In a connection of 10 dry cells each of emf E and internal resistance r in series, two cell is connected opposite. Net emf and net internal resistance will be:

Options

A. `10 E, 10 r `
B. `8 E , 8r `
C. `8 E, 10 r`
D. `8 E, (r )/(10)`

Explanation

To solve the problem of finding the net EMF and net internal resistance of a series connection of 10 dry cells, where two cells are connected in the opposite direction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration of Cells**:
- We have 10 dry cells, each with an EMF of \( E \) and an internal resistance of \( r \).
- Out of these 10 cells, 8 are connected in the same direction, while 2 are connected in the opposite direction.

2. **Calculate the Net EMF**:
- The total EMF contributed by the 8 cells connected in the same direction is \( 8E \).
- The total EMF contributed by the 2 cells connected in the opposite direction is \( -2E \) (since they oppose the direction).
- Therefore, the net EMF (\( E_{net} \)) can be calculated as:
\[ E_{net} = 8E - 2E = 6E \]

3. **Calculate the Net Internal Resistance**:
- The internal resistance of each cell is \( r \).
- Since internal resistances in series add up, the total internal resistance (\( R_{net} \)) for 10 cells is:
\[ R_{net} = 10r \]

4. **Final Results**:
- The net EMF is \( 6E \).
- The net internal resistance is \( 10r \).

### Summary:
- **Net EMF**: \( 6E \)
- **Net Internal Resistance**: \( 10r \)