Equation for the power a light bulb would radiate

In summary, connecting a 2.0 V battery to a 150-Watt light bulb with a resistance of 96 ohms would result in a current flow of approximately 0.02080A. To determine the power dissipated, the formula P=I^2*R can be used, resulting in 0.06514 watts. However, this calculation oversimplifies the situation and does not account for factors such as the temperature of the bulb and the conduction of heat. The true answer is impossible to determine.
  • #1
lampshader
39
0

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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  • #2
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.
 
  • #3
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! :)
 
  • #4
I'm sure your answer is the one expected.
The true answer is impossible to find.
 

Related to Equation for the power a light bulb would radiate

What is the equation for the power a light bulb would radiate?

The equation for the power a light bulb would radiate is P = I x V, where P is the power in watts, I is the current in amperes, and V is the voltage in volts.

How do you calculate the power of a light bulb?

To calculate the power of a light bulb, you can use the equation P = I x V, where I is the current in amperes and V is the voltage in volts. Alternatively, you can use a wattmeter to directly measure the power of the light bulb.

What factors affect the power of a light bulb?

The power of a light bulb is affected by the voltage and current passing through it. It is also affected by the type of bulb, its efficiency, and the temperature of the surrounding environment.

Why is it important to know the power of a light bulb?

Knowing the power of a light bulb is important for several reasons. It can help you determine the energy consumption and cost of using the bulb, as well as its brightness and overall performance. It is also important for safety reasons, as knowing the power can help prevent overloading circuits or causing fires.

Can the power of a light bulb be changed?

Yes, the power of a light bulb can be changed by adjusting the voltage and/or current passing through it. However, this should only be done within the recommended range for the specific bulb to avoid damaging it or creating safety hazards.

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