1. Hey guys! Here's a question:

Water in an electric kettle connected to 220 Volt supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?
A) 5 minutes
B) 6.05 minutes
C) 5.05 minutes
D) 6 minutes

Please explain the answer to this question....I have some doubts regarding it. Thanks for the help!!!😊

2.

3. Originally Posted by Gaurav(-26.7)
I have some doubts regarding it.
What are your doubts regarding the question?

4. Originally Posted by Gaurav(-26.7)
Hey guys! Here's a question:

Water in an electric kettle connected to 220 Volt supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?
A) 5 minutes
B) 6.05 minutes
C) 5.05 minutes
D) 6 minutes

Please explain the answer to this question....I have some doubts regarding it. Thanks for the help!!!
Hey guys, here's my homework for you to do for me? Sorry, no dice. But we are happy to help you learn, hence KJW's question to you.

If you can outline how you would try to answer it, we can guide you. Clue:what is the relevant formula to use for something like this?

5. Ok, sorry. I forgot to post what I thought about it. Initially, what came to my mind was - If the voltage is reduced, then the electric kettle would draw more current, because its power, P=VI is constant. Thus the amount of heat developed, H = VIt would remain the same without the time (t) being affected and the water would get boiled in the same 5 minutes. So, option A.

But then, I asked another person about this question. He told me the answer should be B) 6.05 min. He wrote down the equation for heat developed in terms of V and Resistance (R) as H(1)={V(1)}^2×t/R and then H(2)={V(2)}^2×T/R. He then set H(1)=H(2), V(1)=220 V and V(2)=200 V, t= 5 minutes and solved for T. He got 6.05 minutes.
Now, I am a bit confused as to which point of view is correct, even though I still favour mine, because I think my friend has interfered with the power = VI of the kettle, which should be constant.

6. Originally Posted by Gaurav(-26.7)
Ok, sorry. I forgot to post what I thought about it. Initially, what came to my mind was - If the voltage is reduced, then the electric kettle would draw more current, because its power, P=VI is constant. Thus the amount of heat developed, H = VIt would remain the same without the time (t) being affected and the water would get boiled in the same 5 minutes. So, option A.

But then, I asked another person about this question. He told me the answer should be B) 6.05 min. He wrote down the equation for heat developed in terms of V and Resistance (R) as H(1)={V(1)}^2×t/R and then H(2)={V(2)}^2×T/R. He then set H(1)=H(2), V(1)=220 V and V(2)=200 V, t= 5 minutes and solved for T. He got 6.05 minutes.
Now, I am a bit confused as to which point of view is correct, even though I still favour mine, because I think my friend has interfered with the power = VI of the kettle, which should be constant.
Aha, there's your error. The power rating of an electrical appliance is quoted for a given supply voltage.

What kind of electrical device is the heating element of a kettle? It's just a resistor. So if you reduce the voltage, what happens to the current and hence to the power?

7. Hmm, the resistor has a fixed resistance and therefore with the decrease in voltage, the current decrease, and so does the power. This is why time will have to increase for H=P×t to remain same. Am I right?

I didn't know that the power rating is quoted for a given supply voltage! So my friend's answer is correct.....

8. Originally Posted by Gaurav(-26.7)
Hmm, the resistor has a fixed resistance and therefore with the decrease in voltage, the current decrease, and so does the power. This is why time will have to increase for H=P×t to remain same. Am I right?

I didn't know that the power rating is quoted for a given supply voltage! So my friend's answer is correct.....
Indeed. But regarding power ratings, as the thing is just a resistor, if the voltage drops it has no magical way to increase the current to keep the power constant. To do that it would have to decrease its resistance to increase the current enough to compensate. This would have to be pretty clever - not to mention possibly blowing the fuses due to the extra current! You don't get that sort of thing in a £15 kettle.

If you use Vi = power and consider that, by V=iR, i=V/R, then power can also be expressed as V²/R.

9. Originally Posted by exchemist
Originally Posted by Gaurav(-26.7)
Hmm, the resistor has a fixed resistance and therefore with the decrease in voltage, the current decrease, and so does the power. This is why time will have to increase for H=P×t to remain same. Am I right?

I didn't know that the power rating is quoted for a given supply voltage! So my friend's answer is correct.....
Indeed. But regarding power ratings, as the thing is just a resistor, if the voltage drops it has no magical way to increase the current to keep the power constant. To do that it would have to decrease its resistance to increase the current enough to compensate. This would have to be pretty clever - not to mention possibly blowing the fuses due to the extra current! You don't get that sort of thing in a £15 kettle.

If you use Vi = power and consider that, by V=iR, i=V/R, then power can also be expressed as V²/R.
Now I get it nicely. Thanks for the help, exchemist!

P.S. The friend I have been talking about all this time, is actually my dad...

10. Originally Posted by Gaurav(-26.7)
Originally Posted by exchemist
Originally Posted by Gaurav(-26.7)
Hmm, the resistor has a fixed resistance and therefore with the decrease in voltage, the current decrease, and so does the power. This is why time will have to increase for H=P×t to remain same. Am I right?

I didn't know that the power rating is quoted for a given supply voltage! So my friend's answer is correct.....
Indeed. But regarding power ratings, as the thing is just a resistor, if the voltage drops it has no magical way to increase the current to keep the power constant. To do that it would have to decrease its resistance to increase the current enough to compensate. This would have to be pretty clever - not to mention possibly blowing the fuses due to the extra current! You don't get that sort of thing in a £15 kettle.

If you use Vi = power and consider that, by V=iR, i=V/R, then power can also be expressed as V²/R.
Now I get it nicely. Thanks for the help, exchemist!

P.S. The friend I have been talking about all this time, is actually my dad...
Haha. Glad it makes sense now anyway.

11. Originally Posted by Gaurav(-26.7)
Hey guys! Here's a question:
Originally Posted by Gaurav(-26.7)

Water in an electric kettle connected to 220 Volt supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?
A) 5 minutes
B) 6.05 minutes
C) 5.05 minutes
D) 6 minutes

Please explain the answer to this question....I have some doubts regarding it. Thanks for the help!!!
ohmic heat production is proportional to watts.

I^2R=P
IR=V

assuming 100ohms resistance:

I = 2.2 Amps
R = 100 ohms
V = 220 volts
P= 484 watts

I = 2 amps
R = 100 ohms
V = 200 volts
P = 400 watts

400 w / 484 w = 0.8264 factor of decrease of heat production

assuming change in time to boiling is inversely proportional to change in heat production

1/0.8264 = 1.21 factor of increase of time to boiling

5 minutes original time * 1.21 factor of time increase = 6.05 minutes time to boil

Correct answer: B - 6.05 minutes

12. Originally Posted by devin-m
Originally Posted by Gaurav(-26.7)
Hey guys! Here's a question:
Originally Posted by Gaurav(-26.7)

Water in an electric kettle connected to 220 Volt supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?
A) 5 minutes
B) 6.05 minutes
C) 5.05 minutes
D) 6 minutes

Please explain the answer to this question....I have some doubts regarding it. Thanks for the help!!!
heat production is proportional to amps not watts.

IR=V
V/R=I

I = 2.2 Amps
R = 100 ohms
V = 220 volts

I = 2 amps
R = 100 ohms
V = 200 volts

2 amps / 2.2 amps = 0.90909 factor of decrease of heat production

assuming change in time to boiling is inversely proportional to change in heat production

1/0.90909 = 1.10 factor of increase of time to boiling

5 minutes original time * 1.10 factor of time increase = 5.5 minutes time to boil

Correct answer: E - 5.5 minutes
Nope. Power dissipation in a resistor = i²R = Vi = V²/R. Any 6th former knows that.

Power Dissipation in Resistors

13. ive corrected my post... i’d heard before heat production was proportional to amps not watts, but after re-checking ohmic heating i found it’s proportional to watts.

edit: i remember my source of confusion... it’s the magnetic field strength of a solenoid which is proportional to amps not watts (not the ohmic heating)

”The magnetic field B is proportional to the current I in the coil.”

http://hyperphysics.phy-astr.gsu.edu.../solenoid.html

14. Originally Posted by devin-m
ive corrected my post... i’d heard before heat production was proportional to amps not watts, but after re-checking ohmic heating i found it’s proportional to watts.
Why listen to rumours? And why, knowing that your knowledge is based on such a shaky foundation, would you venture to provide an answer in such authoritative terms?

It's fine to be ignorant, but it's not fine to be ignorant of what you are ignorant of.

edit: i remember my source of confusion... it’s the magnetic field strength of a solenoid which is proportional to amps not watts (not the ohmic heating)

”The magnetic field B is proportional to the current I in the coil.”

Solenoids as Magnetic Field Sources
Why the link? You were the only one confused on this point. The rest of us studied physics.

15. Originally Posted by devin-m
ive corrected my post... i’d heard before heat production was proportional to amps not watts, but after re-checking ohmic heating i found it’s proportional to watts.

edit: i remember my source of confusion... it’s the magnetic field strength of a solenoid which is proportional to amps not watts (not the ohmic heating)

”The magnetic field B is proportional to the current I in the coil.”

Solenoids as Magnetic Field Sources
If you knew even a little bit of basic science, you would know that watts is joules/sec, i.e that "power" is, by its very definition, the rate of energy transfer.

Hence, to say that a heating effect is not proportional to watts is a very silly and obviously wrong statement.

Don't mislead people who have come here to learn. It's a disservice to them and to the rest of us.

16. Originally Posted by exchemist
Originally Posted by devin-m
ive corrected my post... i’d heard before heat production was proportional to amps not watts, but after re-checking ohmic heating i found it’s proportional to watts.

edit: i remember my source of confusion... it’s the magnetic field strength of a solenoid which is proportional to amps not watts (not the ohmic heating)

”The magnetic field B is proportional to the current I in the coil.”

Solenoids as Magnetic Field Sources
If you knew even a little bit of basic science, you would know that watts is joules/sec, i.e that "power" is, by its very definition, the rate of energy transfer.

Hence, to say that a heating effect is not proportional to watts is a very silly and obviously wrong statement.

Don't mislead people who have come here to learn. It's a disservice to them and to the rest of us.
except heating isn’t always proportional to watts.

suppose we have 2 ac transmission lines... both transmitting the same 1,000,000 watts, one is stepped up to 200,000v and one is stepped up to 100,000v... both have the same power but, if i’m not mistaken, not the same rate of heat production... (since different # of amps)

17. Originally Posted by devin-m
except heating isn’t always proportional to watts.
Heating isn't just proportional to watts; heating is EQUAL to watts. In any closed system where only electricity is being added, all the electrical energy is converted to heat.

In open systems (cars, lasers) some energy is converted to other forms (kinetic, optical) and not all is converted to heat. But for a system like an electric heater, it is 100% efficient; 1 kilowatt of electrical energy is converted to 1 kilowatt of heat.

18. Originally Posted by devin-m
Originally Posted by exchemist
Originally Posted by devin-m
ive corrected my post... i’d heard before heat production was proportional to amps not watts, but after re-checking ohmic heating i found it’s proportional to watts.

edit: i remember my source of confusion... it’s the magnetic field strength of a solenoid which is proportional to amps not watts (not the ohmic heating)

”The magnetic field B is proportional to the current I in the coil.”

Solenoids as Magnetic Field Sources
If you knew even a little bit of basic science, you would know that watts is joules/sec, i.e that "power" is, by its very definition, the rate of energy transfer.

Hence, to say that a heating effect is not proportional to watts is a very silly and obviously wrong statement.

Don't mislead people who have come here to learn. It's a disservice to them and to the rest of us.
except heating isn’t always proportional to watts.

suppose we have 2 ac transmission lines... both transmitting the same 1,000,000 watts, one is stepped up to 200,000v and one is stepped up to 100,000v... both have the same power but, if i’m not mistaken, not the same rate of heat production... (since different # of amps)
Oh for Pete's sake, just stop making assertions until you have learnt the basics. You have no business opining anything at all about science because you are woefully ignorant, yet amazingly free of doubt.

Above, you have managed to entangle several errors.

The power delivered to a load is the power delivered to a load. If the powers are equal, they will cause the same heating.

In a transmission line, there will be losses. If two lines suffer the same power losses, they will get equally hot.

In your example, you have confused the power delivered to a load with the power losses suffered by the transmission lines. These are two different things.

Please refrain from offering "help" to students. You are in greater need of education than they are, and only sow confusion as you type.

Listen and learn.

19. To clear up my confusion:

If I have a 100v 10a rms 10 turns sine wave ac primary, transformer core and 2 secondary test conductors. The 2 secondary conductors have the same length, cross section and therefore ohmic resistance (1 ohm).

Assuming 100% transformer efficiency:

Suppose 10 turns of secondary conductor A are wound around the transformer core. What is the wattage & # amps in conductor loop A?

Suppose conductor A is removed and 20 turns of secondary conductor B are wound around the transformer core. What is the wattage & # amps in conductor loop B?

Does ohmic heating in conductors A and B occur at the same rate? (my understanding is these conductors will have different current densities - specifically less amps and current density in conductor B but higher voltage & the same watts)

20. Originally Posted by devin-m
To clear up my confusion:

If I have a 100v 10a rms 10 turns sine wave ac primary, transformer core and 2 secondary conductors. The 2 secondary conductors have the same length, cross section and therefore ohmic resistance (1 ohm).

Assuming 100% transformer efficiency:

Suppose 10 turns of secondary conductor A are wound around the transformer core. What is the wattage & # amps in conductor loop A?

Suppose 20 turns of secondary conductor B are wound around the transformer core. What is the wattage & # amps in conductor loop B?

Does ohmic heating in conductors A and B occur at the same rate? (my understanding is these conductors will have different current densities - specifically less amps and current density in conductor B but higher voltage & the same watts)
Your confusion is indeed profound. Your problem is ill-posed, as it posits counter-scientific assumptions.

If you assume 100% efficiency, there cannot be ohmic heating within the transformer.

If there is 100% efficiency there is no net power in the windings. It is purely reactive. However, your introduction of ohmic resistance is incompatible with your assumption of 100% efficiency.

And on it goes.

I don't know why you would introduce a transformer at all. Start with energy conservation and the definitions of power. Only after you've gotten those solidly understood should you venture into more elaborate constructions.

21. So how does using higher voltage improve ac transmission efficiency if joule heating in the lines is proportional to watts and not amps? My understanding is greater AC power can be transmitted for the same amount of line heating by utilizing higher voltages.

22. Originally Posted by devin-m
To clear up my confusion:

If I have a 100v 10a rms 10 turns sine wave ac primary, transformer core and 2 secondary test conductors. The 2 secondary conductors have the same length, cross section and therefore ohmic resistance (1 ohm).

Assuming 100% transformer efficiency:

Suppose 10 turns of secondary conductor A are wound around the transformer core. What is the wattage & # amps in conductor loop A?

Suppose conductor A is removed and 20 turns of secondary conductor B are wound around the transformer core. What is the wattage & # amps in conductor loop B?

Does ohmic heating in conductors A and B occur at the same rate? (my understanding is these conductors will have different current densities - specifically less amps and current density in conductor B but higher voltage & the same watts)
Yes, it occurs at the same rate - zero - since the transformer is 100% efficient.
So how does using higher voltage improve ac transmission efficiency if joule heating in the lines is proportional to watts and not amps?
Because losses due to both DC and AC resistance is proportional to I^2R. Lower current = lower resistive losses.

23. Originally Posted by devin-m
So how does using higher voltage improve ac transmission efficiency if joule heating in the lines is proportional to watts and not amps? My understanding is greater AC power can be transmitted for the same amount of line heating by utilizing higher voltages.

I already explained it, but you didn't bother to take notice. You must stop confusing the power delivered to a load, with the power lost along the way.

The power LOST in transmission is due to line efficiency. That leaves less for the load.

The power LOST in a resistance is proportional to the SQUARE of current.

The power DELIVERED TO A LOAD is proportional to the PRODUCT of V and I.

You can figure out the answer from those facts, but I'll give it to you: Higher V means less I is needed to deliver that power to a load.

Lower I means less power LOST by the LINE in transmission.

That's why we step up the voltage for long-distance transmission. In Tesla's time that was done most easily with a transformer and thus required AC. Today we can use DC step-up means which are more efficient still.

24. devin-m: your proclivity to make statements without knowing the actual facts is embarrassing - do NOT post in the hard science sub-fora again unless it's to ask questions.

25. I am saying with 2 identical 1 ohm sections of wire used as heaters... one with 100v 10a ac, the other 200v, 5a ac... both have the same 1000w electrical power, but the heating is less with the 200v 5a since heating is proportional to the amps squared... & both wire sections have the same resistance but different amps... this is what I mean by heating proportional to amps rather than watts electrical.

26. Originally Posted by devin-m
I am saying with 2 identical 1 ohm sections of wire used as heaters... one with 100v 10a ac, the other 200v, 5a ac... both have the same 1000w electrical power, but the heating is less with the 200v 5a since heating is proportional to the amps squared... & both wire sections have the same resistance but different amps... this is what I mean by heating proportional to amps rather than watts electrical.
No.

If they both take 1000 watts of power, they both generate 1000 watts of heat. Period.

Your math doesn't work for the first part of your sentence. A 1 ohm heater resistance will not draw 10 amps at 100 volts, nor will it draw 5 amps at 200 volts.

27. In the case of transformers:

If I have 2 secondary test loops (each 1 ohm), wrap 10 turns around primary with one and 20 turns with the other, the electrical power in both secondaries should be the same but different voltages and amperages ac rms and current density and heating. Both secondaries have the same resistance but different amps therefore different heating.

28. Which part of post #23 did you not understand?
See you in 3 days.

29. Originally Posted by devin-m
In the case of transformers:

If I have 2 secondary test loops (each 1 ohm), wrap 10 turns around primary with one and 20 turns with the other, the electrical power in both secondaries should be the same but different voltages and amperages ac rms and current density and heating. Both secondaries have the same resistance but different amps therefore different heating.
While you're on your break, I'll point out again that you shouldn't be fooling around with transformer examples until you learn certain basic concepts first.

I have already taken the pains to identify at least one of your problems, and your response was, unfortunately, completely expected -- you ignored what I said, and you continued to spout ever more stupid nonsense.

So, the very first bit of advice that you must accept is that you don't know what you are talking about. At all. Your ignorance is painful enough, but you compound that problem by fearlessly carrying on as if you had authoritative knowledge. So stop that right now. Be humble, for you have a great deal to be humble about (my apologies to Sir Winston).

Second, go back and READ what I and others have been attempting to teach you. REREAD it until you understand it. Stop insisting that you're right. You're either wrong, or not even wrong in virtually every one of your posts. That's quite a record.

Example: Stop confusing power delivered with power lost. Once you add transformers, you manage to add a third error: You mix up the power RATING of a transformer with the first two quantities. You aren't even able to carry out a calculation using energy conservation principles as a constraint.

You have a seriously bad habit of beginning with a conclusion (based on a faulty gut instinct, evidently), and then searching desperately for phrases that seem to support your beliefs. That was in evidence in your motor example, where you were so blind to reality that you didn't even recognize a dimensional-consistency argument as thoroughly discrediting your assertion at quite fundamental levels.

If you return here and follow the constraints that have now been imposed on you, begin by solving the simplest problem of non-superconducting wires delivering power to a resistance. Once you have managed to do that correctly, then you might be ready to move onto a configuration that involves a transformer. But until you understand the first example, there is absolutely nothing to be gained by leaping to the transformer one. All that does is allow you to flail in deeper waters.

30. *post deleted by author*

31. Originally Posted by billvon
Originally Posted by devin-m
except heating isn’t always proportional to watts.
Heating isn't just proportional to watts; heating is EQUAL to watts. In any closed system where only electricity is being added, all the electrical energy is converted to heat.

In open systems (cars, lasers) some energy is converted to other forms (kinetic, optical) and not all is converted to heat. But for a system like an electric heater, it is 100% efficient; 1 kilowatt of electrical energy is converted to 1 kilowatt of heat.
Originally Posted by billvon
Originally Posted by devin-m
I am saying with 2 identical 1 ohm sections of wire used as heaters... one with 100v 10a ac, the other 200v, 5a ac... both have the same 1000w electrical power, but the heating is less with the 200v 5a since heating is proportional to the amps squared... & both wire sections have the same resistance but different amps... this is what I mean by heating proportional to amps rather than watts electrical.
No.

If they both take 1000 watts of power, they both generate 1000 watts of heat. Period.

Your math doesn't work for the first part of your sentence. A 1 ohm heater resistance will not draw 10 amps at 100 volts, nor will it draw 5 amps at 200 volts.
suppose I have a 50v battery, 100v battery, resistive heating element and mosfet switch with duty cycle % control (pulsed dc on-off time % control)

first I connect the 50v battery to the circuit and adjust the duty cycle % of the mosfet such that 1 amp average per second at 50v (50w) flows from the battery through the heating element.

next I connect the 100v battery to the circuit and adjust the duty cycle % of the mosfet such that 1 amp average per second at 100v (100w) flows from the battery through the heating element.

if it takes 5 minutes for the water to boil with the 50v 1amp (50w) avg heating element, how long does it take with the 100v 1amp (100w) avg identical heating element?

32. Originally Posted by devin-m
suppose I have a 50v battery, 100v battery, resistive heating element and mosfet switch with duty cycle % control (pulsed dc on-off time % control)
OK.
first I connect the 50v battery to the circuit and adjust the duty cycle % of the mosfet such that 1 amp average per second at 50v (50w) flows from the battery through the heating element.
OK. If the duty cycle that happened at was 100%, then your heating element is 50 ohms (R=V/I.) If it happened at 50% duty cycle, it would be 25 ohms (R=((V/I)/2) ). So the most we can say at this point is that the heating element is 50 ohms or less.
next I connect the 100v battery to the circuit and adjust the duty cycle % of the mosfet such that 1 amp average per second at 100v (100w) flows from the battery through the heating element.
OK. So now we still know that the heating element is 50 ohms or less. The duty cycle would have to be significantly lower to achieve the same average current flow at the higher voltage.
if it takes 5 minutes for the water to boil with the 50v 1amp (50w) avg heating element, how long does it take with the 100v 1amp (100w) avg identical heating element?
Pretty easy to figure out. In the first case, average heat output is (50*1) = 50 watts. In the second case, average heat output is (100*1) = 100 watts. In the second case the water would boil roughly 2x faster, although heat loss would prevent it from being exactly 2x.

33. Originally Posted by billvon
Originally Posted by devin-m
suppose I have a 50v battery, 100v battery, resistive heating element and mosfet switch with duty cycle % control (pulsed dc on-off time % control)
OK.
first I connect the 50v battery to the circuit and adjust the duty cycle % of the mosfet such that 1 amp average per second at 50v (50w) flows from the battery through the heating element.
OK. If the duty cycle that happened at was 100%, then your heating element is 50 ohms (R=V/I.) If it happened at 50% duty cycle, it would be 25 ohms (R=((V/I)/2) ). So the most we can say at this point is that the heating element is 50 ohms or less.
next I connect the 100v battery to the circuit and adjust the duty cycle % of the mosfet such that 1 amp average per second at 100v (100w) flows from the battery through the heating element.
OK. So now we still know that the heating element is 50 ohms or less. The duty cycle would have to be significantly lower to achieve the same average current flow at the higher voltage.
if it takes 5 minutes for the water to boil with the 50v 1amp (50w) avg heating element, how long does it take with the 100v 1amp (100w) avg identical heating element?
Pretty easy to figure out. In the first case, average heat output is (50*1) = 50 watts. In the second case, average heat output is (100*1) = 100 watts. In the second case the water would boil roughly 2x faster, although heat loss would prevent it from being exactly 2x.
Assuming the resistor is 50ohms, & average current in both cases (50v @ 100% duty & 100v @ XX.XXX% duty) is 1 amp.

Since thermal heating is calculated via I^2R, isn't thermal heating in both resistors equal to 50w? Shouldn't both heat at the same rate since average current is the same?

50v / 1a / 50w / 50ohm / 100% duty

I = 1a = Average Current

R = 50ohm = Resistance

I^2R

1^2*50 = 50w heating

-------------------

100v / 1a / 100w / 50ohm / XX.XXX% duty

I = 1a = Average Current

R = 50ohm = Resistance

I^2R

1^2*50 = 50w heating ?

----------------

Source: https://en.wikipedia.org/wiki/Joule_heating

"Joule's first law, also known as the Joule–Lenz law,[1] states that the power of heating generated by an electrical conductor is proportional to the product of its resistance and the square of the current"

Source: https://en.wikipedia.org/wiki/Joule_heating

34. Originally Posted by devin-m
[Assuming the resistor is 50ohms, & average current in both cases (50v @ 100% duty & 100v @ XX.XXX% duty) is 1 amp.

Since thermal heating is calculated via I^2R, isn't thermal heating in both resistors equal to 50w? Shouldn't both heat at the same rate since average current is the same?
No.

The two scenarios are quite different. For one, the voltages are not the same! Ignoring that is very strange (and wrong).

Power is the product of voltage and current. Just because the currents are the same does not mean the power is the same.

By obsessively focussing only on current, you keep getting the wrong answer. You are, in essence, not taking into account that the average of a square is not the square of the average. When the switch is closed the resistor dissipates a power of (100v)^2/50ohms = 200W, and the current through the resistor is 2A. When the switch is open, the resistor dissipates nothing. Let's say that the duty cycle is 50%. The average current is 1A, from which your wrong reasoning would say that the average power is (1A)^2*50hms = 50W. However, the average is clearly (200W + 0W)/2 = 100W. I don't know why you keep avoiding doing this simple maths. You seem to trust in your intuition too much. You have none to trust. That's a harsh lesson, I know, but you need to learn it.

Again, the average of a square is not the square of the average!

----------------

Source: https://en.wikipedia.org/wiki/Joule_heating

"Joule's first law, also known as the Joule–Lenz law,[1] states that the power of heating generated by an electrical conductor is proportional to the product of its resistance and the square of the current"

Source: https://en.wikipedia.org/wiki/Joule_heating
Again, I don't know why you feel the need to post these links. We all know this material. You are the one who does not. By combining the Joule-Lenz law with Ohm's law, you could deduce that P = VI. With V = 100 volts and I = 1 ampere, you get 100W. By conservation of energy, that 100W is delivered to the resistor.

[It's faster to get to P=VI from P = dw/dt: V = dw/dq and I = dq/dt, where w is energy and q is charge]

35. Originally Posted by devin-m
Since thermal heating is calculated via I^2R, isn't thermal heating in both resistors equal to 50w? Shouldn't both heat at the same rate since average current is the same?
?? No. You are intentionally increasing power by increasing duty cycle in the case of the 100V supply.

Note that while you can set the AVERAGE current to be 1 amp, the PEAK current in the case of the 100V supply will be much higher. But since you have specified that you are using a PWM scheme to make the average current to be 1 amp, the PWM control will be throttling back the average current.

However, power is still equal to volts times amps. It is also equal to current squared times resistance, but in the 100V case, the peak current (when the switch is on) is much higher - so both equations still return the same numbers when the PWM switch is on.

You are getting yourself all confused due to your PWMing the load. Try it without the PWM and see if it makes more sense to you.

36. @billvon @tk421 I agree with & see what you are both saying with regards to the hypothetical, purely resistive load.

but for PWM driven BLDC motors (which have inductivity from the armature core), I am most curious about the heating component in the windings (hence my entire line of questioning).

My understanding is if the 50 ohm resistor is a stalled BLDC motor, the circuit has a significant amount of inductivity. Also in typical practice, the PWM frequency is high enough that the current (motor amps) seen by the motor windings is effectively constant (current smoothed out by the inductivity from the iron-56 in the stator core and high frequency PWM). The current is inductively recirculating through the motor during duty-cycle-off time.

Simply due to the inductance and high frequency switching, in a stalled motor we don't see on-2amps off-0amps, I think we see effectively 1 amp (with a slight ripple above and below) the whole time, is this correct?

In this case if we use the 50v battery at 100% duty and the 100v battery at 50% duty with the motor stalled, are the following values correct?

------------------

1 motor amp, 1 battery amp, 50v effective pwm voltage, 50w motor heating

------------------

100v battery, 50ohm inductive winding lead to lead, 50% duty (high frequency)

1 motor amp, 0.5 battery amp, 50v effective pwm voltage, 50w motor heating

------------------

Source: https://electronics.stackexchange.co...-a-pulse-train

"...If your switching frequency is fast enough, then the current changes little over a pulse cycle due to the inductance. This is in contrast to a resistor where current is always proportional to the applied voltage.

If you switch slowly compared to the resistive/inductive time constant of the windings, then the resistance dominates. In that case, the RMS formula (after subtracting the back EMF due to motor motion from the applied voltage) tells you how much power you are delivering to the motor, but the torque is still proportional to the average current, which is proportional to the voltage times the duty cycle..."

Source: https://electronics.stackexchange.co...-a-pulse-train

Image Caption: BLDC Electric Motors

37. Originally Posted by devin-m
@billvon @tk421 I agree with & see what you are both saying with regards to the hypothetical, purely resistive load.
ok, that's progress. Just analyze things mathematically. Your intuition is too ill-formed to trust your jumps to conclusions. Work toward the conclusion from the premises. Stop working backwards from a preconceived result.

but for PWM driven BLDC motors (which have inductivity from the armature core), I am most curious about the heating component in the windings (hence my entire line of questioning).
All you need to know is the current through the windings as a function of time and the total resistance through which that current flows. If the current is substantially constant, then square it and multiply by the resistance. That will give you the power absorbed in that resistance.

P = I^2*R

If the current is not substantially constant, then you have to perform an integration.

My understanding is if the 50 ohm resistor is a stalled BLDC motor, the circuit has a significant amount of inductivity.
The word is inductance. And an inductance will be there all the time. That's what coiled wires are. If the flux density is high enough to saturate the core, then the inductance will be lower in the stalled state than when running. But some inductance will always be there.

Also in typical practice, the PWM frequency is high enough that the current (motor amps) seen by the motor windings is effectively constant (current smoothed out by the inductivity from the iron-56 in the stator core and high frequency PWM). The current is inductively recirculating through the motor during duty-cycle-off time.

Simply due to the inductance and high frequency switching, in a stalled motor we don't see on-2amps off-0amps, I think we see effectively 1 amp (with a slight ripple above and below) the whole time, is this correct?
Maybe. It depends on the L/R time constant. If it is large compared to the switching period, then yes. If no, then no. https://en.wikipedia.org/wiki/Time_constant

In this case if we use the 50v battery at 100% duty and the 100v battery at 50% duty with the motor stalled, are the following values correct?

------------------

1 motor amp, 1 battery amp, 50v effective pwm voltage, 50w motor heating

------------------
ok so far.

100v battery, 50ohm inductive winding lead to lead, 50% duty (high frequency)

1 motor amp, 0.5 battery amp, 50v effective pwm voltage, 50w motor heating

------------------
How is the battery current different from the motor current?

You need to be much more precise in your language. You say "1 motor amp". Does that mean 1A average? Peak?

I think one reason you have trouble getting the calculations right is that you don't have the picture right in your mind. If you are less sloppy in your thinking, you will have a better chance of getting the answer right.

If the "battery amp" is an average, then the battery will supply 50W. The motor eats that.

Source: https://electronics.stackexchange.co...-a-pulse-train

"...If your switching frequency is fast enough, then the current changes little over a pulse cycle due to the inductance. This is in contrast to a resistor where current is always proportional to the applied voltage.
Yeah, so what? Again you are posting unnecessarily. You are the only one here who does NOT know these basic facts (which are all easily derived in undergraduate introductory physics), so I ask again that you stop wasting screen space with this idiocy.

You really need to read https://en.wikipedia.org/wiki/Dunnin...3Kruger_effect

"The Dunning–Kruger effect is a cognitive bias wherein people of low ability suffer from illusory superiority, mistakenly assessing their cognitive ability as greater than it is."

38. Originally Posted by tk421

How is the battery current different from the motor current?

During duty cycle ON, mosfets A & F are on.

During duty cycle OFF, mosfets D & F are on, and the current inductively recirculates through the motor during the times that battery current is zero.

Since the motor current continues to circulate while the battery amps are zero, the numerical value of the “motor amps” and “battery amps” are different.

^ This type of electronic speed controller is called a VESC

—————

If you don’t believe me, we can see from these electric skateboard data logging videos that at all times during throttle use the motor current is different from the battery current:

Logging Video 1: https://youtu.be/nGb-zt2Jp9k

Logging Video 2: https://youtu.be/R8FO4BkfFZ4

&

Battery amps = motor amps * (duty cycle %)

for example:

0.5 battery amps = 1 motor amps * (50/100 [50%] duty cycle)

39. Originally Posted by devin-m
Originally Posted by tk421

How is the battery current different from the motor current?

During duty cycle ON, mosfets A & F are on.
Oh good grief. Did you bother to read anything that followed, or did you immediately go to your habit of jumping to conclusions?

Go back and read what I fully wrote.

Originally Posted by what_devin-m_ignored
You need to be much more precise in your language. You say "1 motor amp". Does that mean 1A average? Peak?
I understand how the circuit works. What you continue not to understand is that your sloppiness at language is inhibiting you from understanding what it is going on. That is the point I make in the post. I then proceed to give you an answer to what I think is the best way to interpret your question (but you are making us guess, unnecessarily). You have already multiplied various quantities at random, invented a false conundrum, and misinterpreted the result several times. There is no "the" motor current. It varies with time. And yet you give a single number. That's the problem. You still don't get it, and instead prattle on about irrelevancies.

I can't decide if you are truly this dense, or if you are a troll.

If you paste another giant irrelevant picture, I will reach through the internet and slap you.

You really need to read https://en.wikipedia.org/wiki/Dunnin...3Kruger_effect

"The Dunning–Kruger effect is a cognitive bias wherein people of low ability suffer from illusory superiority, mistakenly assessing their cognitive ability as greater than it is."

40. did you not see the data logging videos? please see where it says “battery current” and “motor current”

these are related via the equation:

battery amps = motor amps * duty cycle %

13.42 battery = 17.34 motor * (77.10 duty / 100)

41. Originally Posted by devin-m
did you not see the data logging videos? please see where it says “battery current” and “motor current”

these are related via the equation:

battery amps = motor amps * duty cycle %

13.42 battery = 17.34 motor * (77.10 duty / 100)
You are beyond hope.

I'm out.

https://en.wikipedia.org/wiki/Resistor

https://en.wikipedia.org/wiki/Inductor

https://en.wikipedia.org/wiki/Platypus

doublefacepalm_Riker_Picard.jpgdoublefacepalm_Riker_Picard.jpgdoublefacepalm_Riker_Picard.jpg

42. Originally Posted by devin-m
In this case if we use the 50v battery at 100% duty and the 100v battery at 50% duty with the motor stalled, are the following values correct?

------------------

1 motor amp, 1 battery amp, 50v effective pwm voltage, 50w motor heating
------------------
100v battery, 50ohm inductive winding lead to lead, 50% duty (high frequency)
1 motor amp, 0.5 battery amp, 50v effective pwm voltage, 50w motor heating
------------------
Sort of. In your example, the motor inverter is effectively reducing the 100V battery to a 50V supply. So you see similar, but not identical, behavior. (For example you will see higher losses in the 100V case due to switching losses.)

43. Originally Posted by tk421
Originally Posted by devin-m
did you not see the data logging videos? please see where it says “battery current” and “motor current”

these are related via the equation:

battery amps = motor amps * duty cycle %

13.42 battery = 17.34 motor * (77.10 duty / 100)
You are beyond hope.

I'm out.

https://en.wikipedia.org/wiki/Resistor

https://en.wikipedia.org/wiki/Inductor

https://en.wikipedia.org/wiki/Platypus

doublefacepalm_Riker_Picard.jpgdoublefacepalm_Riker_Picard.jpgdoublefacepalm_Riker_Picard.jpg
Must say I'm disappointed you did not paste an actual picture of the last item - preferably with a load of extraneous ball bearings, into the bargain.

44. Originally Posted by tk421

How is the battery current different from the motor current?
Originally Posted by billvon
Originally Posted by devin-m
In this case if we use the 50v battery at 100% duty and the 100v battery at 50% duty with the motor stalled, are the following values correct?

------------------

1 motor amp, 1 battery amp, 50v effective pwm voltage, 50w motor heating
------------------
100v battery, 50ohm inductive winding lead to lead, 50% duty (high frequency)
1 motor amp, 0.5 battery amp, 50v effective pwm voltage, 50w motor heating
------------------
Sort of. In your example, the motor inverter is effectively reducing the 100V battery to a 50V supply. So you see similar, but not identical, behavior. (For example you will see higher losses in the 100V case due to switching losses.)
So if my friend has an electric skateboard with a 50v battery, and a 90kv (90 max rpm per volt no-load) hub motor which is 0.1 ohm lead-to-lead, and a reprogrammable VESC electronic speed controller....

& in the reprogrammable settings for the electronic speed controller, my friend chooses the following settings:

Battery Amp Limit: 20a
Motor Amp Limit: 20a

If I stall the motor and use full throttle, how many watts of heating will the motor experience (1000w or 40w)? How many amps will be drawn from the battery? What is the duty cycle? Torque at stall?

Watts heating:

I^2R=W

20a^2*0.1ohm=40w

Battery Amps:

40w/50v=0.8a battery amps

Duty Cycle:

40w / 20a motor amps = 2v effective pwm volts

(2v pwm effective / 50v battery) * 100 = 4% duty cycle

correct so far?

&

battery amps = motor amps * duty cycle %

0.8a [battery amps] = 20a [motor amps] * (4 [duty cycle %] / 100)

Torque at stall:

Kt (torque per amp) = 60/(2*pi*KV)

KT = 0.10610339

20a motor amps * 0.10610339kt torque per amp = 2.122 newton meters torque

correct?

45. Originally Posted by devin-m
If I stall the motor and use full throttle, how many watts of heating will the motor experience (1000w or 40w)? How many amps will be drawn from the battery? What is the duty cycle? Torque at stall?
IF everything is working exactly as you describe then you'd see about 60-80 watts in the motor. 40 watts in ohmic losses, 20-40 watts in core losses. It depends a great deal on the design of the motor. You would also be dissipating a significant amount of power in the controller, since you would be at a very short duty cycle (which it likely isn't designed for.) Battery would see about 3-4 amps being drawn from it (motor losses + controller losses.)

BTW if you want a good motor simulator for bikes, skateboards and similar small EV's check out:

Motor Simulator - Tools

46. Originally Posted by billvon
Originally Posted by devin-m
If I stall the motor and use full throttle, how many watts of heating will the motor experience (1000w or 40w)? How many amps will be drawn from the battery? What is the duty cycle? Torque at stall?
IF everything is working exactly as you describe then you'd see about 60-80 watts in the motor. 40 watts in ohmic losses, 20-40 watts in core losses. It depends a great deal on the design of the motor. You would also be dissipating a significant amount of power in the controller, since you would be at a very short duty cycle (which it likely isn't designed for.) Battery would see about 3-4 amps being drawn from it (motor losses + controller losses.)

BTW if you want a good motor simulator for bikes, skateboards and similar small EV's check out:

Motor Simulator - Tools
correct me if i’m wrong but if the motor is stalled i think there will be no core losses?

47. Originally Posted by devin-m
correct me if i’m wrong but if the motor is stalled i think there will be no core losses?
Why do you think that? To back up a bit - what do you think core loss is?

48. Originally Posted by billvon
Originally Posted by devin-m
correct me if i’m wrong but if the motor is stalled i think there will be no core losses?
Why do you think that? To back up a bit - what do you think core loss is?
the core loss would be eddy currents in the iron 56 core induced by changing magnetic fields and hysterisis while the motor is spinning

however if the motor is stalled the current in the motor effectively isn’t changing - steady 20a dc seen by motor windings (due to high frequency duty cycle and inductive load) so no alternating current induced eddy currents in the iron... keep in mind the current inductively recirculates during duty cycle off — & the switching frequency is high enough the current ripple is minimal

—————

Source: https://electronics.stackexchange.co...-a-pulse-train

"...If your switching frequency is fast enough, then the current changes little over a pulse cycle due to the inductance. This is in contrast to a resistor where current is always proportional to the applied voltage.”

49. I wanted to calculate the joule heating while the motor is in motion and putting out mechanical watts (before calculating iron loss). Are my calculations correct?

If I have a 90kv BLDC hub motor which is 0.1ohm lead to lead, a 50v battery, ESC with 20a motor amp limit, 20a battery amp limit, presently 2000rpm and full throttle

90kv 0.1ohm 50v battery 20a motor limit 20a battery limit 2000rpm

1/90kv = bemf v per rpm = 0.01111111 v

2000rpm * 0.01111111v = 22.2222222v bemf

(A-B)/C=D
A=B+(C*D)

A = pwm effective v
B = bemf v
C = ohms winding resistance
D = motor current

A=B+(C*D)
A=22.2222222v bemf+(0.1ohm *20a motor amps)
A=24.2222222v pwm effective v

24.2222222v pwm effective v * 20a motor current = 484.444444w electrical watts

484.444444w electrical watts / 50v battery = 9.68888888a battery amps

(24.2222222v pwm effective / 50v battery) * 100 = 48.44 % duty cycle

battery amps = motor amps * duty cycle %

9.68888888a battery amps = 20a motor amps * (48.44 [duty cycle %] / 100)

Kt (torque per amp) = 60/(2*pi*KV)

KT = 0.10610339

20a motor amps * 0.10610339kt torque per amp = 2.122 newton meters torque

2.122 newton meters torque * 209.43951 radians per second = 444.4306w mechanical

24.2222222v pwm effective v * 20a motor current = 484.444444w electrical watts

(444.4306w mechanical / 484.44444w electrical ) * 100 = 91.7402% electrical to mechanical conversion efficiency

I^2R=W

(20a motor amps * motor amps 20a)*0.1ohm = 40w joule heating

444.4306w mechanical + 40w joule heating = 484.4306w

9.68888a battery amps * 50v = 484.44w electrical

Simply I calculate that with 20a motor amps at 2000rpm with the 90kv 0.1ohm motor—- of the 484.44 electrical watts drawn from the battery, 40w is converted to joule heating in the motor, while 444.4306w is converted into rotational mechanical watts.

Additionally I observe the battery draw is 9.6888a battery amps @ 50v, motor current is 20a, back emf is 22.22222v, pwm effective v is 24.22222v. Electrical to mechanical conversion efficiency is 91.74% which is numerically the same as the ratio of BEMF to PWM effective V —(22.22222 / 24.22222)*100=91.74 — interestingly before calculating core losses the ratios of mechanical to electrical watts and back emf v to effective v are always the same. correct?

By my calculations the joule heating value (40w) is less than the electrical watts value (484.44w) drawn from battery (9.6888a * 50v) — so does this mean joule heating (for example in a spinning motor) is sometimes proportional to amps (20a motor * 20a motor)*0.1ohm = (40w) not electrical watts (484.44w)?

50. Originally Posted by devin-m
the core loss would be eddy currents in the iron 56 core induced by changing magnetic fields and hysterisis while the motor is spinning

however if the motor is stalled the current in the motor effectively isn’t changing - steady 20a dc seen by motor windings (due to high frequency duty cycle and inductive load) so no alternating current induced eddy currents in the iron... keep in mind the current inductively recirculates during duty cycle off — & the switching frequency is high enough the current ripple is minimal
OK, so let's back up.

A motor controller (inverter) is just a bunch of switches that turn on and off and connect the motor phases to B+ and B- respectively. If connected to a resistive load (as I think you are imagining the motor when stalled) you would get a pure square wave; on or off. The reason you get somewhat even current is because the inductance of the motor windings smooths out the current; it acts as a filter. You can't both discount the motor as an inductor and then also claim to benefit from the filtering effect.

So you have the motor acting as an inductor. That means you get several kinds of losses:

1) Resistive (DC) losses. This you can measure with a multimeter.

2) AC resistive losses, due primarily to skin effect. This is dependent on frequency and to a lesser extent motor geometry. Higher frequencies cause more loss, which is one reason most motor drives aren't very high frequency.

3) Core losses; losses caused by the magnetic materials constraining (and increasing) the flux. These are made of up of:

3a) Hysteresis losses. As you magnetize and demagnetize the stator, there is some loss every cycle; it's not 100% efficient.

3b) Eddy current losses. Any time you magnetize anything you generate eddy currents. In transformers this is largely intentional; you want to induce currents in the secondary. (But still minimize eddy currents in the core material.) In inductors, those eddy currents are all loss. They turn into heat eventually.

On to your second question. (BTW just as a general note - people who use 8 significant digits when talking about physical quantities are often falsely assuming more accuracy than really exists.)
By my calculations the joule heating value (40w) is less than the electrical watts value (484.44w) drawn from battery (9.6888a * 50v) — so does this mean joule heating (for example in a spinning motor) is sometimes proportional to amps (20a motor * 20a motor)*0.1ohm = (40w) not electrical watts (484.44w)?

In general, current is what drives a motor and determines torque and loss. Voltage overcomes back-EMF and allows higher operating speeds. This is not 100% accurate once you introduce an inverter because you get odd relationships between throttle (i.e. PWM) settings and losses due to higher frequency components - but it's a good general rule of thumb.

51. a BLDC motor is a synchronous motor so when it’s stalled there is no AC.

if we take a brushed dc motor with a commutator, stall it, and apply a steady voltage, we get dc through the winding not ac.

bldc is the same except the commutator is switched for an electronic speed controller which uses pwm to control lower effective voltage below pack voltage hence controlling the speed. since steady dc 20a is what the winding sees stalled with the 20a motor amp limit setting, there are no iron losses from ac.

when stalled, the battery sees 20 amps 4% of the time, averaging to 0.8 battery amps, while the winding sees 20a steady dc motor amps the entire time which is not AC.

52. See you in a week.

53. Originally Posted by devin-m
a BLDC motor is a synchronous motor so when it’s stalled there is no AC.
I just showed you how it was AC - because the motor inverter provides AC, not DC.
if we take a brushed dc motor with a commutator, stall it, and apply a steady voltage, we get dc through the winding not ac.
Yes. Now if you take a brushed DC motor with a commutator, stall it, and apply voltage from a motor controller, you get a lot of AC through the windings.
bldc is the same except the commutator is switched for an electronic speed controller which uses pwm to control lower effective voltage below pack voltage hence controlling the speed. since steady dc 20a is what the winding sees stalled with the 20a motor amp limit setting, there are no iron losses from ac. when stalled, the battery sees 20 amps 4% of the time, averaging to 0.8 battery amps, while the winding sees 20a steady dc motor amps the entire time which is not AC.
Sorry, you don't know what you are talking about.

54. Originally Posted by billvon
Originally Posted by devin-m
a BLDC motor is a synchronous motor so when it’s stalled there is no AC.
I just showed you how it was AC - because the motor inverter provides AC, not DC.
if we take a brushed dc motor with a commutator, stall it, and apply a steady voltage, we get dc through the winding not ac.
Yes. Now if you take a brushed DC motor with a commutator, stall it, and apply voltage from a motor controller, you get a lot of AC through the windings.
bldc is the same except the commutator is switched for an electronic speed controller which uses pwm to control lower effective voltage below pack voltage hence controlling the speed. since steady dc 20a is what the winding sees stalled with the 20a motor amp limit setting, there are no iron losses from ac. when stalled, the battery sees 20 amps 4% of the time, averaging to 0.8 battery amps, while the winding sees 20a steady dc motor amps the entire time which is not AC.
Sorry, you don't know what you are talking about.
this information is from a moderator [rew] at the user discussion forum for the VESC hardware devin is talking about. this information is in regards to using the VESC in DC mode to drive a brushed DC motor:

—-

vedder.se/forums/viewtopic.php?f=6&t=883

“Where devin says "coil" he means motor.

The battery, for example sees 10A, 10% of the time, or an average of 1A, while the motor sees the current of 10A running all the time, as the current recirculates through another mosfet. You could think of that 10A being constant due to the inductance of the motor. This is approximated when the inductance of the motor is very high and the PWM frequency too. In reality the current will rise from say 9A to 11A during the 10% on-time, and then decay from 11A to 9A during the 90% off-time.”

55. Originally Posted by bldc-user
this information is from a moderator [rew] at the user discussion forum for the VESC hardware devin is talking about. this information is in regards to using the VESC in DC mode to drive a brushed DC motor:
Sock puppets are frowned upon here.

You don't know very basic electrical theory. You would be well served by going back to the basics, getting a strong foundation in that, and then take up more complex topics like AC losses in motors.

56. Originally Posted by bldc-user
Originally Posted by billvon
Sorry, you don't know what you are talking about.
this information is from a moderator [rew] at the user discussion forum for the VESC hardware devin is talking about. this information is in regards to using the VESC in DC mode to drive a brushed DC motor:
Devin, sock-puppetry is a bannable offense. In many forums, it's an "insta ban" transgression.

As billvon says, you don't have the slightest idea what you are talking about. You are a poster child for Dunning-Kruger syndrome. Several of us have tried to educate you, but it is hopeless. You rudely ignore what people have taken pains to write and continue to prattle on about irrelevancies. You don't even seem to understand that energy is conserved. All the time. If your calculation shows otherwise, you are in error. Full stop.

If you are permitted to return, I'm sure it will be under tightly circumscribed conditions.

57. bldc-user permabanned, devin-m's ban extended to one month.

58. Originally Posted by billvon
Originally Posted by devin-m
a BLDC motor is a synchronous motor so when it’s stalled there is no AC.
I just showed you how it was AC - because the motor inverter provides AC, not DC.
@billvon - i am looking at this BLDC motor commutation chart. to me it appears if the motor is stalled, we get DC. why do you to believe when a BLDC motor is stalled there can be AC?

Originally Posted by billvon
Originally Posted by devin-m
if we take a brushed dc motor with a commutator, stall it, and apply a steady voltage, we get dc through the winding not ac.
Yes. Now if you take a brushed DC motor with a commutator, stall it, and apply voltage from a motor controller, you get a lot of AC through the windings.
@billvon - i am looking at the VESC user discussion forum at the section on using brushed dc motors with the motor controller in "current control" mode:

Link: Brushed DC motor connections - vedder.se forums

are you saying you believe a DC motor can be successfully driven with an AC supply?

----------------

Source: https://en.wikipedia.org/wiki/Brushed_DC_electric_motor

"A brushed DC motor is an internally commutated electric motor designed to be run from a direct current power source."

Source: https://en.wikipedia.org/wiki/Brushed_DC_electric_motor

-------------------------

Originally Posted by billvon
Originally Posted by devin-m
bldc is the same except the commutator is switched for an electronic speed controller which uses pwm to control lower effective voltage below pack voltage hence controlling the speed. since steady dc 20a is what the winding sees stalled with the 20a motor amp limit setting, there are no iron losses from ac. when stalled, the battery sees 20 amps 4% of the time, averaging to 0.8 battery amps, while the winding sees 20a steady dc motor amps the entire time which is not AC.
Sorry, you don't know what you are talking about.
Source: https://en.wikipedia.org/wiki/Buck_converter

"A buck converter (step-down converter) is a DC-to-DC power converter which steps down voltage (while stepping up current) from its input (supply) to its output (load). It is a class of switched-mode power supply (SMPS) typically containing at least two semiconductors (a diode and a transistor, although modern buck converters frequently replace the diode with a second transistor used for synchronous rectification) and at least one energy storage element, a capacitor, inductor, or the two in combination. To reduce voltage ripple, filters made of capacitors (sometimes in combination with inductors) are normally added to such a converter's output (load-side filter) and input (supply-side filter)."

"Switching converters (such as buck converters) provide much greater power efficiency as DC-to-DC converters than linear regulators, which are simpler circuits that lower voltages by dissipating power as heat, but do not step up output current.[2]"

"Continuous modeA buck converter operates in continuous mode if the current through the inductor (IL) never falls to zero during the commutation cycle."

Source: https://en.wikipedia.org/wiki/Buck_converter

-------------------

Originally Posted by devin-m
(24.2222222v pwm effective / 50v battery) * 100 = 48.44 % duty cycle
Originally Posted by billvon
Originally Posted by devin-m
bldc is the same except the commutator is switched for an electronic speed controller which uses pwm to control lower effective voltage below pack voltage hence controlling the speed. since steady dc 20a is what the winding sees stalled with the 20a motor amp limit setting, there are no iron losses from ac. when stalled, the battery sees 20 amps 4% of the time, averaging to 0.8 battery amps, while the winding sees 20a steady dc motor amps the entire time which is not AC.
Sorry, you don't know what you are talking about.

59. I wonder how long the ban will be, this time?

60. Oh look, tons of irrelevant but sparkly images again....

61. Originally Posted by Paleoichneum
Oh look, tons of irrelevant but sparkly images again....
I deplore the absence of ball bearings, however.

62. Originally Posted by exchemist
Originally Posted by Paleoichneum
Oh look, tons of irrelevant but sparkly images again....
I deplore the absence of ball bearings, however.
Agreed. Disappointingly lacking in Heath Robinsonian complexity.

63. Originally Posted by tk421
Originally Posted by exchemist
Originally Posted by Paleoichneum
Oh look, tons of irrelevant but sparkly images again....
I deplore the absence of ball bearings, however.
Agreed. Disappointingly lacking in Heath Robinsonian complexity.
@exchemist - ball bearings just for you...

64. Originally Posted by exchemist
Originally Posted by Paleoichneum
Oh look, tons of irrelevant but sparkly images again....
I deplore the absence of ball bearings, however.
Here you go...

65. Devin, all images and no understanding doesn't help you at all

66. Originally Posted by Paleoichneum
@Paleoichneum - I am looking at the motor simulator (found at bavaria-direct.co.za ). I find fairly close agreement between the numbers I calculated and the simulator. So why do you say I have no understanding?

Originally Posted by devin-m
I wanted to calculate the joule heating while the motor is in motion and putting out mechanical watts (before calculating iron loss). Are my calculations correct?

If I have a 90kv BLDC hub motor which is 0.1ohm lead to lead, a 50v battery, ESC with 20a motor amp limit, 20a battery amp limit, presently 2000rpm and full throttle

90kv 0.1ohm 50v battery 20a motor limit 20a battery limit 2000rpm

1/90kv = bemf v per rpm = 0.01111111 v

2000rpm * 0.01111111v = 22.2222222v bemf

(A-B)/C=D
A=B+(C*D)

A = pwm effective v
B = bemf v
C = ohms winding resistance
D = motor current

A=B+(C*D)
A=22.2222222v bemf+(0.1ohm *20a motor amps)
A=24.2222222v pwm effective v

24.2222222v pwm effective v * 20a motor current = 484.444444w electrical watts

484.444444w electrical watts / 50v battery = 9.68888888a battery amps

(24.2222222v pwm effective / 50v battery) * 100 = 48.44 % duty cycle

battery amps = motor amps * duty cycle %

9.68888888a battery amps = 20a motor amps * (48.44 [duty cycle %] / 100)

Kt (torque per amp) = 60/(2*pi*KV)

KT = 0.10610339

20a motor amps * 0.10610339kt torque per amp = 2.122 newton meters torque

2.122 newton meters torque * 209.43951 radians per second = 444.4306w mechanical

24.2222222v pwm effective v * 20a motor current = 484.444444w electrical watts

(444.4306w mechanical / 484.44444w electrical ) * 100 = 91.7402% electrical to mechanical conversion efficiency

I^2R=W

(20a motor amps * motor amps 20a)*0.1ohm = 40w joule heating

444.4306w mechanical + 40w joule heating = 484.4306w

9.68888a battery amps * 50v = 484.44w electrical

Simply I calculate that with 20a motor amps at 2000rpm with the 90kv 0.1ohm motor—- of the 484.44 electrical watts drawn from the battery, 40w is converted to joule heating in the motor, while 444.4306w is converted into rotational mechanical watts.

Additionally I observe the battery draw is 9.6888a battery amps @ 50v, motor current is 20a, back emf is 22.22222v, pwm effective v is 24.22222v. Electrical to mechanical conversion efficiency is 91.74% which is numerically the same as the ratio of BEMF to PWM effective V —(22.22222 / 24.22222)*100=91.74 — interestingly before calculating core losses the ratios of mechanical to electrical watts and back emf v to effective v are always the same. correct?

By my calculations the joule heating value (40w) is less than the electrical watts value (484.44w) drawn from battery (9.6888a * 50v) — so does this mean joule heating (for example in a spinning motor) is sometimes proportional to amps (20a motor * 20a motor)*0.1ohm = (40w) not electrical watts (484.44w)?

(it appears to me the small difference in mechanical watts (power out) between my calculations and simulator is a result of the simulator subtracting 0.2w iron loss & 8.9w ESC loss from the power out - mechanical watts | 444.43w mechanical watts [the answer i calculated] - 0.2w iron loss - 8.9w esc loss = 435.33w power out [the simulator result])

please note that "power in" appears to be 484.4w electrical and copper loss a.k.a. joule heating appears to be 40w - the same as I had previously calculated. to my eye the heating appears proportional to the amps 20a*20a*0.1ohm=40w and not the electrical watts 24.22222v*20a=484.4w - no?

the "motor efficiency" value I calculated [ 444.4306w mechanical / 484.44444w electrical ) * 100 = 91.7402% electrical to mechanical conversion efficiency ] appears to agree precisely with the simulator.

67. Originally Posted by Zwolver
Originally Posted by exchemist
Originally Posted by Paleoichneum
Oh look, tons of irrelevant but sparkly images again....
I deplore the absence of ball bearings, however.
Here you go...

Very, nice, thank you.

Reminds me of my induction course at Shell in lubrication:"a grease is a sponge......." When you lubricate these things you pack the housing on either side of the race with grease, from which a small amount of oil leaches out to do the actual lubrication, over a long period of time.

68. Devin, I am not an expert in the topic of this thread but I understand the basics, after reading the thread it is clear you don't. I have learned a lot from the posts of tk421 and billvon, it is equally clear you haven't. You seem to be immune to anything that disagrees with your flawed understanding and frankly you are wasting your time, and ours. Bugger off.

69. Originally Posted by PhDemon
Devin, I am not an expert in the topic of this thread but I understand the basics, after reading the thread it is clear you don't. I have learned a lot from the posts of tk421 and billvon, it is equally clear you haven't. You seem to be immune to anything that disagrees with your flawed understanding and frankly you are wasting your time, and ours. Bugger off.
I asked an anonymous rider to send me an acceleration log documenting loaded full throttle acceleration from standstill on an electric skateboard using a BLDC motor & VESC. It clearly shows around 90a motor amps produced in the motor from only 10a battery amps close to full throttle standstill:

This second graph taken by the same rider on the same board clearly shows (@ log 48 sec) that at about 28mph, the same 10a battery amps produces only about 28a motor amps instead of about 90a as in the 1st graph, a result of back emf voltage opposing the battery voltage more and more at faster and faster speeds - no?

The electrical wattage in both cases -- the 90a motor amps @ 10a battery amps -- and the 28a motor amps @ 10 battery amps -- is the same - no?

If he wants to draw 500w electrical (10a battery amps x 50v) full throttle at all physically possible speeds including close to standstill, his motor amp limit setting needs to be at least about 90a motor amps & battery amp limit 10a - no?

-----------

Source: https://en.wikipedia.org/wiki/Buck_converter

"A buck converter (step-down converter) is a DC-to-DC power converter which steps down voltage (while stepping up current) from its input (supply) to its output (load).

Source: https://en.wikipedia.org/wiki/Buck_converter

70. Time wasting muppet :shrug, :

71. Originally Posted by PhDemon
Time wasting muppet :shrug, :
His graphs are just not big enough. When I present at a conference, I use GargantuVision(tm) technology to project my slides. That makes everyone agree with whatever I say, every time.

72. Originally Posted by devin-m
Originally Posted by billvon
Originally Posted by devin-m
a BLDC motor is a synchronous motor so when it’s stalled there is no AC.
I just showed you how it was AC - because the motor inverter provides AC, not DC.
@billvon - i am looking at this BLDC motor commutation chart. to me it appears if the motor is stalled, we get DC. why do you to believe when a BLDC motor is stalled there can be AC?
Because motor inverters, unless they are running flat-out (which is a very bad idea when the motor is stalled) do not provide direct current to the motor. That would be very bad for both motor and battery, resulting in blown fuses, fires etc. Instead they limit current. How do they do that? By PWMing the FETs to control the current. What happens then? AC current flows through the windings. That's why AC losses AND DC losses matter, both at stall and at speed.

Now, you could reduce the voltage to the point where the motor can handle 100% current at stall without overheating. In that condition you might see only DC, and your chart would be valid. (And indeed, that's how very simple stepper motor applications work.) But you're not going to see motors intended for traction applications work like that.

are you saying you believe a DC motor can be successfully driven with an AC supply?
A motor inverter converts DC into AC to drive the motor. That's why it's called an "inverter" - because it regularly inverts the drive waveform to generate AC.
"A brushed DC motor is an internally commutated electric motor designed to be run from a direct current power source."
Yes! Brushed DC motors contain their own "inverter" - the commutator. That converts the DC from the motor terminals to the AC to run the motor.
It clearly shows around 90a motor amps produced in the motor from only 10a battery amps close to full throttle standstill:

Excellent! You may be about to achieve a clue!

If everything is just DC, and the motor is running on DC, and there's no AC - how is the controller generating 90 amps from a 10 amp supply?

73. Originally Posted by billvon
Excellent! You may be about to achieve a clue!

If everything is just DC, and the motor is running on DC, and there's no AC - how is the controller generating 90 amps from a 10 amp supply?
Let's suppose we have a 50v battery, vesc in brushed dc mode (with 2 wires leading to motor), and the resistance from lead to lead of the motor is measured 0.1ohm and we want to put 500w electrical through this stalled motor (100% copper loss).

Stalled there is no back emf voltage, so by ohm's law, if we simply short the 0.1ohm motor leads with 50v battery, we get:

....25000 watts and 500 amps @50v is too much for our goal (more than 500w)...

For a 0.1ohm winding, how many volts do we need to get 500w?

It turns out with a 0.1ohm winding we only need 7.07 volts and 70.71amps for 500w:

But the battery is 50v and therefore 500w is 10a battery amps (10a x 50v = 500w)

So we need to lower the 50v battery volts to 7.07v effective volts to the motor and at the same time we need to increase the amperage from 10a battery amps to 70.71a motor amps to achieve 500w in the winding.

@billvon Your question was "If everything is just DC, and the motor is running on DC, and there's no AC - how is the controller generating 90 amps from a 10 amp supply?"

my understanding on the issue is summarized by 3 posts on the VESC forum:

reposted to illustrate:

^(substituting the load in this diagram for a stalled 0.1ohm brushed dc motor) my understanding is during the on time mosfets A & F are on - and current is drawn from the battery, and during the off time, current isn't drawn from the battery, but the current generated during the on time is inductively circulating via mosfets F & D

my understanding is since the current continues to inductively circulate in the motor via mosfets F & D during the very brief off times when current isn't being drawn from the battery, the amps/current in the motor is greater than what is drawn from the battery, and this has the effect of lowering the effective voltage of the battery while increasing the number of amps in the motor above what was drawn from the battery, no?

74. Post #23. See you in 2 months.

75. Devin-m and his latest sockpuppet now permabanned. Apparently he doesn't learn at all...

76. Originally Posted by devin-m
my understanding is since the current continues to inductively circulate in the motor via mosfets F & D during the very brief off times when current isn't being drawn from the battery, the amps/current in the motor is greater than what is drawn from the battery, and this has the effect of lowering the effective voltage of the battery while increasing the number of amps in the motor above what was drawn from the battery, no?
You are getting closer. But first it is not a "very brief" time. In fact, if the supply is 50V and the motor needs 7V to produce the torque you want with a locked rotor, then the FET is going to be off for most of the time (in fact about 86% of the time.) For the short time the switches are closed, current ramps up rapidly. For the long time the switch is open, the current ramps down more slowly and freewheels through the body diodes of the FETs. (or IGBTs as in your example)

This, you may be interested to note, is the same way a standard buck converter works, which can also reduce voltage while increasing current.

The above action places an AC voltage across the winding. The AC voltage is quite dramatic, slamming from -1 volt to 51 volts in the case of a 50V battery. (You get the extra 1V or so due to the forward voltage of the FETs.) The inductance of the winding reduces the current swing, so if you want 70 amps you might actually see a range of (say) 40 to 100 amps, as the FETs open and close. This occurs too fast (>20KHz) to be noticeable as torque ripple, so for the mechanical purposes of the motor it is 70 amps DC. However the actual motor drive signal has quite a bit of AC current ripple on it and even more AC voltage ripple.

And that's why a BLDC motor sees AC when it is stalled (and being driven by an inverter.)

77. I never thought that a thread on such a small doubt of mine would go such a distance in complexity!! :��

78. Originally Posted by Gaurav(-26.7)
I never thought that a thread on such a small doubt of mine would go such a distance in complexity!! :��
Your perfectly fine OP was unfortunately hijacked by a loud-mouthed dolt with delusions of competence. Now that he's been properly dealt with, we'll have peace until the next Dunning-Kruger poster child makes an inevitable appearance.

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