Calculate Power Consumption of 60 W Lightbulb | Resistance and Voltage Equations

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Homework Help Overview

The discussion revolves around calculating the power consumption of a 60 W lightbulb at different temperatures, specifically when it is cold and when it is hot. The problem involves understanding the relationship between resistance, voltage, and power in an electrical circuit.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the power formula P=V^2/R and the implications of using different resistance values for the cold and hot states of the lightbulb. There is questioning regarding the correct interpretation of resistance units and the need for RMS voltage in AC circuits. Some participants express confusion about determining the resistance at the moment the lightbulb is turned on and how temperature affects this.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the units of measurement and the nature of AC voltage. Some guidance has been offered regarding the calculation of power for different resistance states, but there is still uncertainty about how to approach the initial power consumption when the lightbulb is first turned on.

Contextual Notes

There is a mention of the need to consider the timing of the experiment, with a specific duration noted for measuring the transition from cold to hot states. The original poster's understanding of the term "few" in relation to time is also being explored.

NikkiNik
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Homework Statement



(a)A 60 W, 115 V lightbulb has a resistance of 10.9 W when cold (20.0°C) and 132 W when on (hot). Calculate its power consumption at the instant it is turned on.

(b)Calculate its power consumption after a few moments when it is hot


Homework Equations



P=V^2/R

The Attempt at a Solution



I tried adding and subtracting the resistances then dividing the voltage squared by either values but that is incorrect. I don't know how to determine how much power is used when the switch is turned on. Since I don't know the first answer I can't solve the second part which I assume I would use the Power found in part a and divide the voltage squared by that number
 
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NikkiNik said:
...resistance of 10.9 W when cold (20.0°C) and 132 W when on (hot).
P=V^2/R

I think you mean 10.9 Ω and 132 Ω right? You can just write "ohms" if you don't have the symbol for it, but it is very important to get the units right. Resistance is not the same thing as power, and therefore, they are not measured using the same units!

The equation you posted gives you the power dissipated in a resistor having resistance R and voltage V applied across it. Period.

In this situation, you have two different situations, each with a different R, but the same V (because the resistance of the lightbulb changes with temperature, but the voltage being applied across it is being kept steady). Can you calculate the power for each of these two different situations?
 
For V, I think you will want to use the Vrms for figuring power. This is AC current. If it was DC you could use 115 straight.
 
AC power is always quoted as RMS (e.g. the 230 VAC or 120 VAC) anyway. Furthermore, although we know that the voltage is AC, this is not stated explicitly in the question, suggesting that the OP isn't supposed to worry about it either way.
 
I didn't realize I posted W instead of omega. Yes I know how to calculate individual powers but I don't understand how I could figure out the power consumed the instant it is turned on. How am I supposed to know the resistance? How do I what the temperature is?
 
The lightbulb is "cold" at the instant it is turned on, and "hot" after being on a few moments.
 
NikkiNik said:
...(b)Calculate its power consumption after a few moments when it is hot.

Be sure to properly time your experiment.
One moment = 1.5 minutes.
http://www.unc.edu/~rowlett/units/dictM.html"

I'm searching for the meaning of 'few'.
Be back in three shakes.
 
Last edited by a moderator:
Man I was really overthinking the problem! Thank you for your help
 

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