Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then, the current in the external circuit is

Options

A. `(2E)/(4r+R)`
B. `(3E)/(4r+R)`
C. `(3E)/(3r+R)`
D. `(2E)/(3r+R)`

Answer

To solve the problem of finding the current in the external circuit when four cells are connected in series, with one cell connected in reverse, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have four cells, each with an emf (E) and internal resistance (r), connected in series.
- One of the cells is connected in reverse.

2. **Determine the Total EMF**:
- The total emf (E_total) of the series circuit can be calculated considering the orientation of the cells.
- Three cells are connected in the same direction, contributing positively to the total emf, while one cell is connected in the opposite direction, contributing negatively.
- Therefore, the total emf is:
\[ E_{\text{total}} = 3E - E = 2E \]

3. **Calculate the Total Internal Resistance**:
- Since all four cells have the same internal resistance (r) and are in series, the total internal resistance (R_internal) is:
\[ R_{\text{internal}} = r + r + r + r = 4r \]

4. **Set Up the Circuit Equation**:
- According to Kirchhoff's loop rule, the sum of the potential differences in the circuit must equal zero.
- The equation can be set up as follows:
\[ E_{\text{total}} - I \cdot R_{\text{internal}} - I \cdot R = 0 \]
- Substituting the values we have:
\[ 2E - I \cdot (4r + R) = 0 \]

5. **Solve for Current (I)**:
- Rearranging the equation gives us:
\[ I \cdot (4r + R) = 2E \]
- Therefore, the current (I) in the circuit is:
\[ I = \frac{2E}{4r + R} \]

### Final Answer:
The current in the external circuit is:
\[ I = \frac{2E}{4r + R} \]