Ohms law, Understanding formula

[brother]

Just brushing up on my math. E=I*R I=E/R R=E/I So in a motor load, that has alot(increase) of resistance (on a rail that needs oiling) , and the breaker trips. How does this fit with the formula though?? Just thinking out loud.

 

[steve66]

Ohms law works for linear loads (constant resistance or impedence) which your motor is not.

Better to think in terms of power P=E*I. The motor requires more power to overcome the resistance of the rails. Since E can't go up, I has to increase.

The intersting thing is that for a constant load, if you lower the voltage to a motor, the current goes up. That's the opposite of what you would expect from ohms law. But if you look at P=E*I, if P stays constant, and E goes down, I has to go up.

Edit: Someone will probably give me grief for not including power factor, since motors came into the discussion. But the basic relationships stay the same even if we start adding the difference between KVA and P.

Steve

 

[drbond24]

Ohm's law only works right with resistive loads. A motor is an inductive load, so it doesn't necessarily observe this law.

Other than to tell you that, I can't help because I don't understand your question. :smile:


(edit: I didn't type fast enough! Steve beat me. )

 

[brother]

Ohms law works for linear loads (constant resistance or impedence) which your motor is not.

Better to think in terms of power P=E*I. The motor requires more power to overcome the resistance of the rails. Since E can't go up, I has to increase.

The intersting thing is that for a constant load, if you lower the voltage to a motor, the current goes up. That's the opposite of what you would expect from ohms law. But if you look at P=E*I, if P stays constant, and E goes down, I has to go up.

Edit: Someone will probably give me grief for not including power factor, since motors came into the discussion. But the basic relationships stay the same even if we start adding the difference between KVA and P.

Steve

ok so a motor would be an inductive load. I can see that, but so are transformers. P= power or watts, volt amperes right?? so should I just look at a motor as a watt?? and the 'resistance' on the rails as more watts added??

 

[raider1]

ok so a motor would be an inductive load. I can see that, but so are transformers. P= power or watts, volt amperes right?? so should I just look at a motor as a watt?? and the 'resistance' on the rails as more watts added??

Transformers are not a load. Transformers work on the principals of induction, but Ohm's law does work for transformer calculations.

Chris

 

[dereckbc]

As stated Ohm's Law is for pure resistance loads, and does not iclude a 5th element in AC inductive loads known as inductive reactance. Ohm;s law will get you in the ball park but is not accurate with AC circuits unless it is purely a passive resistance circuit like a heat strip.

 

[steve66]

There is an ohms law that works for AC circuits that have capacitance and inductance. It is:

V=I * Z

V, I, and Z are all phasors, and the * is an operation a lot like (but not exactly) multiplication.

So you can use ohms law to find out that 120 volts across a inductive reactance of 120 ohms is 1 amp. But you have to have a single frequency, constant voltage source. And you have to keep careful track of angles for anything more complex than the simplest problems.

The big difference with the motor problem is that a motor isn't linear. As the load it sees changes (more resistance on the rails) its resistance and inductance changes.

So if you are just looking at the direction the numbers go (and not actually trying to calculate exactly how many amps there are), you can follow the chain:

1. More rail resistance = more watts to the motor
2. More watts (with a voltage that stays the same) = more current. (from P=I*E)
3. Then from Ohms law (E=I * Z), since I went up, and E stayed the same, the impedence (Z) of the motor that the source sees had to go down.

Of course, we are still ingoring a lot of details, like power factor.

Steve

 

[Rockyd]

This might help...​

AC has Z for total reactance, instead of R. The wheel is a teaser for the rest of the formulas to look at

 

[brother]

This might help...​

AC has Z for total reactance, instead of R. The wheel is a teaser for the rest of the formulas to look at

I already know about the power wheel, thanks for the info though. but we have to remember that the basic wheel that works for DC also works for AC , just like residential/commercial calculations, 2400 VA / 120volts = 20 amps. (this is just a calc i did for receptacles in commercial (180 va * 20amp breaker).