = \frac{K^2 r}{N} \]

### Step 4: Total E.M.F. of the Configuration
The total e.m.f. of the configuration (since all sets are identical) remains:

\[ E_{\text{total}} = K \cdot E \]

### Step 5: Applying Maximum Power Transfer Theorem
According to the maximum power transfer theorem, maximum power is delivered to the load when the load resistance \( R \) is equal to the equivalent internal resistance \( R_{\text{eq}} \):

\[ R = R_{\text{eq}} = \frac{K^2 r}{N} \]

### Step 6: Solving for K
To find the value of K that maximizes power transfer, we can rearrange the equation:

\[ K^2 = \frac{R \cdot N}{r} \]

Taking the square root gives:

\[ K = \sqrt{\frac{R \cdot N}{r}} \]

### Step 7: Maximum Power Calculation
The maximum power \( P_{\text{max}} \) delivered to the load can be calculated using:

\[ P_{\text{max}} = \frac{E_{\text{total}}^2}{(R_{\text{eq}} + R)^2} \cdot R \]

Substituting \( R_{\text{eq}} = R \):

\[ P_{\text{max}} = \frac{(K \cdot E)^2}{(2R)^2} \cdot R = \frac{K^2 E^2}{4R} \]

Substituting \( K^2 \) from earlier:

\[ P_{\text{max}} = \frac{R \cdot N}{r} \cdot \frac{E^2}{4R} = \frac{N \cdot E^2}{4r} \]

### Conclusion
The derived expressions show that the value of K for maximum power transfer is:

\[ K = \sqrt{\frac{R \cdot N}{r}} \]

And the maximum power delivered to the load is:

\[ P_{\text{max}} = \frac{N \cdot E^2}{4r} \]