N identical cells, each of emf `epsilon` and internal resistances r are joined in series. Out of these, n cells are wrongly connected, i.e. their terminals are connected in reverse order as required for series connection. If `epsilon_(o)` be the emf of the resulting battery and `r_(o)` be its internal resistance then for n `lt N//2`

Options

A. `epsilon_(0) = (N-n)epsilon,r_(o) = (N-n)r`
B. `epsilon_(0 )=(N-n)epsilon, r_(o) = (N-2n) r`
C. v `epsilon_(o)= (N-2n) epsilon, r_(o) = Nr`
D. `epsilon_(o) = (N-n)epsilon, r_(o) = Nr`

Answer

To solve the problem, we need to determine the effective electromotive force (emf) and internal resistance of a battery made up of N identical cells, where n cells are connected in reverse order.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have N identical cells, each with an emf of \( \epsilon \) and an internal resistance of \( r \).
- Out of these N cells, n cells are connected in reverse order.

2. **Calculating the Effective EMF**:
- The total emf contributed by the correctly connected cells is \( (N - n) \epsilon \) (since N - n cells are connected in the correct direction).
- The total emf contributed by the n incorrectly connected cells is \( -n \epsilon \) (since they are connected in reverse).
- Therefore, the effective emf \( \epsilon_0 \) of the resulting battery can be calculated as:
\[ \epsilon_0 = (N - n) \epsilon - n \epsilon = (N - 2n) \epsilon \]

3. **Calculating the Total Internal Resistance**:
- The internal resistance of each cell is \( r \).
- Since all cells are connected in series, the total internal resistance \( r_0 \) of the battery is simply the sum of the internal resistances of all N cells:
\[ r_0 = N \cdot r \]

4. **Final Expressions**:
- Thus, we have:
\[ \epsilon_0 = (N - 2n) \epsilon \]
\[ r_0 = N \cdot r \]

### Conclusion:
For \( n < \frac{N}{2} \), the effective emf and internal resistance of the battery are given by:
- Effective emf: \( \epsilon_0 = (N - 2n) \epsilon \)
- Internal resistance: \( r_0 = N \cdot r \)