Calculate percentage change in heat output

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SUMMARY

The discussion focuses on calculating the percentage change in heat output of a 500W heating unit when the voltage drops from 115V to 110V. The resistance of the heater is first determined using the formula R=V^2/P, which is applied with the initial voltage to find the resistance. The new power output is then calculated using the reduced voltage, leading to a final percentage drop in heat output of 8.5%. This method effectively demonstrates the relationship between voltage, resistance, and power output in electrical systems.

PREREQUISITES
  • Understanding of Ohm's Law and power calculations
  • Familiarity with the formula R=V^2/P
  • Basic knowledge of electrical resistance and its impact on power
  • Ability to perform percentage calculations
NEXT STEPS
  • Study the implications of voltage drops on electrical appliances
  • Learn about the relationship between resistance and power in AC circuits
  • Explore advanced power calculation techniques in electrical engineering
  • Investigate the effects of varying voltage on heating elements
USEFUL FOR

Physics students, electrical engineering students, and anyone interested in understanding the effects of voltage changes on heating elements and power output calculations.

physicist93
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Homework Statement


A 500W heating unit is designed to operate on a 115V line. If the line voltage drops to 110V, calculate the percentage drop in heat output.

Homework Equations


R=V^2/P

The Attempt at a Solution


P=V^2/R

Use V=115, then use V=110, use Power P=500...but not sure how

p.s. I am a UK Physics student. I was wondering could you also say whether you are in the UK or US or...
 
Last edited:
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Find the resistance of the heater. Use that value of resistance and 110 volts to find new power output. %=(initial-finial)*100/initial
 
Thanks that's great! I've done it now
 
I got an answer of 8.5%! sounds right...

Btw what do people think about the questions on online-exam-solutions?
 
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?

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