A set of `'n'` equal resistor, of value of `'R'` each are connected in series to a battery of emf `'E'` and internal resistance `'R'`. The current drawn is `I`. Now, the `'n'` resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10.1. The value of `'n'` is

Options

A. 20
B. 11
C. 10
D. 9

Explanation

To solve the problem, we need to analyze the two scenarios: when the resistors are connected in series and when they are connected in parallel.

### Step-by-Step Solution:

1. **Series Connection**:
- When `n` resistors of value `R` each are connected in series, the total resistance \( R_s \) is given by:
\[ R_s = nR + R = (n + 1)R \]
- The current \( I \) drawn from the battery with EMF \( E \) and internal resistance \( R \) is given by Ohm's law:
\[ I = \frac{E}{R_s} = \frac{E}{(n + 1)R} \]
- This is our **Equation (1)**.

2. **Parallel Connection**:
- When the same `n` resistors are connected in parallel, the equivalent resistance \( R_p \) is:
\[ R_p = \frac{R}{n} \]
- The total resistance in this case, including the internal resistance of the battery, is:
\[ R_{total} = R_p + R = \frac{R}{n} + R = \left(\frac{1}{n} + 1\right)R = \frac{(n + 1)R}{n} \]
- The current \( I' \) drawn from the battery is:
\[ I' = \frac{E}{R_{total}} = \frac{E}{\frac{(n + 1)R}{n}} = \frac{nE}{(n + 1)R} \]
- This is our **Equation (2)**.

3. **Relating the Currents**:
- According to the problem, when the resistors are connected in parallel, the current drawn becomes \( I' = 10I \).
- Substituting the expressions for \( I \) and \( I' \) from Equations (1) and (2):
\[ \frac{nE}{(n + 1)R} = 10 \cdot \frac{E}{(n + 1)R} \]
- Since \( E \) and \( R \) are common in both sides, we can cancel them out:
\[ n = 10 \]

### Final Answer:
Thus, the value of \( n \) is \( 10 \).

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