Two electric bulbs, each designed to operate with a power of 500W in 220V line, are in series with a `110 V` line. What will be the power generated by each bulb?

Options

A. `31.25 W`
B. `21.25 W`
C. `11.25 W`
D. `9.25W`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the resistance of each bulb
The power \( P \) of each bulb is given as 500 W and the voltage \( V \) is 220 V. We can use the formula for power:

\[ P = \frac{V^2}{R} \]

Rearranging this formula to find the resistance \( R \):

\[ R = \frac{V^2}{P} \]

Substituting the values:

\[ R = \frac{220^2}{500} \]

Calculating \( R \):

\[ R = \frac{48400}{500} = 96.8 \, \Omega \]

### Step 2: Calculate the total resistance in series
Since the two bulbs are connected in series, the total resistance \( R_{net} \) is the sum of the individual resistances:

\[ R_{net} = R + R = 96.8 + 96.8 = 193.6 \, \Omega \]

### Step 3: Calculate the power generated by both bulbs in the 110 V line
Now, we need to find the power generated by the bulbs when connected to a 110 V line. We can use the power formula again:

\[ P' = \frac{V'^2}{R_{net}} \]

Where \( V' = 110 \, V \). Substituting the values:

\[ P' = \frac{110^2}{193.6} \]

Calculating \( P' \):

\[ P' = \frac{12100}{193.6} \approx 62.5 \, W \]

### Step 4: Calculate the power generated by each bulb
Since the total power \( P' \) is shared equally between the two bulbs, the power generated by each bulb \( P_{each} \) is:

\[ P_{each} = \frac{P'}{2} = \frac{62.5}{2} = 31.25 \, W \]

### Final Answer
The power generated by each bulb is **31.25 W**.

---