3 identical bulbs are connected in series and these together dissipate a power P . If now the bulbs are connected in parallel, then the power dissipated will be

Options

A. `(P)/(3)`
B. `3P`
C. `9P`
D. `(P)/(9)`

Answer

To solve the problem, we need to analyze the power dissipation of identical bulbs when they are connected in series and then in parallel. Let's break down the solution step by step.

### Step 1: Understand the Series Connection
When the three identical bulbs (resistors) are connected in series, the total resistance \( R_s \) is the sum of the individual resistances. If each bulb has a resistance \( R \), then:
\[ R_s = R + R + R = 3R \]
The power \( P \) dissipated in a series circuit can be expressed using the formula:
\[ P = \frac{V^2}{R_s} = \frac{V^2}{3R} \]

### Step 2: Analyze the Parallel Connection
When the same bulbs are connected in parallel, the equivalent resistance \( R_p \) can be calculated using the formula for resistors in parallel:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_p = \frac{R}{3} \]
Now, using the same voltage \( V \) across the parallel combination, we can calculate the power \( P' \) dissipated:
\[ P' = \frac{V^2}{R_p} = \frac{V^2}{\frac{R}{3}} = 3 \frac{V^2}{R} \]

### Step 3: Relate the Power in Parallel to the Power in Series
From the power dissipation in series, we have:
\[ \frac{V^2}{R} = 3P \]
Thus, substituting this into the power equation for the parallel connection:
\[ P' = 3 \frac{V^2}{R} = 3 \times 3P = 9P \]

### Conclusion
The power dissipated when the bulbs are connected in parallel is:
\[ P' = 9P \]

### Final Answer
The power dissipated when the bulbs are connected in parallel is \( 9P \).

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