Can You Use P = VI to Calculate Power Output in a Load with Changing Current?

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Homework Help Overview

The discussion revolves around calculating power output in a load with changing current, specifically questioning the applicability of the formula P = VI. The original poster presents a scenario where a load rated at 160 W with a current of 6.0 A is examined for power output when the current increases to 15 A.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between power, voltage, and current, questioning whether voltage remains constant as current changes. There is also discussion about the nature of the load, whether it is resistive or not, and how that affects the calculations.

Discussion Status

The discussion is ongoing, with participants providing insights into the assumptions regarding voltage and resistance. Some guidance has been offered regarding the need for more details about the load type, and the implications of assuming constant resistance or voltage.

Contextual Notes

There is a noted deficiency in the posting template used by the original poster, which may affect the clarity of the problem statement. Additionally, the nature of the load (resistor vs. motor) is under consideration, impacting the assumptions made in the calculations.

Coco12
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A load has a power rating of 160 w when the current is 6.0 A. What will be the power output if the current increases to 15 A?

I know how to do it : use the formula v=I^2R
In this case 15/6 = 2.5^2 = 6.25
160 * 6.25 = 1000W.
My question is why can't you use p = VI ? Because in this case , it would be 160* 2.5?
 
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Coco12, please be sure to use the posting template provided when a new thread is started here in the homework sections of Physics Forums. Otherwise you're setting yourself up to incur infraction points...

Coco12 said:
A load has a power rating of 160 w when the current is 6.0 A. What will be the power output if the current increases to 15 A?

I know how to do it : use the formula v=I^2R
In this case 15/6 = 2.5^2 = 6.25
160 * 6.25 = 1000W.
My question is why can't you use p = VI ? Because in this case , it would be 160* 2.5?

You cannot assume that V remains unchanged when I changes.
 
gneill said:
Coco12, please be sure to use the posting template provided when a new thread is started here in the homework sections of Physics Forums. Otherwise you're setting yourself up to incur infraction points...

You cannot assume that V remains unchanged when I changes.
Thank you very much. I posted this from an ipad, and the template did not come up. I wonder why?
 
Coco12 said:
Thank you very much. I posted this from an ipad, and the template did not come up. I wonder why?

Ah. It's a known deficiency in the app, and is being looked into. I suggest manually formatting your initial posts along the lines of the approved manner until the deficiency is corrected.

1) Problem Statement
2) Relevant Equations
3) Attempt at Solution
 
gneill said:
Ah. It's a known deficiency in the app, and is being looked into. I suggest manually formatting your initial posts along the lines of the approved manner until the deficiency is corrected.

1) Problem Statement
2) Relevant Equations
3) Attempt at Solution

Thank you.
 
I believe you will also need to provide more details of the load. Is it safe to assume it's a resistor? The answer would be different for a motor.
 
gneill said:
Coco12, please be sure to use the posting template provided when a new thread is started here in the homework sections of Physics Forums. Otherwise you're setting yourself up to incur infraction points...

You cannot assume that V remains unchanged when I changes.
Another question: so you can assume that resistance doesn't change?
 
You tell us. Does the problem statement describe the load?

If the load is a resistor then yes it's probably safe to assume that the resistance doesn't change. In which case the if the current changes that implies a change in the voltage.

However if the load is a motor then the above isn't necessarily true. A DC motor operating on a fixed 12V could draw a wide range of currents (and hence power) depending on the load on the motor shaft.

If the problem statement doesn't describe the load in any more detail then you should probably state that you are assuming the load is resistive, and not something like a motor, then proceed accordingly.
 
CWatters said:
You tell us. Does the problem statement describe the load?

If the load is a resistor then yes it's probably safe to assume that the resistance doesn't change. In which case the if the current changes that implies a change in the voltage.

However if the load is a motor then the above isn't necessarily true. A DC motor operating on a fixed 12V could draw a wide range of currents (and hence power) depending on the load on the motor shaft.

If the problem statement doesn't describe the load in any more detail then you should probably state that you are assuming the load is resistive, and not something like a motor, then proceed accordingly.
OK thank you.
 

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