The emf of a cell is `epsilon` and its internal resistance is r. its termnals are cannected to a resistance R. The potential difference between the terminals is `1.6V for R = 4 Omega, and 1.8 V for R = 9 Omega. Then,

Options

A. `epsilon =1 V, r = 1 Omega`
B. `epsilon =2V, r = 1 Omega`
C. `epsilon =2 V, r =2Omega`
D. `epsilon =2.5 V, r = 0.5 Omega`

Explanation

To solve the problem, we will use the concepts of electromotive force (emf), internal resistance, and Ohm's law. Let's break down the solution step by step.

### Step 1: Write the equations for the two cases
We know that the potential difference (V) across the external resistor (R) can be expressed as:
\[ V = \epsilon - I \cdot r \]
Where:
- \( \epsilon \) is the emf of the cell,
- \( I \) is the current flowing through the circuit,
- \( r \) is the internal resistance of the cell.

The current \( I \) can be calculated using Ohm's law:
\[ I = \frac{V}{R} \]

### Step 2: Set up the equations for the two given scenarios
For the first case where \( R = 4 \, \Omega \) and \( V = 1.6 \, V \):
\[ 1.6 = \epsilon - I \cdot r \]
Substituting \( I \):
\[ 1.6 = \epsilon - \left(\frac{1.6}{4}\right) r \]
This simplifies to:
\[ 1.6 = \epsilon - 0.4r \]
Rearranging gives us:
\[ \epsilon = 1.6 + 0.4r \quad \text{(Equation 1)} \]

For the second case where \( R = 9 \, \Omega \) and \( V = 1.8 \, V \):
\[ 1.8 = \epsilon - I \cdot r \]
Substituting \( I \):
\[ 1.8 = \epsilon - \left(\frac{1.8}{9}\right) r \]
This simplifies to:
\[ 1.8 = \epsilon - 0.2r \]
Rearranging gives us:
\[ \epsilon = 1.8 + 0.2r \quad \text{(Equation 2)} \]

### Step 3: Equate the two equations
Since both expressions equal \( \epsilon \), we can set them equal to each other:
\[ 1.6 + 0.4r = 1.8 + 0.2r \]

### Step 4: Solve for \( r \)
Rearranging the equation:
\[ 0.4r - 0.2r = 1.8 - 1.6 \]
This simplifies to:
\[ 0.2r = 0.2 \]
Dividing both sides by 0.2 gives:
\[ r = 1 \, \Omega \]

### Step 5: Substitute \( r \) back to find \( \epsilon \)
Now we can substitute \( r = 1 \, \Omega \) back into either Equation 1 or Equation 2 to find \( \epsilon \). Using Equation 1:
\[ \epsilon = 1.6 + 0.4(1) \]
This simplifies to:
\[ \epsilon = 1.6 + 0.4 = 2.0 \, V \]

### Conclusion
Thus, the values we have found are:
- The emf of the cell \( \epsilon = 2.0 \, V \)
- The internal resistance \( r = 1 \, \Omega \)