A cell of emf `epsilon` and internal resistance `r` drives a current `i` through an extermal resistance `R`
(i) The cell suppllied `epsilon i` power
(ii) Heat is produced in `R` at the rate `epsilon i`
(iii) Heat is produced in `R` at the rate `epsilon I ((R )/(R+r))`
(iv) Heat is produced in the cell at the rate `epsilon i(r /(R+r))`

Answer

To solve the problem, we need to analyze the power supplied by the cell, the heat produced in the external resistance \( R \), and the heat produced in the internal resistance \( r \) of the cell.

### Step-by-Step Solution:

1. **Power Supplied by the Cell**:
The power supplied by the cell can be expressed as:
\[ P_{\text{supplied}} = \epsilon i \]
where \( \epsilon \) is the EMF of the cell and \( i \) is the current flowing through the circuit.

**Hint**: Remember that the total power supplied by the cell is the product of EMF and the current.

2. **Power Dissipated in the External Resistance \( R \)**:
The heat produced in the external resistance \( R \) is given by Joule's law:
\[ P_R = i^2 R \]
This represents the rate of heat produced in the resistor due to the current flowing through it.

**Hint**: Use the formula \( P = I^2 R \) to find the power dissipated in the resistor.

3. **Power Dissipated in the Internal Resistance \( r \)**:
The power dissipated in the internal resistance of the cell is:
\[ P_r = i^2 r \]
This is the heat produced in the internal resistance due to the current flowing through it.

**Hint**: Similar to the external resistance, apply \( P = I^2 R \) for the internal resistance as well.

4. **Total Power Supplied**:
The total power supplied by the cell is used to overcome both the external and internal resistances:
\[ P_{\text{total}} = P_R + P_r = i^2 R + i^2 r = i^2 (R + r) \]
Thus, we can express the current \( i \) in terms of \( \epsilon \) and the resistances:
\[ i = \frac{\epsilon}{R + r} \]

**Hint**: The total power supplied by the cell equals the sum of the power dissipated in both resistances.

5. **Heat Produced in the External Resistance \( R \)**:
Substituting \( i \) in the expression for \( P_R \):
\[ P_R = i^2 R = \left(\frac{\epsilon}{R + r}\right)^2 R = \frac{\epsilon^2 R}{(R + r)^2} \]

**Hint**: Substitute the expression for current \( i \) into the power formula for the external resistance.

6. **Heat Produced in the Internal Resistance \( r \)**:
Similarly, substituting \( i \) in the expression for \( P_r \):
\[ P_r = i^2 r = \left(\frac{\epsilon}{R + r}\right)^2 r = \frac{\epsilon^2 r}{(R + r)^2} \]

**Hint**: Again, substitute the expression for current \( i \) into the power formula for the internal resistance.

### Conclusion:
From our analysis, we can conclude that:
- The power supplied by the cell is \( \epsilon i \).
- The heat produced in the external resistance \( R \) is \( \frac{\epsilon^2 R}{(R + r)^2} \).
- The heat produced in the internal resistance \( r \) is \( \frac{\epsilon^2 r}{(R + r)^2} \).

Thus, the correct options based on the statements provided in the question are:
- (iii) Heat is produced in \( R \) at the rate \( \epsilon i \frac{R}{R+r} \).
- (iv) Heat is produced in the cell at the rate \( \epsilon i \frac{r}{R+r} \).