Was Ohm wrong?

[ShawnD1]

You have probably all seen the equation I = V/R.
I is current, V is voltage and R is resistance.

Technically speaking, as resistance increases, current should decrease. Strangely, I've seen it where the exact opposite is true.
If you plug an air compressor into 200 feet of extension cord and turn it on, it will trip the breaker. Having a 200 foot extension cord adds a lot of resistance into the system. The breaker trips when too much current flows through the circuit. If having a 200 foot extension cord puts lots of resistance into the circuit, why does the current somehow INCREASE?

 

[glugglug]

Not all the current is going to the air compressor.

Most of it is probably in fact going between the 2 wires in the extension cord. The insulation of the cord itself has gone down and it is close to having a short.

 

[FrustratedUser]

Originally posted by: ShawnD1
You have probably all seen the equation I = V/R.
I is current, V is voltage and R is resistance.

Technically speaking, as resistance increases, current should decrease. Strangely, I've seen it where the exact opposite is true.
If you plug an air compressor into 200 feet of extension cord and turn it on, it will trip the breaker. Having a 200 foot extension cord adds a lot of resistance into the system. The breaker trips when too much current flows through the circuit. If having a 200 foot extension cord puts lots of resistance into the circuit, why does the current somehow INCREASE?

The 200 feet cord also has a big inductance. As you might know an inductance is working 'against' voltage changes. A positive pulse of voltage travels along the cord, the inductance counteracts the change causing a resulting current peak higher than the compressor actually uses ---> breaker trips.

 

[blahblah99]

Originally posted by: FrustratedUser

Originally posted by: ShawnD1
You have probably all seen the equation I = V/R.
I is current, V is voltage and R is resistance.

Technically speaking, as resistance increases, current should decrease. Strangely, I've seen it where the exact opposite is true.
If you plug an air compressor into 200 feet of extension cord and turn it on, it will trip the breaker. Having a 200 foot extension cord adds a lot of resistance into the system. The breaker trips when too much current flows through the circuit. If having a 200 foot extension cord puts lots of resistance into the circuit, why does the current somehow INCREASE?

The 200 feet cord also has a big inductance. As you might know an inductance is working 'against' voltage changes. A positive pulse of voltage travels along the cord, the inductance counteracts the change causing a resulting current peak higher than the compressor actually uses ---> breaker trips.

What the guy above me said.

Ohm is never wrong!

 

[glugglug]

Inductance works against CURRENT changes not voltage changes. Inductance would NOT cause a current spike, it smooths the current out.

Also, if the cord had a high inductance it wouldn't be very useful since your electric outlets are AC not DC.

 

[syberscott]

There is no inductance increasing the current or whatever.
The answer is...
The long cord will cause a voltage drop that will decrease the rpm of the electric motor in the compressor. An electric motor running at a slower speed will draw more current. Therefore the breaker trips.

 

[rjain]

Erm, I think everyone missed the question. A higher resistance with the same voltage across the whole system will cause a higher current. The OP basically observed Ohm's law in action and wondered if that violates Ohm's law...

 

[Stealth1024]

you guys are trying to apply v = IR from DC current to AC current...

 

[glugglug]

Originally posted by: Stealth1024
you guys are trying to apply v = IR from DC current to AC current...

And it still applies. Only for AC, the values have to be treated as imaginary numbers, for example the value for R has resistance as it's real component and inductance as the imaginary component.

rjain:

Higher resistance causes LOWER current in Ohms law, not HIGHER current.

syberscott's explanation makes sense, it could be that since the cord and the compressor are in series, in order to get enough energy to the compressor the current through the cord combined with the resistance of the cord means too much power is being dissipated by the cord.

To check whether the problem is from a near-short in the cord or from high resistance in the cord, plug the cord in without the compressor. Walk along the cord WITH SNEAKERS. When you step on the bad part of the cord if there is one, it will blow the fuse, and likely also melt/burn the shielding around that spot.

 

[sgtroyer]

rjain I think it's you that has it backwards. For a given voltage, current will decrease with increasing resistance.

It's important to note that Ohm's law is not an absolute. There are many materials which don't follow ohm's law. Capacitors and inductors don't follow it, their "resistance" varies with frequency. Transistors really don't follow it. Ohm's law isn't a law that universally applies, really it's more of a definition (defining resistance). Some materials can be approximated as having a fixed resisistance, and you can use Ohm's law. Others don't, and you can't.

I don't understand the specifics of this situation, but syberscott's idea sounds promising. In a complicated system, the resistance of a wire can have wierd end effects, causing a non-intuitive result. Although the wire may be approximated as a simple resistor, the motor certainly can't.

By the way, I have a friend that ruined his new planer by plugging it into a long extension cord. I wouldn't recommend this type of experiment if you want your air compressor to last very long.

 

[Trianon]

Yeap, motors at stall(rotor is still) draw more current then when spinning. I would not be surprised that this same compressor would trip the breaker when plugged in directly... Adding extension cable with proper insulation should not cause overload.

 

[Mday]

V=IR can be applied to AC, if you make R an impedance, which can be a complex value.

the problem here is that ohm's law doesnt apply here. the motor will draw as much current as it can or want to given the setting that it is at.

the extension cord would not kill any device if it is properly used. that is, dont use a 16gauge extension cord when you need 12 or 14.

 

[rjain]

Hmm... I remember that people add resistors to circuits to pump more current through them. Maybe this only applies to transistors.

 

[Peter]

The answer to the original question:
Electrical motors draw a HUGE current when they start spinning. The slower they ramp up, the higher the current. Now, with the extra resistance of the long cord decreasing the voltage at the motor, it starts very reluctantly, with a much higher current than at its rated voltage.

This is one of the reasons why higher supply voltages are preferrable - things like these don't happen at all when you're on 230 Volts.

 

[Mark R]

The answer has already been given - the lower voltage available to the motor, causes lower speed with subsequent higher current consumption.

This is the reason why 'brownouts' are so undesirable - the current drawn by motors increases, and motors under heavy load (e.g. AC or refrigeration) can stall blowing breakers or even burning out.

 

[Trianon]

Originally posted by: Mark R
The answer has already been given - the lower voltage available to the motor, causes lower speed with subsequent higher current consumption.

This is the reason why 'brownouts' are so undesirable - the current drawn by motors increases, and motors under heavy load (e.g. AC or refrigeration) can stall blowing breakers or even burning out.

I concur, that's the answer... Motor winding current is dependant of RL constant for that motor, at low speed L is smaller, which would cause larger current draw by the motor.. Either way, check extension cord or get another one with smaller gauge number(higher ampacity). Mark, you are hired

 

[ShawnD1]

Ok so if I understand correctly, you are saying that because the motor is not spinning fast enough, there is not enough back current from the motor and that is what causes so much power to flow?

 

[Mark R]

Ok so if I understand correctly, you are saying that because the motor is not spinning fast enough, there is not enough back current from the motor and that is what causes so much power to flow?

That's basically it.

It's actually a bit more subtle though. When the motor is first switched on, there is a massive surge of current, as the motor offers no back-EMF to limit the current. This current can be several hundred amps. As torque is related to current, the motor develops massive torque and accelerates rapidly to operating speed.

With a resistance in the power supply, the motor starting current is externally limited. The motor cannot develop as much torque and starts much slower. Although the peak current may be lower in this case, the higher current is drawn for a longer time period - circuit breakers and fuses are frequently designed to tolerate very short periods of overload precisely to withstand motor starts.

The other problem, particularly with compressors, is that if the motor's torque is reduced, then it may not be able to start at all - if the compressor is already charged then the motor has to act against the stored air as well as it's own inertia. It may simply stall resulting in prompt tripping of protection systems.

 

[Trianon]

That's basically it.

It's actually a bit more subtle though. When the motor is first switched on, there is a massive surge of current, as the motor offers no back-EMF to limit the current. This current can be several hundred amps. As torque is related to current, the motor develops massive torque and accelerates rapidly to operating speed.

With a resistance in the power supply, the motor starting current is externally limited. The motor cannot develop as much torque and starts much slower. Although the peak current may be lower in this case, the higher current is drawn for a longer time period - circuit breakers and fuses are frequently designed to tolerate very short periods of overload precisely to withstand motor starts.

The other problem, particularly with compressors, is that if the motor's torque is reduced, then it may not be able to start at all - if the compressor is already charged then the motor has to act against the stored air as well as it's own inertia. It may simply stall resulting in prompt tripping of protection systems.

Spoken like a true power electronics engineer.

 

[PowerEngineer]

Originally posted by: Trianon

That's basically it.

It's actually a bit more subtle though. When the motor is first switched on, there is a massive surge of current, as the motor offers no back-EMF to limit the current. This current can be several hundred amps. As torque is related to current, the motor develops massive torque and accelerates rapidly to operating speed.

With a resistance in the power supply, the motor starting current is externally limited. The motor cannot develop as much torque and starts much slower. Although the peak current may be lower in this case, the higher current is drawn for a longer time period - circuit breakers and fuses are frequently designed to tolerate very short periods of overload precisely to withstand motor starts.

The other problem, particularly with compressors, is that if the motor's torque is reduced, then it may not be able to start at all - if the compressor is already charged then the motor has to act against the stored air as well as it's own inertia. It may simply stall resulting in prompt tripping of protection systems.

Spoken like a true power electronics engineer.

Spoken like a true electrical power engineer!

Yes, the apparent impedance of an electrical motor is not constant, and will actually be less at lower voltages (and at slower speeds during start-up). Because the extenstion cord adds resistance that reduces the actual voltage being applied to the motor, its impedance is lowered by more than the resistance added by the cord; this means the current is actually higher. At least in this case.

This characteristic of electric motors is also why large ones are often started (i.e. spun up to speed with out load on them) with resistors intentionally inserted in series to reduce that initial "inrush" current. As the motor reaches speed (and its impedance increases and the current drawn decreases), these resistors are then "shorted out" of the circuit before the load is applied to the motor.

It's also the reason why trying to reduce customer loads during peak periods through voltage reductions don't work as well as one might think. The motors tend to draw more current to at least partially offset the drop in voltage, meaning their power consumption doesn't drop as much as a resistive load (and the increased current increases losses).