voltage vs current with resistive loading

[mull982]

In most cases I typicall deal with raising the voltage for a particular load will reduce the current draw, and vise versa. I have always understood this to be due to the fact that the load kW or power stayed the same and just the voltage and current were changing to adjust to this load power. I have always understood this to be the case for inductive loads such as motors and transformers.

With resistive loads however I am having a hard time understanding this due to the fact that Ohms law does not agree. If I have a resistive load such as a water heater, or oven and I raise the voltage per say, I would expect the current to increase in proportion with the voltage based upon V=IR. So to use an example of a resistive load and saying that the with keeping the kW or Power the same and changing the voltge would lead to a decrease in current, would not agree with ohms law. Can someone help explain?

Thanks

 

[dereckbc]

Mull you need to examine the other 11 formuals of Ohm' law, especially the 3 that pertain to Power like P=IE, P=I^2R, and E^2/R. the key in a resistive load is R is the constant. So in a purely resitive circuit, if voltage rise, then so does current, P=EI, while R is constant or a fixed value

 

[drbond24]

Ohm's law as you stated it (V=IR) is only true for resistive loads. When capacitance and/or inductance are introduced, more complicated equations are required.

If you came across a situation where an increase in voltage did not create a proportional increase in current, you were dealing with reactance (i.e. inductance and/or capacitance) and not just resistance. At that point, the correct term is 'impedance' and the rules change.

 

[wasasparky]

Depends if you are talking about the same animal at a completely different voltage, or small voltage changes around it?s rating.

You have to differentiate operating voltage options, from small changes in supply voltage.

ALL LOADS will have a lower current at higher voltage, or higher current at lower voltage assuming the power requirement/ horsepower rating stays the same. (1hp motor at 240V vs. 480V, 5kW heater at 240V vs. 480V)

Now lets say you have a 240V load. If the supply voltage is slightly higher, the resistive device will draw a bit more current and power, a motor however will draw a bit less current to maintain ?constant? power. (not sure of the limits of "constant" power, and how linear in nature?but an accepted generaliziation...)

You would not take a 240V water heater and simply supply it with 480V, but you may connect it to a 208V source...

 

[weressl]

Mull you need to examine the other 11 formuals of Ohm' law, especially the 3 that pertain to Power like P=IE, P=I^2R, and E^2/R. the key in a resistive load is R is the constant. So in a purely resitive circuit, if voltage rise, then so does current, P=EI, while R is constant or a fixed value

The only thing you have to remember that in the case of heater resistance the R value will change as the resistor heats up, so the current change will not be linear with the voltage. (This one of the reasions why at calculation the tables specify that the wire resistance is given at 20Centigrades and correction factors are applicable at other temperatures.)

 

[coulter]

guys and girls -
I am hearing a lot of things that don't look right. Here's how I see the basic science. I am absolutely open to being corrected.

Ohm's law as you stated it (V=IR) is only true for resistive loads. When capacitance and/or inductance are introduced, more complicated equations are required.

The equation is still V=IR. V, I, R are vectors with phase angles. The algebra gets messy, but I wouldn't call it more complicated.

ALL LOADS will have a lower current at higher voltage, or higher current at lower voltage assuming the power requirement/ horsepower rating stays the same. (1hp motor at 240V vs. 480V, 5kW heater at 240V vs. 480V)

First, unless you have someting really strange, heaters are not constant power. The are very close to zero temperature coeficient, constant R. A 10% increase in voltage will give you a 21% increse in power.

Now for the motors: I know poeple say this about motors because of the graph in NEMA MG-1. However, very few mechanical loads are constant power. For a load to be constant power if the rpm went up a little, the torque required would have to drop a porportional amount. Very few loads do that - I can't think of any that aren't electronically controlled.

So, standard induction motors, with real life load (not an MG-1 load suitable for making graphs), if the voltage goes up a bit, the speed will increase a bit, the load will require a bit more torque, motor current will increase.

...raising the voltage for a particular load will reduce the current draw, and vise versa. I have always understood this to be due to the fact that the load kW or power stayed the same ...

...If I have a resistive load such as a water heater, or oven and I raise the voltage per say, I would expect the current to increase in proportion with the voltage based upon V=IR. So to use an example of a resistive load and saying that the with keeping the kW or Power the same and changing the voltge would lead to a decrease in current, would not agree with ohms law ..{/quote]

mull -
Ohms law absolutely works - doesn't matter if we are talking heaters, motors, reactive loads or even light bulbs.

Motors are a little wierd in they have a speed/torque/voltage/current curve, and the load has a speed/torque curve. And since the two are connected, they interact. Motors just don't operate at 100% with a constant power load that is shown in the NEMA MG-1 curves.


carl (I'm listening)

 

[bcorbin]

I always like general forms of equations much better. What is typically referred to as "Ohm's Law" is only true when you have a purely resistive load. R may have a phase angle, but it's always zero, period. If you get used to thinking V = I*Z, you can't go wrong.

 

[drbond24]

Coulter:

I don't disagree with anything you said, except to point out that R with a phase angle isn't R anymore, it is Z. Like bcorbin said, R has a phase angle of zero.

I was thinking of the calculus and differential equations I did in college when I said the equations were more complicated. I tend to forget that they didn't necessarily show me the easiest way to do things. Plus, geeks like me just enjoy the math. You can't spell geek without EE!!

 

[ctrane]

You are correct about the change in voltage, if R is constant, then an increase in Voltage is directionally proportional, by R times, to an increase in Current. What you have to look at is the KW reading of your equipment. The KW reading is the most the equipment can take without overheating, breaking down, etc...So, if you increase the voltage, you must physically de-rate the current accordingly to keep the specified KW. So a heater running off of 120V will take half as much current as 240V to keep its KW rating, i.e. breaker size will be halved.

 

[infinity]

ALL LOADS will have a lower current at higher voltage, or higher current at lower voltage assuming the power requirement/ horsepower rating stays the same. (1hp motor at 240V vs. 480V, 5kW heater at 240V vs. 480V).

You might want to rethink that statement. Comparing purely resistive loads to inductive loads is like comparing apples to oranges.

 

[ctrane]

I don't think he needs to rethink that statement. The relationship in DC, or purely resistive loads is: V=IR, when you get into AC and include inductive-capacitive elements the relationship gets restated V=IZ. The difference is that in V=IR, it is a linear relationship, in V=IZ, it is no longer linear. Your solutions for most equations in the V=IZ world will now be solutions to (generally) 2nd order differential equations. So, yes at certain points you may have unusual phenomena, i.e. at poles and zeroes, but the basic realtionship holds true.

 

[wasasparky]

You might want to rethink that statement. Comparing purely resistive loads to inductive loads is like comparing apples to oranges.

Don't read too much or too little into the above.
I'm pointing out the big-picture little-picture stuff.
A 1hp motor at 240V is a different animal than a 1hp motor at 480V.
I contend the 480V motor will draw fewer amps (as simple as that).

 

[LarryFine]

Mull, when you have the time, read through these two threads and see if they help explain anything (especially my posts :roll:

http://forums.mikeholt.com/showthread.php?t=85729
http://forums.mikeholt.com/showthread.php?t=79786

 

[infinity]

Don't read too much or too little into the above.
I'm pointing out the big-picture little-picture stuff.
A 1hp motor at 240V is a different animal than a 1hp motor at 480V.
I contend the 480V motor will draw fewer amps (as simple as that).

Yes and I agree with that part of your post, it was the contention that a 5kw heater would have a lower current at a higher voltage, (480 v. 240)that simply is not true. That heater would have twice the current and 4X the output in watts.

 

[dbuckley]

The bottom line is that the OPs original assertion, that power is constant with change of voltage, is a flawed assertion. Very very few loads exhibit that behaviour, and those that do only do so over a limited range of voltages.

For resistive loads, the resistance changes with voltage, due to the heating effect and temperature coefficients. It's easy to demonstrate with lamps.

This is for a 500W linear tube lamp, rated at 240V.

R constant? I dont think so...

 

[coulter]

...This is for a 500W linear tube lamp, rated at 240V.

R constant? I dont think so...

db -
What you said is true for incandescant lamps. However, not all resistive loads are lamps. If you do the same tests on a standard electric heater, usually nichrome (sp?) or similar, the temperature coeficient is close to zero. So yes,there are loads that are constant R. Although, not for an unlimited range - the 240V heating element in your dryer, while pretty flat, certainly gets non-linear at 4160V

edited to add: About the constant power loads, I agree, as I said in post 6. Which, of course, has nothing to do with constant R.

carl

 

[rattus]

Beg to Differ:

Beg to Differ:

Ohm's law as you stated it (V=IR) is only true for resistive loads. When capacitance and/or inductance are introduced, more complicated equations are required.

If you came across a situation where an increase in voltage did not create a proportional increase in current, you were dealing with reactance (i.e. inductance and/or capacitance) and not just resistance. At that point, the correct term is 'impedance' and the rules change.

Ohm's law works with impedance as well. Of course, the voltages and currents must be sinusoidal. That is, reactance is only defined for sinusoids.

 

[jrannis]

Resistive Load Water Heater

Resistive Load Water Heater

This is a useful experiment you can try out at home.

1. Measure your water heater amperage with a clamp type meter.
2. Turn off your waterheater breaker
3. take one leg off of the breaker and
4. connect it to your neutral bar

Should draw about half of the amps

Useful information if you have a power outage and only have a small generator and want a warm shower.

 

[LarryFine]

Should draw about half of the amps.

And, of course, 1/4 the power.

 

[ron]

We've heard of brownouts during the summer, when overloaded transmission lines and feeders experience high voltage drop due to high current draw, but in NYC, Con Ed has been known on occasion to purposefully reduce the voltage on their low voltage grid to reduce power consumption. I guess the load in NYC will respond with lower P when V is lowered.
So is it mostly resistive or inductive? ......