Small drop in voltage change in power?

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SUMMARY

The discussion focuses on calculating the voltage drop required for a 60W light bulb to operate like a 50W bulb during a brownout. The relevant equation is P=V^2/R, where resistance remains constant. The user proposes using P=(V-h)^2/R to represent power under a small voltage drop, but struggles to eliminate the variable h. A suggested approach is to use P_2 = V_2^2/R, where P_2 equals 5/6 of P_1, to find the ratio of V2 to V1.

PREREQUISITES
  • Understanding of electrical power equations, specifically P=V^2/R
  • Basic knowledge of voltage, current, and resistance
  • Familiarity with the concept of brownouts and their effects on electrical devices
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Research the implications of voltage drops on electrical appliances
  • Learn about the effects of brownouts on lighting systems
  • Explore advanced electrical power calculations involving varying resistance
  • Study the relationship between power, voltage, and resistance in AC circuits
USEFUL FOR

Electrical engineers, physics students, and anyone interested in understanding the effects of voltage changes on electrical devices during brownouts.

boderam
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In a brownout the power company's voltage drops. Assuming the voltage drop is small, how would I figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb? The relevant equation I am guessing would be P=V^2/R. I also know the resistance doesn't change. I tried P=(V-h)^2/R for small h, representing the power in a small voltage drop. thus P is approximately V-2Vh/R but I don't think this helps me any since I haven't eliminated h. What do you think?
 
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boderam said:
In a brownout the power company's voltage drops. Assuming the voltage drop is small, how would I figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb? The relevant equation I am guessing would be P=V^2/R. I also know the resistance doesn't change. I tried P=(V-h)^2/R for small h, representing the power in a small voltage drop. thus P is approximately V-2Vh/R but I don't think this helps me any since I haven't eliminated h. What do you think?
Your idea is right. Use P_2 = V_2^2/R where P_2 = 5P_1/6 =\frac{5}{6}V_1^2/R and find the ratio of V2 to V1.

AM
 

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