`n` identical light bulbs, each designed to draw `P` power from a certain voltage supply , are joined in series across that supply. The total power which they will draw is

Options

A. `nP`
B. `P`
C. `P//n`
D. `P//n^(2)`

Explanation

To solve the problem, we need to determine the total power drawn by `n` identical light bulbs, each designed to draw `P` power when connected to a voltage supply, when they are connected in series.

### Step-by-Step Solution:

1. **Understand the Power of a Single Bulb**:
Each bulb is designed to draw a power `P` when connected to a voltage `V`. The relationship between power, voltage, and resistance is given by the formula:
\[ P = \frac{V^2}{R} \]
where `R` is the resistance of one bulb.

2. **Calculate the Resistance of One Bulb**:
Rearranging the formula gives us the resistance:
\[ R = \frac{V^2}{P} \]

3. **Connect `n` Bulbs in Series**:
When `n` identical bulbs (resistors) are connected in series, the total or equivalent resistance `R_eq` is the sum of the individual resistances:
\[ R_{eq} = n \cdot R \]
Substituting the value of `R` from step 2:
\[ R_{eq} = n \cdot \frac{V^2}{P} \]

4. **Calculate the Total Power in Series**:
The total power `P'` consumed by the series connection of the bulbs can be calculated using the power formula:
\[ P' = \frac{V^2}{R_{eq}} \]
Substituting the expression for `R_eq`:
\[ P' = \frac{V^2}{n \cdot \frac{V^2}{P}} = \frac{P}{n} \]

5. **Conclusion**:
The total power drawn by `n` identical light bulbs connected in series is:
\[ P' = \frac{P}{n} \]

### Final Answer:
The total power which they will draw is \( \frac{P}{n} \).

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