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_AB) to the electromotive force (EMF, E) of the cell when it is connected in series with an external resistance (nR) and has an internal resistance (r). Given that n = 9, we can follow these steps:

### Step-by-Step Solution:

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

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

3. **Determine the Potential Drop Across the Internal Resistance**:
The potential drop (V_drop) across the internal resistance (r) can be calculated as:
\[ V_{\text{drop}} = I \cdot r = \left(\frac{E}{10r}\right) \cdot r = \frac{E}{10} \]

4. **Calculate the Terminal Potential Difference (V_AB)**:
The terminal potential difference (V_AB) is given by the EMF minus the potential drop across the internal resistance:
\[ V_{AB} = E - V_{\text{drop}} = E - \frac{E}{10} = \frac{10E}{10} - \frac{E}{10} = \frac{9E}{10} \]

5. **Find the Ratio of Terminal Potential Difference to EMF**:
Now, we can find the ratio of the terminal potential difference (V_AB) to the EMF (E):
\[ \text{Ratio} = \frac{V_{AB}}{E} = \frac{\frac{9E}{10}}{E} = \frac{9}{10} = 0.9 \]

### Final Answer:
The ratio of the terminal potential difference to the EMF when n = 9 is **0.9**.