Equation for the power a light bulb would radiate

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Homework Help Overview

The discussion revolves around the calculation of current and power for a 150-Watt light bulb connected to a 2.0 V battery, with a focus on understanding the underlying electrical principles and equations involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the current and power using Ohm's Law and the power formula, while expressing uncertainty about the appropriate equations for power dissipation in this context.
  • Some participants question the assumptions made regarding the light bulb's behavior when connected to a low voltage battery and the implications of resistance changes when the bulb is cold versus hot.
  • Others suggest considering alternative formulas for power based on the relationships between voltage, current, and resistance.

Discussion Status

Participants are engaging in a productive discussion, with some providing clarifications and alternative perspectives on the calculations. There is an acknowledgment of the oversimplified nature of the problem, and the original poster continues to seek guidance on determining the power radiated by the bulb.

Contextual Notes

There is a mention of the potential for confusion due to differing variable conventions in physics texts, and participants note the limitations of the problem setup, particularly regarding the light bulb's operational characteristics at low voltage.

lampshader
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Homework Statement



An ordinary 150-Watt light bulb has a resistance of about 96 ohms.
Suppose you were to connect a 2.0 V battery to such a light bulb. How much current would flow? Determine how much power the bulb would radiate? (probably too little for a person to detect). Use you own calculations for this question. The lab can be used to test your results by comparing a similar model.


Homework Equations



I = E (base eq) / R (base eq);
P (base w) = I^2*R.

I don't know the equation to find the power a light bulb would radiate
in a given electrical circuit.


The Attempt at a Solution


If I hooked a 150 W bulb (about 96 Ohms) to a 2V Battery then:

E (base eq) = 2V
R (base eq) = 96 Ohms.

I = E (base eq) / R (base eq) = 2V / 96 Ohms =~ 0.02080A

0.02080 A is the current that a 150 Watt Light Bulb would draw from a 2V Battery.

To find the power that would dissipate:

P (base w) = I^2*R = (0.02080A)^2 * 150 Ohms =~ 0.06514;

is the amount of power dissipated
 
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Your calculation is correct. Curious - I haven't seen E used for potential since the 1960's - everyone seems to be using V now. If you put I = E/R into P=IE you get P = E²/R, which is the most efficient formula for this question.

You are quite right to be puzzled by this question, which is much oversimplified. The bulb will not light up so all the power will be converted to heating and only a part of that will be radiated in the infrared - the rest will conduct away into the wires and fixture. Also, in the lab it will be important to know that the resistance of a light bulb is much lower when it is cold than when it is hot.
 
Thanks Delphi,

The E is used for Potential in, "College Physics" by Giambattista. I'm learning out of his text therefore using his assigned varaibles. And yes, I agree, he should use the more recent varaible names.

Anyway, I do appreciate your reply, but I am still wondering how to determine the power the bulb would radiate, and what equation I should use..


Again, thanks for your reply! :)
 
I'm sure your answer is the one expected.
The true answer is impossible to find.
 

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