There are two electric bulbs of `40 W` and `100 W`. Which one will be brighter when first connected in series and then in parallel ?

Options

A. 40 W in series and 100 W in parallel.
B. 100 W in series and 40 W in parallel
C. 40 W both in series and parallel will be uniform
D. 100W both in series and parallel will be uniform

Answer

To determine which bulb will be brighter when connected in series and in parallel, we can follow these steps:

### Step 1: Understand the Power Ratings
We have two electric bulbs with power ratings:
- Bulb 1: \( P_1 = 40 \, \text{W} \)
- Bulb 2: \( P_2 = 100 \, \text{W} \)

We will assume that both bulbs are rated for the same voltage \( V \).

### Step 2: Calculate the Resistance of Each Bulb
Using the formula for power:
\[ P = \frac{V^2}{R} \]
We can rearrange this to find the resistance \( R \):
\[ R = \frac{V^2}{P} \]
Thus, the resistances for the two bulbs are:
- Resistance of Bulb 1:
\[ R_1 = \frac{V^2}{40} \]
- Resistance of Bulb 2:
\[ R_2 = \frac{V^2}{100} \]

### Step 3: Analyze the Series Connection
In a series connection, the same current \( I \) flows through both bulbs. The power consumed by each bulb can be expressed as:
- Power for Bulb 1:
\[ P_1 = I^2 R_1 = I^2 \left(\frac{V^2}{40}\right) \]
- Power for Bulb 2:
\[ P_2 = I^2 R_2 = I^2 \left(\frac{V^2}{100}\right) \]

### Step 4: Compare the Powers in Series
Since \( I^2 \) and \( V^2 \) are the same for both bulbs, we can compare the powers:
\[ P_1 = \frac{I^2 V^2}{40}, \quad P_2 = \frac{I^2 V^2}{100} \]
Since \( \frac{1}{40} > \frac{1}{100} \), it follows that:
\[ P_1 > P_2 \]
Thus, **Bulb 1 (40 W)** is brighter when connected in series.

### Step 5: Analyze the Parallel Connection
In a parallel connection, the voltage across both bulbs is the same. The power consumed by each bulb can be expressed as:
- Power for Bulb 1:
\[ P_1 = \frac{V^2}{R_1} = \frac{V^2}{\frac{V^2}{40}} = 40 \]
- Power for Bulb 2:
\[ P_2 = \frac{V^2}{R_2} = \frac{V^2}{\frac{V^2}{100}} = 100 \]

### Step 6: Compare the Powers in Parallel
Since the voltage is the same for both bulbs, we can directly compare:
\[ P_1 = 40, \quad P_2 = 100 \]
Thus, **Bulb 2 (100 W)** is brighter when connected in parallel.

### Conclusion
- In series, **Bulb 1 (40 W)** is brighter.
- In parallel, **Bulb 2 (100 W)** is brighter.

### Final Answer
- **Bulb 1 is brighter in series.**
- **Bulb 2 is brighter in parallel.**