The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected `(220 xx 0.8)` volt sources, then the actual power would be

Options

A. `100 xx 0.8`W
B. `100 xx (0.8)^(2)` W
C. Between `100 xx 0.8` W and `100 xx 0.8` W
D. Between `100 xx (0.8)^(2)` W and `100 xx 0.8`W

Answer

To solve the problem, we need to find the actual power consumed by the electric bulb when it is connected to a voltage source of \( 220 \times 0.8 \) volts. The bulb is rated at 220 volts and 100 watts.

### Step-by-Step Solution:

1. **Identify the rated power and voltage**:
The electric bulb is rated at \( P = 100 \, \text{W} \) and \( V = 220 \, \text{V} \).

2. **Calculate the resistance of the bulb**:
The power formula is given by:
\[ P = \frac{V^2}{R} \]
Rearranging this formula to find the resistance \( R \):
\[ R = \frac{V^2}{P} \]
Substituting the known values:
\[ R = \frac{220^2}{100} \]
Calculating \( 220^2 \):
\[ 220^2 = 48400 \]
Therefore,
\[ R = \frac{48400}{100} = 484 \, \Omega \]

3. **Determine the new voltage**:
The new voltage connected to the bulb is:
\[ V' = 220 \times 0.8 = 176 \, \text{V} \]

4. **Calculate the actual power at the new voltage**:
Using the power formula again with the new voltage:
\[ P' = \frac{(V')^2}{R} \]
Substituting the values:
\[ P' = \frac{(176)^2}{484} \]
Calculating \( 176^2 \):
\[ 176^2 = 30976 \]
Now substituting back into the power formula:
\[ P' = \frac{30976}{484} \]
Performing the division:
\[ P' \approx 64 \, \text{W} \]

Thus, the actual power consumed by the bulb when connected to a \( 220 \times 0.8 \) volt source is approximately **64 watts**.