A 100 watt bulb is rated for use in a 120 volt line. During a brownout the actual voltage drops to 100 V. The actual power dissipated in the bulb under these conditions is O 48.2 W O 83.3 W O 120 W. O 100 W. O 69.4 W
A 100 watt bulb is rated for use in a 120 volt line. During a brownout the actual voltage drops to 100 V. The actual power dissipated in the bulb under these conditions is O 48.2 W O 83.3 W O 120 W. O 100 W. O 69.4 W
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![**Problem Statement:**
A 100 watt bulb is rated for use in a 120 volt line. During a brownout, the actual voltage drops to 100 V. The actual power dissipated in the bulb under these conditions is:
- ○ 48.2 W
- ○ 83.3 W
- ○ 120 W
- ○ 100 W
- ○ 69.4 W
**Explanation:**
This problem involves determining the power dissipation of a bulb when the voltage supply drops from its rated voltage. The options given are potential power values under the reduced voltage condition. This is a typical scenario to understand the effects of voltage change on electrical appliances.
For a bulb, the power \( P \) is related to the voltage \( V \) and resistance \( R \) by the formula:
\[ P = \frac{V^2}{R} \]
First, solve to find the resistance \( R \) using the initial conditions:
\[ 100 \, \text{W} = \frac{(120 \, \text{V})^2}{R} \]
Then, recalculate the power using the lowered voltage of 100 V:
\[ P' = \frac{(100 \, \text{V})^2}{R} \]
Substitute the known resistance from the first equation into the second to find the new power dissipation. This approach allows calculation and understanding of power variations during voltage fluctuations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3e6b254a-d25d-479c-b42b-752b8b5bbcbb%2F3f83f609-e689-4210-ac26-6ec6408c68a4%2Flhhv6a_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 100 watt bulb is rated for use in a 120 volt line. During a brownout, the actual voltage drops to 100 V. The actual power dissipated in the bulb under these conditions is:
- ○ 48.2 W
- ○ 83.3 W
- ○ 120 W
- ○ 100 W
- ○ 69.4 W
**Explanation:**
This problem involves determining the power dissipation of a bulb when the voltage supply drops from its rated voltage. The options given are potential power values under the reduced voltage condition. This is a typical scenario to understand the effects of voltage change on electrical appliances.
For a bulb, the power \( P \) is related to the voltage \( V \) and resistance \( R \) by the formula:
\[ P = \frac{V^2}{R} \]
First, solve to find the resistance \( R \) using the initial conditions:
\[ 100 \, \text{W} = \frac{(120 \, \text{V})^2}{R} \]
Then, recalculate the power using the lowered voltage of 100 V:
\[ P' = \frac{(100 \, \text{V})^2}{R} \]
Substitute the known resistance from the first equation into the second to find the new power dissipation. This approach allows calculation and understanding of power variations during voltage fluctuations.
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