Solving a 1.50E2-W Heater Problem

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SUMMARY

The discussion centers on calculating the change in power for a 1.50E2-W heater operating at an EMF of 120V when the potential difference changes by 2.6%. The relationship between voltage (E), current (I), resistance (R), and power (P) is established using the formulas E=I*R and P=E*I. The resistance remains constant, allowing for the determination of how power varies with voltage changes through differential analysis.

PREREQUISITES
  • Understanding of electrical power formulas (P=E*I)
  • Basic knowledge of Ohm's Law (E=I*R)
  • Familiarity with differential calculus
  • Concept of fixed resistance in electrical circuits
NEXT STEPS
  • Study the application of differential calculus in physics problems
  • Learn about the implications of Ohm's Law in circuit design
  • Explore power calculations in AC circuits
  • Investigate the effects of resistance changes on power output
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Students studying electrical engineering, physics enthusiasts, and anyone interested in understanding the relationship between voltage and power in electrical systems.

cseet
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Hi all,

I really need somebody to explain to me how to work the following problem...

Q:
A 1.50E2 - W heater is designed to operate with an EMF of 120V. If the potential difference across the heater changes by a small amount such as 2.6%, approximately by what factor does the power change? (Hint: approximately the changes with differentials) (unit %).

can somebody kindly explain to to go about it?? thanks
cseet
 
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Ok, how about trying this.

E=I*R
P=E*I

E is voltage or potential
I is current
R is resistance
P is power

The resistance should be fixed. Solve the latter eq to establish the relationship between E and P.

Cliff
 

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