A cell supplies a current `i_(1)` trhough a resistnace `R_(1)` and a current `i_(2)` through a resistance `R_(2)`. The internal resistance of this cell is

Options

A. `R_(2)-R_(1)`
B. `(i_(1)R_(2)-i_(2)R_(1))/(i_(1)-i_(2))`
C. `(i_(2)R_(2)-i_(1)R_(1))/(i_(1)-i_(2))`
D. `((i_(1)+i_(2))/(i_(1)-i_(2)))sqrt(R_(1)R_(2))`

Answer

To find the internal resistance of the cell, we can start by using the relationships given for the currents \(i_1\) and \(i_2\) through the resistances \(R_1\) and \(R_2\) respectively.

### Step-by-Step Solution:

1. **Understand the circuit configuration**:
- When the cell with EMF \(E\) is connected to resistance \(R_1\), the total resistance in the circuit is \(R_1 + r\) (where \(r\) is the internal resistance of the cell).
- The current flowing through this circuit is given as \(i_1\).

2. **Write the equation for the first circuit**:
\[ i_1 = \frac{E}{R_1 + r} \]
Rearranging gives:
\[ E = i_1(R_1 + r) \]
This is our **Equation (1)**.

3. **Repeat for the second circuit**:
- When the same cell is connected to resistance \(R_2\), the current flowing through this circuit is \(i_2\).
\[ i_2 = \frac{E}{R_2 + r} \]
Rearranging gives:
\[ E = i_2(R_2 + r) \]
This is our **Equation (2)**.

4. **Set the two expressions for \(E\) equal**:
Since both equations represent the same EMF \(E\), we can set them equal to each other:
\[ i_1(R_1 + r) = i_2(R_2 + r) \]

5. **Expand and rearrange**:
Expanding both sides gives:
\[ i_1 R_1 + i_1 r = i_2 R_2 + i_2 r \]
Rearranging terms to isolate \(r\):
\[ i_1 R_1 - i_2 R_2 = i_2 r - i_1 r \]
\[ i_1 R_1 - i_2 R_2 = r(i_2 - i_1) \]

6. **Solve for the internal resistance \(r\)**:
Dividing both sides by \((i_2 - i_1)\):
\[ r = \frac{i_1 R_1 - i_2 R_2}{i_2 - i_1} \]

### Final Answer:
The internal resistance of the cell is given by:
\[ r = \frac{i_1 R_1 - i_2 R_2}{i_2 - i_1} \]