Understanding the Voltage Across a Broken Lamp

Click For Summary
SUMMARY

The voltage across a broken lamp in a circuit can be measured as 12V due to the assumption of infinite resistance for the broken lamp. In this scenario, the intact lamp has a significantly lower resistance, causing the majority of the voltage to appear across the broken lamp. The first voltmeter reads 0V because there is no current flowing through the broken lamp, while the second voltmeter measures the full supply voltage of 12V across the broken lamp. This understanding is crucial for analyzing voltage distribution in series circuits.

PREREQUISITES
  • Basic understanding of electrical circuits
  • Knowledge of Ohm's Law
  • Familiarity with voltage measurement techniques
  • Concept of resistance in series circuits
NEXT STEPS
  • Study the principles of series and parallel circuits
  • Learn about Ohm's Law and its applications
  • Explore the concept of infinite resistance in circuit analysis
  • Investigate voltage division in electrical circuits
USEFUL FOR

Students studying electrical engineering, educators teaching circuit analysis, and hobbyists interested in understanding voltage behavior in electrical systems.

Icy98
Messages
17
Reaction score
1

Homework Statement


Why is the voltage across the broken lamp 12V?

Homework Equations


Can I assume the broken lamp is of infinite resistance?

The Attempt at a Solution


I understand the reading for the first voltmeter is 0V. But can anyone explain to me why the reading of the second voltmeter is 12V?[/B]
 
Physics news on Phys.org
That would be a good way to look at it yes. How would the voltages then divide over the two light bulbs if there was a very small current present and the other intact light bulb has a much smaller resistance?
 
  • Like
Likes Icy98
Most of the voltage will be across the broken one?
 
Correct.
 
Thank you![emoji3]
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
14
Views
701
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
Replies
7
Views
4K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 8 ·
Replies
8
Views
5K