There are n cells, each of emf E and internal resistance r, connected in series with an external resistance R. One of the cells is wrongly connected, so that it sends current in the opposite direction. The current flowing in the circuit is

Options

A. `((n-1)E)/((n+1)r+2)`
B. `((n-1)E)/(nr+R)`
C. `((n-2)E)/(nr+R)`
D. `((n-2)E)/((n-2)r+R)`

Answer

To solve the problem, we need to determine the current flowing in a circuit with n cells, each with an emf (E) and internal resistance (r), where one cell is connected in reverse. Here’s a step-by-step solution:

### Step 1: Understand the configuration of the circuit
We have n cells connected in series. Each cell has an emf of E and an internal resistance of r. One cell is connected in the opposite direction, which affects the total emf of the circuit.

**Hint:** Visualize the circuit with n cells in series, marking the direction of the current from the correctly connected cells and the opposite direction for the wrongly connected cell.

### Step 2: Calculate the total emf of the circuit
If all cells were connected correctly, the total emf would be the sum of the emfs of all n cells, which is \( nE \). However, since one cell is connected in reverse, its contribution to the total emf will be negative. Therefore, the net emf (E_net) can be calculated as follows:

\[ E_{\text{net}} = nE - E = (n - 2)E \]

**Hint:** Remember that the reversed cell subtracts its emf from the total.

### Step 3: Calculate the total resistance of the circuit
The total resistance in the circuit consists of the internal resistances of the n cells and the external resistance R. The total internal resistance from n cells is \( nr \). Thus, the total resistance (R_total) in the circuit is:

\[ R_{\text{total}} = R + nr \]

**Hint:** Consider all resistances in series add up to give the total resistance.

### Step 4: Apply Ohm’s Law to find the current
Using Ohm's Law, the current (I) flowing in the circuit can be calculated using the formula:

\[ I = \frac{E_{\text{net}}}{R_{\text{total}}} \]

Substituting the values we derived:

\[ I = \frac{(n - 2)E}{R + nr} \]

**Hint:** Ensure you substitute the correct expressions for both the net emf and the total resistance.

### Final Answer
The current flowing in the circuit is:

\[ I = \frac{(n - 2)E}{R + nr} \]

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