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

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SUMMARY

The discussion centers on calculating power output in a load with varying current using the formula P = VI. A load rated at 160 W with a current of 6.0 A was analyzed, revealing that if the current increases to 15 A, the power output would be 1000 W, calculated using the formula V = I²R. Participants emphasized that voltage (V) cannot be assumed constant when current (I) changes, particularly in non-resistive loads like motors, which can draw varying currents at a fixed voltage.

PREREQUISITES
  • Understanding of Ohm's Law and power calculations
  • Familiarity with electrical load types (resistive vs. non-resistive)
  • Knowledge of DC motor characteristics and behavior
  • Ability to manipulate and rearrange formulas in physics
NEXT STEPS
  • Study the relationship between voltage, current, and resistance in Ohm's Law
  • Learn about the characteristics of different electrical loads, including resistors and motors
  • Explore the implications of varying current on power output in electrical circuits
  • Investigate the behavior of DC motors under different load conditions
USEFUL FOR

Electrical engineers, physics students, and anyone involved in circuit design or analysis will benefit from this discussion, particularly those interested in understanding power calculations in varying load conditions.

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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