n identical cells, each of emf `epsilon` and internal resistance r, are joined in series to form a closed cirucit. One cell A is joined with reversed polarity. The potential difference across any one cell is

Options

A. zero
B. `epsilon`
C. `2(epsilon)/(n)`
D. `(n -1)/(n)epsilon`

Explanation

To solve the problem step by step, we need to analyze the circuit formed by the n identical cells, each with an emf (ε) and internal resistance (r), where one cell is connected with reversed polarity.

### Step 1: Determine the total emf in the circuit
When n identical cells are connected in series, the total emf (E_net) can be calculated by considering the effect of the reversed cell. If one cell is connected in reverse, it subtracts its emf from the total.

For n cells, if one cell is reversed, the total emf can be expressed as:
\[ E_{net} = (n-1) \epsilon - \epsilon = (n-2) \epsilon \]

### Step 2: Calculate the total internal resistance
The total internal resistance (R_net) of n cells connected in series is simply the sum of the internal resistances of each cell:
\[ R_{net} = n \cdot r \]

### Step 3: Find the current in the circuit
Using Ohm's law, the current (I) flowing through the circuit can be calculated as:
\[ I = \frac{E_{net}}{R_{net}} = \frac{(n-2) \epsilon}{n \cdot r} \]

### Step 4: Calculate the potential difference across one cell
The potential difference (V) across one cell can be calculated using the formula:
\[ V = \epsilon - I \cdot r \]
Substituting the value of I from Step 3:
\[ V = \epsilon - \left(\frac{(n-2) \epsilon}{n \cdot r}\right) \cdot r \]

### Step 5: Simplify the expression for V
Now, simplifying the expression:
\[ V = \epsilon - \frac{(n-2) \epsilon}{n} \]
\[ V = \epsilon \left(1 - \frac{(n-2)}{n}\right) \]
\[ V = \epsilon \left(\frac{n - (n-2)}{n}\right) \]
\[ V = \epsilon \left(\frac{2}{n}\right) \]

### Final Result
Thus, the potential difference across one cell is:
\[ V = \frac{2\epsilon}{n} \]