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

In summary: I will assume it's resistive.OK thank you. I will assume it's resistive.In summary, the power output of a load with a power rating of 160 w at 6.0 A will be 1000W when the current is increased to 15 A. This is calculated using the formula v=I^2R, where 15/6 = 2.5^2 = 6.25 and 160 * 6.25 = 1000W. However, it is important to note that this assumes the load is resistive and the resistance remains constant. If the load is a motor or some other type of load, this may not be a valid assumption.
  • #1
Coco12
272
0
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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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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.
 

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

1. What factors affect the power output of a system?

The power output of a system is affected by several factors, including the type and size of the energy source, the efficiency of the system, and any external factors such as temperature or humidity. Additionally, the design and maintenance of the system can also impact the power output.

2. How is power output calculated?

Power output is typically calculated by multiplying the voltage and current of a system, using the equation P = VI. However, this calculation may vary depending on the type of system and the units being used.

3. Can power output be increased or improved?

Yes, power output can be increased or improved through various means such as using more efficient components, optimizing the design of the system, or using renewable energy sources. Regular maintenance and upgrades can also help improve the power output of a system.

4. How does power output affect energy production?

Power output is directly related to energy production, as it determines the amount of energy that can be generated by a system. A higher power output means more energy can be produced, while a lower power output will result in less energy produced.

5. What are some common methods for measuring power output?

Some common methods for measuring power output include using an ammeter and voltmeter to measure current and voltage, respectively, and then using the P = VI equation to calculate power output. Other methods may include using power meters or energy monitors specifically designed for measuring power output in various systems.

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