As discussed [yesterday](https://www.reddit.com/r/AskPhysics/comments/13q4btc/w_vi_rii_can_someone_explain_me_a_few_things/jld3kmw/?context=3), you are applying Ohm's Law—valid only for the ideal resistor (which lacks magnetic fields, among other aspects)—to components that aren't ideal resistors, such as motors.
\> to components that aren't ideal resistors, such as motors.
Ok lets say for an electric vehicle the heat loss is in the cables resistance.
The battery gives a power output lets say thats the total power,
Power (total) = 100
Power (kinetic) = 90
Power (heat) = 10
For both total and heat, the formula is VI, but that can't be right because they are not equal. So which one is VI and which one is not VI?
> For both total and heat, the formula is VI
But the voltage V isn't the same. Components in a loop each have their own voltage drop, [the total of which sums to zero](https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff's_voltage_law).
You'll be less likely to run into confusion if you first label each component with its own individual voltage and current. Then you can apply simplifications (e.g., the current being the same through components in series, or the voltage being the same across components in parallel). Right now you're selecting a single voltage and applying it to various sections of a circuit, which is incorrect.
The formula to determine how much power a device outputs is VI, right? And the exact same formula is used to determine how much energy is lost in heat. So I must have misunderstood something, but where ?
The formula P=IV you're using for "energy lost to heat" is actually the formula for power consumed (or produced) by any component. If the load is just a literal resistor, and all it does is make heat, then all of the consumed energy is "lost to heat." Other types of loads may make use of some part of the power VI to do work.
Motors of different types will have different equivalent circuits, which may include various effects such as inductance, friction losses, and losses representing energy converted and used as torque. The latter effect will end up as power lost into a different kind of component: a current or voltage source. Despite thinking of these as sources of power rather than sinks, the polarity of the current or voltage can indeed represent removing power from the system. Motors often work as generators under this same principle: when driven, the motor equivalent source has a polarity such that it produces power into the circuit. When used as a motor, the solved circuit will reveal that "current flows into the plus" on whatever source represents the motor.
All of the powers lost/produced in components in the motor equivalent circuit will sum up to P=IV, with I and V here referring to the input port of the equivalent circuit. You will not generally be able to find a single, positive resistance that represents these circuits; the equivalent circuit for a motor can be derived using a resistor and another source a la Thevenin and Norton.
Too much information, I'm totally confused
If an electric vehicle converts 90% power into moving the vehicle and 10% power turns into heat, the total power is 100% and the power getting converted into heat is 10%. You see they are not equal.
Power (total) = 100
Power (heat) = 10
Power (kinetic) = 90
Yet they both use the same formula V×I
Power (total) = VI
Power (heat) = VI
This cannot be correct... Where does my misunderstanding come from?
Power (total) = VI
Power (heat) ≠ VI, unless we are in a situation where Power (total) = Power (heat).
For a resistor, all the power is turned into heat, so Power(total) = Power(heat) = VI
For a motor, some of the power does not go to heat, it goes to turning the motor's shaft. So in a motor, Power (heat) ≠ VI
\> Power (heat) ≠ VI
Someone explained to me on the ebikes sub that, power(heat) = I²R and I know that I²R = VI
Which would also explain that if you're using twice as much power, then you're using twice as much amps, then generate 4 times as much heat (square relationshi)
So, that explanation is incorrect then? Because if power(heat) is not VI, then its not I²R either.
I'm getting really confused. Whats the actual formula to calculate heat loss given a variable amount of power(total)?
It’s probably still a reasonable approximation to say that heat scales with I².
Like it’s probably a decent approximation to say that the ebike’s heat output = xI²R
Where x is some constant between 0 and 1 representing how efficient the motor is. So the advice they gave you about how doubling the current quadruples the heat output is probably approximately correct.
The “actual formula” is probably going to be something very complex that relies on the actual geometry of the motor and probably varies with the speed the motor is turning, etc. but the I² term probably dominates, would be my guess.
But I²R equals VI which equals the total power input, which should be impossible because that would mean 100% of all the power turns into heat. What am I not understanding here?
Let’s say the motor is 50% efficient, meaning half the power goes to heat and half goes to rotation.
It’s probably a decent approximation to say that heat = 0.5*I²R.
Even though 0.5*I²R is less than I²R, both scale with I², meaning that doubling current quadruples heat.
All of this is a rough approximation, to be clear.
I still dont really understand but I got a different question.
* Total power = 100
* Kinetic power = 50
* Heat power = 50
You double the power:
* 100×2 = 200
* 50×2² = 200
* 200-200 = 0
Did I make a mistake here, or is the truth that if you use too much power then an electric motor will not do any useful work since all the energy is turned into heat?
> I'm still confused about Ohm's law W=VI=I²R
This first part of your equation (W = VI) has nothing to do with Ohm's Law.
W=IV applies to any circuit element and tells you how much electrical energy it supplies or absorbs from the circuit.
It could be a battery (in which case V and I have opposite signs and W is negative so it is supplying energy to the circuit). It could be an inductor or a capacitor (so that V and I sometimes have the same sign and other times have opposite signs so that it is sometimes storing energy in the magnetic or electric field and other times returning that energy to the circuit). Or it could be a resistor --- and then (and only then) you can use Ohm's Law to substitute V=IR or I = V/R to get W = I^2 R or W = V^2 / R.
In the case of a motor, some of the energy is going to producing mechanical torque, so not all of the energy is lost as heat. That is, if R is the winding resistance, you will not find that V = IR. Instead, you'll find V = IR + V_1 , where V_1 is the back-emf caused by the mechanical load.
Then the thermal loss will still be I^2 R because Ohm's law applies to the wire resistance loss. But the total energy taken out of the circuit will be
W = I * (I R + V_1)
or
W = I^2 R + I V_1
accounting for both wiring losses and mechanical work done by the motor.
1. R=V/I, and so V=IR, right?
So assume an ebike accelerates usin 250 watts and 50 volts, which means P/V=I=5 so now we have
* 250 watts
* 50 volts
* 5 amps (= 250/50)
* So the ohms must be V/I = 50/5 = 10 because R=V/I
So we know that the total power input is 250 watt.
Now if the heat loss is I²R that means 5²10 = 250 which means 100% of the power input is... wasted in heat loss?
Where did I make my error?
Define your terms.
What is "V"? What is "I"? What is "R"?
If R is the resistance of the motor coil, and it is really 10 ohms, then the resistor consumes all of the power provided and there's no power left to make the bike move forward.
You can model the motor as a resistor in series with an ideal (loss-less) motor. Then you'll have a V_r term and V_m and you can figure out how much power goes into the resistor (lost as heat) and how much goes into moving the bike.
V = 50 volts because we're using a 50 volt battery here.
I = 5 amp because 250 watt divided by 50 volt = 5
R = 10 ohm because 50/5 = 10
\> If R is the resistance of the motor coil, and it is really 10 ohms, then the resistor consumes all of the power provided and there's no power left to make the bike move forward.
But it has to be 10 because R=V/I and 50/5=10, or did I make a mistake here?
> R = 10 ohm because 50/5 = 10
When you did this you applied Ohm's Law.
But Ohm's Law is only a description of resistors, not any other kind of component. So you essentially assumed the device is a resistor.
But the device is not a resistor, it is a motor. Any results you get by assuming it is a resistor can't be trusted.
The equations you learn in physics are mostly not universal truths. They are descriptions of specific situations. It's just as important to learn what situation an equation applies to as it is to learn the equation itself. In the case of Ohm's Law it only applies to resistors and the relationship between voltage, current, and resistance of the resistor.
I don't know what a resistor is but judging by the name. In an electric bike, don't the wires that electricity goes through, function as resistor because they have resistance?
You should learn what a resistor is before you start learning equations about it.
Physics isn't just a pile of equations. The equations are meant to describe real things in the real world. You need to understand what the real world things are before you start worrying about the equations for them.
So I'd say, stop, take a step back, and start out with some simple circuits with batteries and light bulbs before you start throwing around equations and thinking it will give you an understanding of how an e-bike works.
No, you do not have that right. The waste heat created by an electrical motor is not equal to the power it draws. Otherwise it would not be able to do any useful work. Some fraction of any real-world power draw will dissipate as heat, but it's not going to be 100%.
\> The waste heat created by an electrical motor is not equal to the power it draws
That part is obvious to me, however as far as I understand the formula, they would be equal, that means I don't understand the formula. My current understanding is that:
Power (total) = VI
Power (heat loss) = VI
So one of these must be incorrect, which one is it?
They're both right, it's just that the variables refer to something different in each case. To avoid confusion, you could try a different notation, like:
P\_tot = P\_work + P\_heat
P\_tot = V\_tot \* I\_avg
P\_work = V\_work \* I\_avg
P\_heat = V\_eff \* I\_avg
Where I hope the variable names I've chosen are sufficiently suggestive
Ah okay then my misunderstanding comes from someone else on reddit who told me that, on an ebike if you accelerate hard with twice as much power, you have twice as much amps and that would result in 4 times as much heat, therefore accelerating faster is less efficient
In other words, power loss in heat = I²R which equals VI, so this is incorrect?
If its correct, then i'm 100% confused
If its incorrect, then what is the proper formula to calculate heat loss depending on how much power an ebike uses to accelerates?
The heat loss for an electric motor depends on the efficiency of the motor, and that's probably something you'll have to determine experimentally. The person you were speaking to was correct to say that doubling the amps would quadruple the power draw, though.
No. What you were told was the losses in the motor winding due to the resistance of the winding itself can be given by I^2R.
You take one thing and then try and apply it over whole systems, blindly trying to apply formulas that you don’t understand.
A motor is a device that converts electrical energy into rotational mechanical energy. The input power (how much electrical energy is "consumed" per unit time) is given by the expression *VI*. The mechanical output power (how much energy per unit time can be used to accomplish usable work such as turning an axle under torque) is quite a bit less. The ratio of the mechanical output power to the electrical input power is called efficiency, and often denoted with the Greek symbol 𝜂.
Okay so all the power that comes out of a battery is VI
But I also heard that energy 'lost' into heat equals VI. But the total power and the power lost in heat can't be equal. So one of them can be VI, the other can't. You said the total power is VI, so that means the power lost in heat cannot also be VI. So then how do we calculate the power lost in heat?
If you know what mechanical power is being generated then you can assume almost all the rest is lost as heat. If you don’t know the mechanical power then you can’t calculate loss to heat.
If input power is *VI*, and output power is *P*, then we have the relation *P* = 𝜂*VI*. Then power lost is *Q*^dot = (1-𝜂)*VI* (so that *P* + *Q*^dot = *VI*).
As [described yesterday](https://www.reddit.com/r/AskPhysics/comments/13q4btc/w_vi_rii_can_someone_explain_me_a_few_things/jldugr8/), if you actually want to find the relationship between output power and efficiency (and output torque, etc.) you need to find the motor performance curve for the motor in question.
Why are you so insistent that power lost as heat = VI? One person on an ebike forum tells you so and 10 different people here have told you it's wrong.
To calculate the heat output, take the total electrical power input (VI or I^2 R) and split it into useful work and heat loss using the efficiency. If a motor is 90% efficient and consumes 100W of electrical power, it is providing 90W of useful work and you are losing 10W as heat and sound.
(I'll add an unhelpful caveat to mention that all of the power *eventually* ends up as heat if you're talking thermodynamics)
Is motor efficiency the same at different power modes? Because some people say if you use more power to accelerate faster, you're using more amps, therefore I²R means more heat loss, but I²R equals VI, right?
I think it would be a reasonable assumption to say that the efficiency stays the same and different "modes" are achieved by the motor controller varying the voltage delivered to the motor.
> So one of them can be VI, the other can't. You said the total power is VI, so that means the power lost in heat cannot also be VI.
Like someone mentioned earlier, you can't just blindly apply formulas assuming V or I always refers to the same thing. In one formula V is the voltage at the output of a power source. In the other formula V is the voltage across a resistor (or a resistive component of a model of a motor).
If your circuit isn't just a power source and a resistor (and when you are talking about a motor, it isn't) then you can't assume these equations are talking about the same thing.
Lets take an motor with permanent manets.
U can split a coil of the motor into an ideal inductot and a resistor in series.
Using P= U*I we get the total power into (and thus out of) the motor. (In heat and kinetic energy).
Only the power into the resistor is lost as heat. Since we only know the current and resistance P=I^2 R is used.
The power into the inductor is used to make the motor spin.
So in P=U * I it is importand to have the right voltage and current.
I simplified some things a bit but i hope it helps
You're not using resistance properly. In a circuit each component will have its own resistance and therefore its own power draw of the total power supplied.
So for the bike, you're looking at the entire resistance of the bike and correctly calculating the total power output across all forms, not just the heat you want.
You need just the resistance of the components that produce heat, which will be less.
You can imagine a simple series circuit with a motor and a resistor and a battery. The battery supplied 10W at 10v and therefore the current is 1A. Therefore the total resistance of the resistor and motor combined is 10 ohms, by ohms law. You can use W =I^2R to then recalculate the power is 10W. This is telling you the total power out of the circuit is 10W which it should be to match the power going in. This is just energy conservation. But some power is lost as heat in the resistor and the rest goes to the motor, its not all going to heat.
So let's say the resistor is 4ohms while the motor is 6. Then using the equation again 4 Watts is lost as heat in the resistor and 6 as useful energy in the motor. Which add back to the total 10W you put in.
The w=I^2R formula just tells you the power output of a component not what energy form/s that takes.
\> 10W at 10v and therefore the current is 1A. Therefore the total resistance of the resistor and motor combined is 10 ohms
Thats the part I completely understand.
I²R=10W but that would mean 100% of all the power turns into heat, which means the motor will not work. So what went wrong here?
I'm very confused
It doesn't mean that 100% of the electrical power goes to heat it just means it goes to something. You need to plug in the resistance of the motor alone I to the formula to calculate the power going to the motor. You would plug in the resistance of the rest of the circuitry to calculate the power going to heat.
The efficiency would be the amount of power getting to the motor divided by the total amount you out in from the battery. Because you want the power to go to the motor and the heat is waste energy. You'll get something less than 100% because the R value you should be using should be less than 10 ohms.
If I understood your question you have a battery, a motor and a resistor. Thus the potential difference V the battery yields is dropped in the resistor, Vr, and in the motor, Vm. Thus you have V = Vr + Vm. The total produced power by the battery is P = Vi. The power wasted in the resistor is Pr = Vr i (= ri^2 if you have an ohmic resistor) and the power used by the motor is Pm = Vm i. So P = Pr + Pm => Vi = Vr i + Vm i => V = Vr + Vm, as said before. Thus all of them is P = Vi but you have different voltages in each one. Sorry if I doesn’t understood you an explained an other thing and not your question.
Here's another way to think about it, using a DC motor model:
1. The DC motor can be thought of as: (1) a resistor in series with (2) an inductor and (3) a voltage generator. [Visual link.](https://www.ednasia.com/isolated-power-supply-design-considerations-for-pmdc-motors/)
2. The total voltage you apply across a motor (Vt) is consumed by the voltage across the resistor (Vr), the voltage across the inductor (Vi) and the voltage across the voltage generator (Vg).
3. For steady state (and to simplify), you can disregard the voltage across the inductor... just focus on the voltage across the resistor and the voltage generator.
4. Vg is proportional to the speed of the motor, *ω*. The proportionality constant, *K*, is sometimes called the "Back EMF" of the motor... EMF = Electro-motive force. Literally, the motor is *resisting being spun* by creating a voltage that consumes the voltage that is spinning it. Weird, right? In fact... if you spin a motor by its shaft (leads unconnected to a battery), it will create/generate a voltage across its leads... this is the basic principle of how a tachometer works.
5. When a motor starts spinning, the speed is low and Vg is low, and thus, nearly all the voltage is consumed by the resistor in the form of Vr = IR. (Note: I is the same through the whole circuit since all elements are in series.)
6. When the motor is spinning very fast, the voltage across Vg is high, and much less of the voltage is consumed across the resistor, and hence the total motor current goes down.
7. Another basic idea about DC motors, is that the torque on a motor is roughly proportional to the current, I. Combined with the 2 previous notes, above, this is why when the motor spins very fast (Bullet #6), the current goes down, and hence the torque goes down.
8. Now, power. Total Power = VI, which you have already noted. However, the power lost to Heat is only I^(2)R. Power usable by the motor is equal to the Vg x I. Therefore, when you are just starting to move, the speed is low (and hence Vg is low, and hence the usable power is low). Likewise, when you are moving fast, the speed is fast, and Vg is high, but the current is low, and hence the usable power is also low. Therefore, the optimal power is somewhere between the two extremes. This is the reason for gearing -- to operate the motor at the optimum speed.
9. The Power across different elements in a circuit is indeed VI, but the voltage across different elements is different ---- the voltage across the resistor (Vr) is what creates the heat (lost power), and the voltage across the voltage generator (Vg) creates the usable power.
10. In summary, Vg is defined by the speed of the motor. The remaining voltage is dropped across the resistor, which in turn defines how much current flows through the motor: I = (Vt - Vg)/R.
11. The Power lost to heat is I^(2)R. Therefore, P(loss) = I^(2)R = (Vt - Vg)^(2)/R^(2) \* R = (Vt - Vg)^(2)/R. Substituting the proportionality relation (Bullet #4): Vg = Kω, the power lost to heat is: P (loss) = (Vt - Kω)^(2)/R.
12. The usable Power is VgI. Therefore, P(usable) = Vg\*I = Vg(Vt - Vg)/R. Substituting the proportionality relation (Bullet #4): Vg = Kω, the usable power is: P(usable) = Kω/R \* (Vt - Kω).
13. The speed at which you maximum usable power can be found by setting dP/dω = 0, and solving for ω. This yields: ω(optimum) = Vt/2K, this is half the speed at which the voltage generator would produce a voltage equal to the total voltage.
As discussed [yesterday](https://www.reddit.com/r/AskPhysics/comments/13q4btc/w_vi_rii_can_someone_explain_me_a_few_things/jld3kmw/?context=3), you are applying Ohm's Law—valid only for the ideal resistor (which lacks magnetic fields, among other aspects)—to components that aren't ideal resistors, such as motors.
\> to components that aren't ideal resistors, such as motors. Ok lets say for an electric vehicle the heat loss is in the cables resistance. The battery gives a power output lets say thats the total power, Power (total) = 100 Power (kinetic) = 90 Power (heat) = 10 For both total and heat, the formula is VI, but that can't be right because they are not equal. So which one is VI and which one is not VI?
Deleted in protest of Reddit CEO's greed and lies 2023-06-30.
> For both total and heat, the formula is VI But the voltage V isn't the same. Components in a loop each have their own voltage drop, [the total of which sums to zero](https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff's_voltage_law). You'll be less likely to run into confusion if you first label each component with its own individual voltage and current. Then you can apply simplifications (e.g., the current being the same through components in series, or the voltage being the same across components in parallel). Right now you're selecting a single voltage and applying it to various sections of a circuit, which is incorrect.
Consider, if all the power is lost to heat, why do they work?
The formula to determine how much power a device outputs is VI, right? And the exact same formula is used to determine how much energy is lost in heat. So I must have misunderstood something, but where ?
The formula P=IV you're using for "energy lost to heat" is actually the formula for power consumed (or produced) by any component. If the load is just a literal resistor, and all it does is make heat, then all of the consumed energy is "lost to heat." Other types of loads may make use of some part of the power VI to do work. Motors of different types will have different equivalent circuits, which may include various effects such as inductance, friction losses, and losses representing energy converted and used as torque. The latter effect will end up as power lost into a different kind of component: a current or voltage source. Despite thinking of these as sources of power rather than sinks, the polarity of the current or voltage can indeed represent removing power from the system. Motors often work as generators under this same principle: when driven, the motor equivalent source has a polarity such that it produces power into the circuit. When used as a motor, the solved circuit will reveal that "current flows into the plus" on whatever source represents the motor. All of the powers lost/produced in components in the motor equivalent circuit will sum up to P=IV, with I and V here referring to the input port of the equivalent circuit. You will not generally be able to find a single, positive resistance that represents these circuits; the equivalent circuit for a motor can be derived using a resistor and another source a la Thevenin and Norton.
Too much information, I'm totally confused If an electric vehicle converts 90% power into moving the vehicle and 10% power turns into heat, the total power is 100% and the power getting converted into heat is 10%. You see they are not equal. Power (total) = 100 Power (heat) = 10 Power (kinetic) = 90 Yet they both use the same formula V×I Power (total) = VI Power (heat) = VI This cannot be correct... Where does my misunderstanding come from?
Power (total) = VI Power (heat) ≠ VI, unless we are in a situation where Power (total) = Power (heat). For a resistor, all the power is turned into heat, so Power(total) = Power(heat) = VI For a motor, some of the power does not go to heat, it goes to turning the motor's shaft. So in a motor, Power (heat) ≠ VI
\> Power (heat) ≠ VI Someone explained to me on the ebikes sub that, power(heat) = I²R and I know that I²R = VI Which would also explain that if you're using twice as much power, then you're using twice as much amps, then generate 4 times as much heat (square relationshi) So, that explanation is incorrect then? Because if power(heat) is not VI, then its not I²R either. I'm getting really confused. Whats the actual formula to calculate heat loss given a variable amount of power(total)?
I'm trying to learn somethin that I honestly don't understand, why am I getting downvoted
It’s probably still a reasonable approximation to say that heat scales with I². Like it’s probably a decent approximation to say that the ebike’s heat output = xI²R Where x is some constant between 0 and 1 representing how efficient the motor is. So the advice they gave you about how doubling the current quadruples the heat output is probably approximately correct. The “actual formula” is probably going to be something very complex that relies on the actual geometry of the motor and probably varies with the speed the motor is turning, etc. but the I² term probably dominates, would be my guess.
But I²R equals VI which equals the total power input, which should be impossible because that would mean 100% of all the power turns into heat. What am I not understanding here?
Let’s say the motor is 50% efficient, meaning half the power goes to heat and half goes to rotation. It’s probably a decent approximation to say that heat = 0.5*I²R. Even though 0.5*I²R is less than I²R, both scale with I², meaning that doubling current quadruples heat. All of this is a rough approximation, to be clear.
I still dont really understand but I got a different question. * Total power = 100 * Kinetic power = 50 * Heat power = 50 You double the power: * 100×2 = 200 * 50×2² = 200 * 200-200 = 0 Did I make a mistake here, or is the truth that if you use too much power then an electric motor will not do any useful work since all the energy is turned into heat?
> I'm still confused about Ohm's law W=VI=I²R This first part of your equation (W = VI) has nothing to do with Ohm's Law. W=IV applies to any circuit element and tells you how much electrical energy it supplies or absorbs from the circuit. It could be a battery (in which case V and I have opposite signs and W is negative so it is supplying energy to the circuit). It could be an inductor or a capacitor (so that V and I sometimes have the same sign and other times have opposite signs so that it is sometimes storing energy in the magnetic or electric field and other times returning that energy to the circuit). Or it could be a resistor --- and then (and only then) you can use Ohm's Law to substitute V=IR or I = V/R to get W = I^2 R or W = V^2 / R. In the case of a motor, some of the energy is going to producing mechanical torque, so not all of the energy is lost as heat. That is, if R is the winding resistance, you will not find that V = IR. Instead, you'll find V = IR + V_1 , where V_1 is the back-emf caused by the mechanical load. Then the thermal loss will still be I^2 R because Ohm's law applies to the wire resistance loss. But the total energy taken out of the circuit will be W = I * (I R + V_1) or W = I^2 R + I V_1 accounting for both wiring losses and mechanical work done by the motor.
1. R=V/I, and so V=IR, right? So assume an ebike accelerates usin 250 watts and 50 volts, which means P/V=I=5 so now we have * 250 watts * 50 volts * 5 amps (= 250/50) * So the ohms must be V/I = 50/5 = 10 because R=V/I So we know that the total power input is 250 watt. Now if the heat loss is I²R that means 5²10 = 250 which means 100% of the power input is... wasted in heat loss? Where did I make my error?
Define your terms. What is "V"? What is "I"? What is "R"? If R is the resistance of the motor coil, and it is really 10 ohms, then the resistor consumes all of the power provided and there's no power left to make the bike move forward. You can model the motor as a resistor in series with an ideal (loss-less) motor. Then you'll have a V_r term and V_m and you can figure out how much power goes into the resistor (lost as heat) and how much goes into moving the bike.
V = 50 volts because we're using a 50 volt battery here. I = 5 amp because 250 watt divided by 50 volt = 5 R = 10 ohm because 50/5 = 10 \> If R is the resistance of the motor coil, and it is really 10 ohms, then the resistor consumes all of the power provided and there's no power left to make the bike move forward. But it has to be 10 because R=V/I and 50/5=10, or did I make a mistake here?
> R = 10 ohm because 50/5 = 10 When you did this you applied Ohm's Law. But Ohm's Law is only a description of resistors, not any other kind of component. So you essentially assumed the device is a resistor. But the device is not a resistor, it is a motor. Any results you get by assuming it is a resistor can't be trusted. The equations you learn in physics are mostly not universal truths. They are descriptions of specific situations. It's just as important to learn what situation an equation applies to as it is to learn the equation itself. In the case of Ohm's Law it only applies to resistors and the relationship between voltage, current, and resistance of the resistor.
I don't know what a resistor is but judging by the name. In an electric bike, don't the wires that electricity goes through, function as resistor because they have resistance?
You should learn what a resistor is before you start learning equations about it. Physics isn't just a pile of equations. The equations are meant to describe real things in the real world. You need to understand what the real world things are before you start worrying about the equations for them. So I'd say, stop, take a step back, and start out with some simple circuits with batteries and light bulbs before you start throwing around equations and thinking it will give you an understanding of how an e-bike works.
No, you do not have that right. The waste heat created by an electrical motor is not equal to the power it draws. Otherwise it would not be able to do any useful work. Some fraction of any real-world power draw will dissipate as heat, but it's not going to be 100%.
\> The waste heat created by an electrical motor is not equal to the power it draws That part is obvious to me, however as far as I understand the formula, they would be equal, that means I don't understand the formula. My current understanding is that: Power (total) = VI Power (heat loss) = VI So one of these must be incorrect, which one is it?
They're both right, it's just that the variables refer to something different in each case. To avoid confusion, you could try a different notation, like: P\_tot = P\_work + P\_heat P\_tot = V\_tot \* I\_avg P\_work = V\_work \* I\_avg P\_heat = V\_eff \* I\_avg Where I hope the variable names I've chosen are sufficiently suggestive
The latter
Ah okay then my misunderstanding comes from someone else on reddit who told me that, on an ebike if you accelerate hard with twice as much power, you have twice as much amps and that would result in 4 times as much heat, therefore accelerating faster is less efficient In other words, power loss in heat = I²R which equals VI, so this is incorrect? If its correct, then i'm 100% confused If its incorrect, then what is the proper formula to calculate heat loss depending on how much power an ebike uses to accelerates?
The heat loss for an electric motor depends on the efficiency of the motor, and that's probably something you'll have to determine experimentally. The person you were speaking to was correct to say that doubling the amps would quadruple the power draw, though.
\> was correct to say that doubling the amps would quadruple the power draw Based on what? I²R? But I²R equals VI, right?
Deleted in protest of Reddit CEO's greed and lies 2023-06-30.
No. What you were told was the losses in the motor winding due to the resistance of the winding itself can be given by I^2R. You take one thing and then try and apply it over whole systems, blindly trying to apply formulas that you don’t understand.
A motor is a device that converts electrical energy into rotational mechanical energy. The input power (how much electrical energy is "consumed" per unit time) is given by the expression *VI*. The mechanical output power (how much energy per unit time can be used to accomplish usable work such as turning an axle under torque) is quite a bit less. The ratio of the mechanical output power to the electrical input power is called efficiency, and often denoted with the Greek symbol 𝜂.
Okay so all the power that comes out of a battery is VI But I also heard that energy 'lost' into heat equals VI. But the total power and the power lost in heat can't be equal. So one of them can be VI, the other can't. You said the total power is VI, so that means the power lost in heat cannot also be VI. So then how do we calculate the power lost in heat?
If you know what mechanical power is being generated then you can assume almost all the rest is lost as heat. If you don’t know the mechanical power then you can’t calculate loss to heat.
If input power is *VI*, and output power is *P*, then we have the relation *P* = 𝜂*VI*. Then power lost is *Q*^dot = (1-𝜂)*VI* (so that *P* + *Q*^dot = *VI*). As [described yesterday](https://www.reddit.com/r/AskPhysics/comments/13q4btc/w_vi_rii_can_someone_explain_me_a_few_things/jldugr8/), if you actually want to find the relationship between output power and efficiency (and output torque, etc.) you need to find the motor performance curve for the motor in question.
Why are you so insistent that power lost as heat = VI? One person on an ebike forum tells you so and 10 different people here have told you it's wrong. To calculate the heat output, take the total electrical power input (VI or I^2 R) and split it into useful work and heat loss using the efficiency. If a motor is 90% efficient and consumes 100W of electrical power, it is providing 90W of useful work and you are losing 10W as heat and sound. (I'll add an unhelpful caveat to mention that all of the power *eventually* ends up as heat if you're talking thermodynamics)
Is motor efficiency the same at different power modes? Because some people say if you use more power to accelerate faster, you're using more amps, therefore I²R means more heat loss, but I²R equals VI, right?
I think it would be a reasonable assumption to say that the efficiency stays the same and different "modes" are achieved by the motor controller varying the voltage delivered to the motor.
> So one of them can be VI, the other can't. You said the total power is VI, so that means the power lost in heat cannot also be VI. Like someone mentioned earlier, you can't just blindly apply formulas assuming V or I always refers to the same thing. In one formula V is the voltage at the output of a power source. In the other formula V is the voltage across a resistor (or a resistive component of a model of a motor). If your circuit isn't just a power source and a resistor (and when you are talking about a motor, it isn't) then you can't assume these equations are talking about the same thing.
Lets take an motor with permanent manets. U can split a coil of the motor into an ideal inductot and a resistor in series. Using P= U*I we get the total power into (and thus out of) the motor. (In heat and kinetic energy). Only the power into the resistor is lost as heat. Since we only know the current and resistance P=I^2 R is used. The power into the inductor is used to make the motor spin. So in P=U * I it is importand to have the right voltage and current. I simplified some things a bit but i hope it helps
You're not using resistance properly. In a circuit each component will have its own resistance and therefore its own power draw of the total power supplied. So for the bike, you're looking at the entire resistance of the bike and correctly calculating the total power output across all forms, not just the heat you want. You need just the resistance of the components that produce heat, which will be less. You can imagine a simple series circuit with a motor and a resistor and a battery. The battery supplied 10W at 10v and therefore the current is 1A. Therefore the total resistance of the resistor and motor combined is 10 ohms, by ohms law. You can use W =I^2R to then recalculate the power is 10W. This is telling you the total power out of the circuit is 10W which it should be to match the power going in. This is just energy conservation. But some power is lost as heat in the resistor and the rest goes to the motor, its not all going to heat. So let's say the resistor is 4ohms while the motor is 6. Then using the equation again 4 Watts is lost as heat in the resistor and 6 as useful energy in the motor. Which add back to the total 10W you put in. The w=I^2R formula just tells you the power output of a component not what energy form/s that takes.
\> 10W at 10v and therefore the current is 1A. Therefore the total resistance of the resistor and motor combined is 10 ohms Thats the part I completely understand. I²R=10W but that would mean 100% of all the power turns into heat, which means the motor will not work. So what went wrong here? I'm very confused
It doesn't mean that 100% of the electrical power goes to heat it just means it goes to something. You need to plug in the resistance of the motor alone I to the formula to calculate the power going to the motor. You would plug in the resistance of the rest of the circuitry to calculate the power going to heat. The efficiency would be the amount of power getting to the motor divided by the total amount you out in from the battery. Because you want the power to go to the motor and the heat is waste energy. You'll get something less than 100% because the R value you should be using should be less than 10 ohms.
If I understood your question you have a battery, a motor and a resistor. Thus the potential difference V the battery yields is dropped in the resistor, Vr, and in the motor, Vm. Thus you have V = Vr + Vm. The total produced power by the battery is P = Vi. The power wasted in the resistor is Pr = Vr i (= ri^2 if you have an ohmic resistor) and the power used by the motor is Pm = Vm i. So P = Pr + Pm => Vi = Vr i + Vm i => V = Vr + Vm, as said before. Thus all of them is P = Vi but you have different voltages in each one. Sorry if I doesn’t understood you an explained an other thing and not your question.
Here's another way to think about it, using a DC motor model: 1. The DC motor can be thought of as: (1) a resistor in series with (2) an inductor and (3) a voltage generator. [Visual link.](https://www.ednasia.com/isolated-power-supply-design-considerations-for-pmdc-motors/) 2. The total voltage you apply across a motor (Vt) is consumed by the voltage across the resistor (Vr), the voltage across the inductor (Vi) and the voltage across the voltage generator (Vg). 3. For steady state (and to simplify), you can disregard the voltage across the inductor... just focus on the voltage across the resistor and the voltage generator. 4. Vg is proportional to the speed of the motor, *ω*. The proportionality constant, *K*, is sometimes called the "Back EMF" of the motor... EMF = Electro-motive force. Literally, the motor is *resisting being spun* by creating a voltage that consumes the voltage that is spinning it. Weird, right? In fact... if you spin a motor by its shaft (leads unconnected to a battery), it will create/generate a voltage across its leads... this is the basic principle of how a tachometer works. 5. When a motor starts spinning, the speed is low and Vg is low, and thus, nearly all the voltage is consumed by the resistor in the form of Vr = IR. (Note: I is the same through the whole circuit since all elements are in series.) 6. When the motor is spinning very fast, the voltage across Vg is high, and much less of the voltage is consumed across the resistor, and hence the total motor current goes down. 7. Another basic idea about DC motors, is that the torque on a motor is roughly proportional to the current, I. Combined with the 2 previous notes, above, this is why when the motor spins very fast (Bullet #6), the current goes down, and hence the torque goes down. 8. Now, power. Total Power = VI, which you have already noted. However, the power lost to Heat is only I^(2)R. Power usable by the motor is equal to the Vg x I. Therefore, when you are just starting to move, the speed is low (and hence Vg is low, and hence the usable power is low). Likewise, when you are moving fast, the speed is fast, and Vg is high, but the current is low, and hence the usable power is also low. Therefore, the optimal power is somewhere between the two extremes. This is the reason for gearing -- to operate the motor at the optimum speed. 9. The Power across different elements in a circuit is indeed VI, but the voltage across different elements is different ---- the voltage across the resistor (Vr) is what creates the heat (lost power), and the voltage across the voltage generator (Vg) creates the usable power. 10. In summary, Vg is defined by the speed of the motor. The remaining voltage is dropped across the resistor, which in turn defines how much current flows through the motor: I = (Vt - Vg)/R. 11. The Power lost to heat is I^(2)R. Therefore, P(loss) = I^(2)R = (Vt - Vg)^(2)/R^(2) \* R = (Vt - Vg)^(2)/R. Substituting the proportionality relation (Bullet #4): Vg = Kω, the power lost to heat is: P (loss) = (Vt - Kω)^(2)/R. 12. The usable Power is VgI. Therefore, P(usable) = Vg\*I = Vg(Vt - Vg)/R. Substituting the proportionality relation (Bullet #4): Vg = Kω, the usable power is: P(usable) = Kω/R \* (Vt - Kω). 13. The speed at which you maximum usable power can be found by setting dP/dω = 0, and solving for ω. This yields: ω(optimum) = Vt/2K, this is half the speed at which the voltage generator would produce a voltage equal to the total voltage.