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 find the current flowing in a circuit with n cells, each having an emf \(E\) and an internal resistance \(r\), where one cell is connected in reverse. Here’s a step-by-step solution:

### Step 1: Determine the total emf of the cells
When n cells are connected in series, the total emf contributed by the cells is given by:
\[ \text{Total EMF} = nE \]
However, since one cell is connected in reverse, it effectively reduces the total emf by \(E\). Therefore, the net emf in the circuit becomes:
\[ \text{Net EMF} = nE - E = (n - 1)E \]

### Step 2: Calculate the total internal resistance
The internal resistance of each cell is \(r\). Since there are n cells in series, the total internal resistance is:
\[ \text{Total Internal Resistance} = nr \]
Since one cell is connected in reverse, it does not change the total internal resistance. Thus, the total internal resistance remains:
\[ \text{Total Internal Resistance} = nr \]

### Step 3: Determine the total resistance in the circuit
The total resistance in the circuit is the sum of the internal resistance of the cells and the external resistance \(R\):
\[ \text{Total Resistance} = nr + R \]

### Step 4: Apply Ohm's Law to find the current
Using Ohm's Law, the current \(I\) flowing in the circuit can be calculated as:
\[ I = \frac{\text{Net EMF}}{\text{Total Resistance}} = \frac{(n - 1)E}{nr + R} \]

### Final Result
Thus, the current flowing in the circuit is:
\[ I = \frac{(n - 1)E}{nr + R} \]