A cell of emf E and internal resistance r is connected in series with an external resistance nr. Than what will be the ratio of the terminal potential difference to emf, if n=9.

Answer

To solve the problem, we need to find the ratio of the terminal potential difference (V) to the electromotive force (E) of a cell with internal resistance (r) connected in series with an external resistance (nr), where n = 9.

### Step-by-Step Solution:

1. **Identify the Components**:
- The cell has an electromotive force (emf) of E and an internal resistance of r.
- The external resistance is given as nr, where n = 9.

2. **Total Resistance in the Circuit**:
- The total resistance (R_total) in the circuit is the sum of the internal resistance and the external resistance:
\[ R_{\text{total}} = r + nr = r + 9r = 10r \]

3. **Current in the Circuit**:
- According to Ohm's Law, the current (I) flowing through the circuit can be calculated using the formula:
\[ I = \frac{E}{R_{\text{total}}} = \frac{E}{10r} \]

4. **Terminal Potential Difference (V)**:
- The terminal potential difference (V) can be calculated using the formula:
\[ V = I \cdot (nr) = I \cdot (9r) \]
- Substituting the expression for I:
\[ V = \left(\frac{E}{10r}\right) \cdot (9r) = \frac{9E}{10} \]

5. **Ratio of Terminal Potential Difference to EMF**:
- Now, we need to find the ratio \( \frac{V}{E} \):
\[ \frac{V}{E} = \frac{\frac{9E}{10}}{E} = \frac{9}{10} \]

6. **Final Result**:
- Therefore, the ratio of the terminal potential difference to the emf is:
\[ \frac{V}{E} = \frac{9}{10} \]