Power PLEASE Help Test THURSDAY

  • Thread starter BuBbLeS01
  • Start date
  • Tags
    Power Test
In summary, the conversation discusses the confusion surrounding power equations for finding the power dissipated by light bulbs. The equations P = IV, I^2R, and V^2/R are mentioned, but the speaker is having trouble determining which one to use for different problems. They have a test on Thursday and give an example of two 97 W light bulbs wired in series. The only equation that gives the correct answer is P = V^2/R, which results in a power of 24.3 W for each bulb. The reason for this is that the current reduces through each bulb when connected in series.
  • #1
BuBbLeS01
602
0
Power...PLEASE Help...Test THURSDAY!

I am having a hard time understanding why some power equations for finding the power dissipated by light bulb(s) don't work. I have...
P = IV = I^2R = V^2/R
But for some of the problems these equations are getting me different numbers for each one! I am getting confused and I now don't know when to use which one and I have a test on Thursday! One example...

Two 97 W (114 V) lightbulbs are wired in series, then the combination is connected to a 114 V supply. How much power is dissipated by each bulb?

I found I using P/V = .851 then when I go to find the power the only equation that gets me the right answer is P = V^2/R then divide by 2 = 24.3 W
 
Physics news on Phys.org
  • #2
The current reduces through each when connected in series, so that the power is not just 2*97 W. You have to find the V across each or find I.
 
  • #3


First of all, it is important to remember that power is the rate at which energy is transferred or converted. In the context of electrical circuits, power is the rate at which electrical energy is converted into other forms of energy, such as heat or light. Therefore, it is crucial to use the correct equations and understand their applications in order to accurately calculate power in a circuit.

The equations P = IV, P = I^2R, and P = V^2/R are all valid equations for calculating power in a circuit. However, each equation has a specific application and should be used accordingly.

P = IV is the most general equation for calculating power in a circuit. It can be used in any circuit, regardless of whether the components are in series or parallel. This equation simply states that power is equal to the product of current and voltage.

P = I^2R is the equation to use when calculating power dissipated by a resistor. This equation is derived from Ohm's Law (V = IR) and the definition of power (P = IV). It is important to note that this equation is only applicable to resistors and cannot be used for other components, such as light bulbs.

P = V^2/R is the equation to use when calculating power dissipated by a resistor in a series circuit. In a series circuit, the same current flows through all components, so the power dissipated by each resistor is proportional to its resistance. This equation is derived from the equation P = I^2R and the fact that in a series circuit, the total voltage (V) is equal to the sum of the individual voltages across each component.

For the specific example given, where two light bulbs are wired in series, it is important to recognize that light bulbs are not purely resistive components. They have a non-linear relationship between voltage and current, which means that Ohm's Law does not apply. Therefore, P = I^2R cannot be used to calculate the power dissipated by the light bulbs. Instead, P = V^2/R should be used, as it takes into account the non-linear relationship between voltage and current in the light bulbs.

In summary, it is important to understand the different equations for calculating power in a circuit and their applications. When dealing with non-linear components, such as light bulbs, it is best to use P = V^2/R to accurately calculate power. I recommend practicing with different types of circuits and components to
 

1. What is the purpose of "Power PLEASE Help Test THURSDAY"?

The purpose of "Power PLEASE Help Test THURSDAY" is to assess individuals' knowledge and understanding of power in a specific subject matter before an important test on Thursday.

2. Who should participate in "Power PLEASE Help Test THURSDAY"?

Anyone who will be taking the test on Thursday and wants to gauge their understanding of power in the subject matter should participate.

3. How can "Power PLEASE Help Test THURSDAY" benefit me?

"Power PLEASE Help Test THURSDAY" can help you identify any gaps in your knowledge and understanding of power, allowing you to focus your studying efforts and potentially improve your test performance on Thursday.

4. Is "Power PLEASE Help Test THURSDAY" mandatory?

No, "Power PLEASE Help Test THURSDAY" is not mandatory. It is meant to be a helpful tool for individuals who want to test their understanding of power before an important test.

5. How can I access "Power PLEASE Help Test THURSDAY"?

The details for accessing "Power PLEASE Help Test THURSDAY" will be provided by your instructor or posted in a designated location. It may be in the form of a study guide, practice test, or online quiz.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
690
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
664
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
249
  • Introductory Physics Homework Help
Replies
9
Views
958
  • Introductory Physics Homework Help
Replies
1
Views
757
  • Introductory Physics Homework Help
Replies
18
Views
959
Back
Top