There are N identical cells of emf E and internal resistance r connected in series with each other. But n cells are connected with reverse polarity and `n lt N//2`. Effective emf and internal resistance of this combination are

Options

A. `(n - 2n)e, Nr`
B. `(N- 2n)E, nr`
C. `(N -n) E, Nr`
D. `(N - n) E, nr`

Explanation

To solve the problem, we need to find the effective EMF and internal resistance of a combination of N identical cells connected in series, with n cells connected in reverse polarity. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the configuration of cells
We have N identical cells, each with an EMF of E and an internal resistance of r. Among these, n cells are connected in reverse polarity.

### Step 2: Calculate the effective EMF
The effective EMF (E_eff) of the combination can be calculated by considering the contribution of all cells:
- The total EMF from N cells connected in series would be \( N \times E \).
- However, since n cells are connected in reverse polarity, they effectively subtract from the total EMF. Thus, the contribution of these n cells is \( -n \times E \).

Putting this together, the effective EMF can be expressed as:
\[ E_{\text{eff}} = (N - n)E - nE = (N - 2n)E \]

### Step 3: Calculate the total internal resistance
The internal resistance of cells connected in series adds up. Since all cells have the same internal resistance r, the total internal resistance (R_eff) is given by:
\[ R_{\text{eff}} = N \times r \]

### Final Results
Thus, the effective EMF and internal resistance of the combination are:
- Effective EMF: \( E_{\text{eff}} = (N - 2n)E \)
- Effective Internal Resistance: \( R_{\text{eff}} = N \times r \)