Voltage, current, and power

[Shaneyj]

I'm embarrassed that I don't have this nailed down but I came across a scenario today for which I have no clear understanding.
It has been my understanding that when the voltage to a load is increased, the current will decrease.
P=VA.
Unless I don't understand and it is the power consumed that fluctuates.
It seems intuitive that given any load the load resistance (ignoring changes due to heat or other factors that may be negligible) is fixed.
Like a heater, or a bulb. Fixed resistance, no?
In that instance, if ohms law is used to get either current or voltage, with a fixed resistance, and increase in one means an increase in the other. V=IR
So based on my understanding, those two scenarios contradict each other.
Can someone enlighten me?

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[infinity]

If the load is a purely resistive load like a heater an increase in voltage will cause an increase in current. For a inductive load like a motor the opposite is true.

 

[GoldDigger]

If the load is a purely resistive load like a heater an increase in voltage will cause an increase in current. For a inductive load like a motor the opposite is true.

Not trying to quibble, but I feel that the distinction is quite important:

The reason that the current to a motor decreases as voltage increases is that to a first approximation a motor is a constant power load. (Whatever the motor is driving does not change as the voltage changes, and the motor is typically running close enough to synchronous speed that any speed increase is also unimportant.)
If you have a purely inductive circuit the current will increase linearly as you increase the voltage, just as it does for a resistive load.

And when you are looking at large voltage differences, like driving a water heater from 120 or from 240, the two input power possibilities configure a different load resistance, so that once again the power is constant and the resistance is not.

 

[ActionDave]

I agree with goldy. It's a small point but an important one that helped me get my head around theory better.

 

[Shaneyj]

Not trying to quibble, but I feel that the distinction is quite important:

The reason that the current to a motor decreases as voltage increases is that to a first approximation a motor is a constant power load. (Whatever the motor is driving does not change as the voltage changes, and the motor is typically running close enough to synchronous speed that any speed increase is also unimportant.)
If you have a purely inductive circuit the current will increase linearly as you increase the voltage, just as it does for a resistive load.

And when you are looking at large voltage differences, like driving a water heater from 120 or from 240, the two input power possibilities configure a different load resistance, so that once again the power is constant and the resistance is not.

Is voltage and current being proportional to one another a function of what portion of the load is inductive vs resistive?
If I understand correctly, voltage and current are proportional in a purely inductive load AND a purely resistive load.
But a motor, which is somewhere in between (and also has a fixed power rating?), the voltage and current are inversely proportional.
Is it the fixed power out that results in V or I increasing as the other decreases?

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[GoldDigger]

Is voltage and current being proportional to one another a function of what portion of the load is inductive vs resistive?
If I understand correctly, voltage and current are proportional in a purely inductive load AND a purely resistive load.
But a motor, which is somewhere in between (and also has a fixed power rating?), the voltage and current are inversely proportional.
Is it the fixed power out that results in V or I increasing as the other decreases?

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You have hit a critical point in understanding what is going on and have asked the right questions!

1. Any combination of inductive, resistive and capacitive elements is still a linear load and current will be proportional to applied voltage. And the phase angle of the current will remain unchanged as voltage changes.
2. One key difference for the motor is that in addition to resistive and inductive elements, a motor equivalent diagram contains a voltage source (back EMF) which is directly proportional to the motor speed and will typically also be proportional in some way to the current flowing in the motor.
The equilibrium operating point for a motor is where the difference between the back EMF and the applied voltage is sufficient to drive the required current through the motor to produce the mechanical power needed by the load.

 

[Shaneyj]

You have hit a critical point in understanding what is going on and have asked the right questions!

1. Any combination of inductive, resistive and capacitive elements is still a linear load and current will be proportional to applied voltage. And the phase angle of the current will remain unchanged as voltage changes.
2. One key difference for the motor is that in addition to resistive and inductive elements, a motor equivalent diagram contains a voltage source (back EMF) which is directly proportional to the motor speed and will typically also be proportional in some way to the current flowing in the motor.
The equilibrium operating point for a motor is where the difference between the back EMF and the applied voltage is sufficient to drive the required current through the motor to produce the mechanical power needed by the load.

To sum up... any linear load (any composition of resistive, capacitive, inductive) will result in voltage and current being proportional.

Constant power loads are the only time voltage and current will have an inverse relationship.

Are these statements accurate?

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[GoldDigger]

To sum up... any linear load (any composition of resistive, capacitive, inductive) will result in voltage and current being proportional.

Constant power loads are the only time voltage and current will have an inverse relationship.

Are these statements accurate?

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Pretty much.
Constant power loads will have an exact inverse relationship.
More complicated near constant power loads will have an opposite variation but not mathematically inverse (product not exactly constant)