Higher the voltage the less the amperage

Scottywatt

Member
Location
Mass
Occupation
Electrician/ Construction
This is what was taught to me I'm looking at unit 9 in his theory course and he is saying the opposite am I missing something I feel stupid I can't go any further on that course because now I'm all messed up
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
Retired Electrical Engineer - Power Systems
You need to provide a few more details.

If you are looking at formulas for VA or Watts, then you are correct.
However if you are looking at the characteristics of a motor during starting then your instructor is correct.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
This is what was taught to me I'm looking at unit 9 in his theory course and he is saying the opposite am I missing something I feel stupid I can't go any further on that course because now I'm all messed up
Allow me:

For a given, fixed load impedance (AC resistance), the resultant current will rise and fall proportionately with the applied voltage.

For a desired resultant power, the load impedance must be changed so, as available voltage rises, a lower current will result.

Let me know if you need more clarification.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
To expand on Larry's reply:

It explicitly depends on what you are looking at.

If you have a load which is designed to consume constant power, then as voltage goes up current goes down. Examples of such loads are wide range lighting drivers, which consume constant power from say 100V to 277V, or spinning motors over their design voltage range.

If you are designing a power distribution system to deliver a certain amount of power, then as voltage goes up current goes down.

In both of these situations power is the constant and voltage and current are inversely related. In both of these situations the load impedance (resistance) is changing.

But in many situations power is _not_ constant, and instead it is the load impedance (resistance) that is constant. The prime example of this is a resistance heater. If you increase the voltage applied to a heater then the current goes up. When resistance is constant, current is directly related to voltage and power changes as the square of the voltage.

Motors are more complex beasties, with characteristics that change with rotor speed.

Jon
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
You need to provide a few more details.

If you are looking at formulas for VA or Watts, then you are correct.
However if you are looking at the characteristics of a motor during starting then your instructor is correct.
Oopsie.
Small error here.
If a motor drives a constant power load, then it will draw less current at equilibrium speed if the voltage is higher (up to the point that saturation occurs in the magnetic circuit of the motor).
However when the motor is not yet rotating (locked rotor or starting amps) the current will be greater for a higher applied voltage. This is why reduced voltage starting is sometimes used to limit starting current for a large motor.
 

mikeames

Senior Member
Location
Germantown MD
Occupation
Teacher - Master Electrician - 2017 NEC
What you were taught is based on getting a certain amount of work done.
Example 12v @ 10 amps = 120 watts of power
120v@ 1 amp = 120 watts of power

Same work done with different current and voltage characteristics. I think you understand this........ Your right if you increase the voltage you can get the same power out with less current. But different loads behave differently (Reactive loads and resistive loads).

Now look at it differently. Take a purely resistive circuit such as an incandescent light bulb. The greater the voltage you apply to the light bulb the more current you will force through the static resistance. So as voltage goes up, the load will actually consume more current. In the case of an incandescent light bulb or resistive heating element, the current will get to a point where it will burn up the load.

Example:

An incandescent light bulb is rated at 60 watts at 120 volts. That means its normal operating current will be .5 amps. If you apply 120v to it.
Given that, we can calculate that its normal operating resistance is 240 ohms. Generally speaking that resistance is fixed. It technically increases as it heats up but for this lets ignore that.


So if it has 240 ohms of resistance what happens if we apply 240v instead of the 120 its designed for? Well 240v / 240 ohms = 1 amp. 240volts X 1 amp = 240 watts

The light bulb was designed for 1/2 amp and 60 watts of heat, if we double the voltage it will now have 1 amp flowing through it and be putting out 240 watts of heat. The result will be a bright shorth lived bulb.


So for resistive loads if the voltage goes up so does the current.
 
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