A question about the power in RL circuit

[Heidi]

Dear friends,

In a circuit in which an AC source, R,L are connected in series, the average/real/active power delivered to combination of R and L is, according to its definition, the average of product of the instantaneous source voltage and the source current in one period of the instantaneous source voltage/current.

Is it correct that the average power can also be obtained by calculating the average of product of the instantaneous voltage across the RESISTOR and the current through the RESISTOR in one period?

Thank you!

 

[MrAl]

Hello there,

The short answer is "yes". That's when we do the calculation in the time domain as suggested by your query containing the instantaneous voltage and current. That would be because the only power lost is in the resistor. The way to prove this would be to do the calculation in the time domain using the two different methods in question evaluated at some distant time in the future where the exponential part goes to zero leaving only the sinusoidal parts.

But if we did this in the frequency domain we'd have to use the voltage across the resistor not the voltage across both L and R or we'd get a result that was not correct. That would involve finding the RMS voltage and current and since the voltage is higher across the entire network than across the resistor alone we'd get a different result.

 

[Heidi]

But in the VoltageSource-R-L series circuit, when the value of the voltage source is positive, the value of the inductor voltage is negative, and the value of the current is positive and decreasing, specifically, roughly between the 6ms and 8.3ms time interval in the graph attached below, indicating that the inductor is delieving energy. Where does the energy the inductor supplies/releases go? Is it delieved to the resistor and dissipated as heat? how do we know?

Thank you!

 

[MrAl]

Hello,

We are talking about average here, not instantaneous. We need the instantaneous response in order to calculate the average power response, but the average is a much different quantity than the instantaneous. If we loose more energy than we store then there's still a net loss. But the energy we store is not lost, it's saved for later use. At some points the source will also absorb energy, and that happens for a short time just after the source voltage goes through zero. The inductor supplies all the energy when the source is at or near zero. When the source absorbs energy it may be considered a loss or a storage, depending on what kind of source it is.

Also of interest might be that we are only looking at the AC response, which does not include the exponential start up part. If we consider the average power there too we get an extra roughly 100 watts. Some of that is going to be stored by the inductor and some dissipated.

To see the difference you are talking about however all you have to do is include a few more waveforms in your simulation or analysis. Simply take the product of Vs*I and call that PT (power of the total circuit), then take the product of Vs*IR call that PR (power in the resistor). When you compare the two you'll see that PT goes through zero while PR does not, and the initial PT is higher that PR (by say 100 watts) but the averages are the same after a reasonable time period where the exponential part is allowed to die out.
So what we see at that point is:
average(PT)=average(PT)

Try that and see what you can make of it. Feel free to ask more questions about this if you still are not satisfied.

Another way to look at this is to simply calculate the average power using both the source voltage and the resistor voltage, then compare the two. Again we find that the average PT is equal to the average PR after a reasonable time period of maybe 50 cycles.

Still yet another way to look at this is to realize that the ideal inductor stores energy and does not consume it like a resistor. Since the ideal source would absorb the energy and dissipate it, the source itself is a loss of energy too unless it is capable of storage (such as a real life line or battery system converter, but there we might have a partial loss anyway depending on the type of source and it's charge storage efficiency). But for these simple exercises usually the source is not considered a load when it is being SUPPLIED with energy and not actually delivering it as it normally does.

I might add that the only time we have a problem with this kind of analysis is when we try to take short cuts. Short cuts include a frequency domain analysis where we have to be careful how we evaluate the voltages which ultimately are made up of single numbers with no time associated with them. It's only when we do a complete time domain analysis that we can do it in a more straightforward way and the results absolutely have to come out right (unless we make a math error of course).

Maybe your question is hinting at what happens when the source absorbs some energy, but that's not during the period you are holding in question presently. If you do the power wave you'll see it go negative for a short period during each half cycle.

BTW, when you plot the waveforms it is much clearer to do so at 50Hz instead of 60Hz.

 

[Heidi]

At some points the source will also absorb energy, and that happens for a short time just after the source voltage goes through zero. The inductor supplies all the energy when the source is at or near zero. When the source absorbs energy it may be considered a loss or a storage, depending on what kind of source it is.

In this source-R-L series circuit (I redraw a circuit with different parameters attached below), assuming v1(t) for the instantaneous voltage across the source, vR(t) for the voltage across the resistor, vL(t) for the voltage across the inductor. By KVL, we have
vR(t)=v1(t)-vL(t), and
i(t)*vR(t) = i(t)*v1(t) - i(t)*vL(t).
So I thought that when the instantaneous power supplied by the source [i(t)*v1(t)] is positive, the power absorbed by the inductor [i(t)*vL(t)] is negative (indicating that the inductor is supplying energy), then the instantaneous power dissipated by the resistor is supplied by BOTH the voltage source and the inductor, is this correct?

But when i(t)*v1(t) and i(t)*vL(t) are negative, and | i(t)*vL(t) | > | i(t)*v1(t) |, then where does the power | i(t)*vL(t) |- i(t)*vR(t) go? Is it "consumed" by the ideal voltage source?

I used to connect a 3V battery and a 1.5V one in parallel, but I didn't feel any noticeable warm on the connecting wire. Is this connection a complete circuit? How is the electric energy consumed?

Thank you!

 

[Ratchit]

Heidi,

In a circuit in which an AC source, R,L are connected in series, the average/real/active power delivered to combination of R and L is, according to its definition, the average of product of the instantaneous source voltage and the real part of the source current in one period of the instantaneous source voltage/current.

In a series LR circuit, the current will lag the voltage, possibly up to 90° depending the the ratio of L to R. Therefore, you have to make sure you take the real part of current before multiplying it with the voltage.

Is it correct that the average power can also be obtained by calculating the average of product of the instantaneous voltage across the RESISTOR and the current through the RESISTOR in one period?

Yes, the voltage and current are in phase with each other in the resistor.

Ratch

 

[Ratchit]

Heidi,

So I thought that when the instantaneous power supplied by the source [i(t)*v1(t)] is positive, the power absorbed by the inductor [i(t)*vL(t)] is negative (indicating that the inductor is supplying energy), then the instantaneous power dissipated by the resistor is supplied by BOTH the voltage source and the inductor, is this correct?

The perfect inductor stores and releases energy back to the circuit. The resistor always dissipates energy as heat whenever voltage is across it or current exists through it. Whether you consider energy absorption/dissipation as positive/negative depends on how you set up and interpret your equations.

But when i(t)*v1(t) and i(t)*vL(t) are negative, and | i(t)*vL(t) | > | i(t)*v1(t) |, then where does the power | i(t)*vL(t) |- i(t)*vR(t) go? Is it "consumed" by the ideal voltage source?

The voltage source in the RL circuit only supplies power. The resistor always dissipates power. The inductor stores and releases power.

I used to connect a 3V battery and a 1.5V one in parallel, but I didn't feel any noticeable warm on the connecting wire. Is this connection a complete circuit? How is the electric energy consumed?

Yes, you are forcing a current through the lower voltage battery. Without knowing the resistance of the wire is and how much current existed in the wire, how can anyone tell? Within the chemistry of the smaller voltage battery and the internal resistances.

The following attachment shows a voltage (red), a 30° lagging current (green), and the power product (blue) of the two. Whenever the blue line goes below the horizontal axis, the power is being returned back to the circuit. If both current and voltage were in phase, the blue power curve would never go below zero. If the current and voltage were 90° apart, the blue power curve would be below and above zero an equal amount of time.

Ratch

 

[Heidi]

Thank you, MrAl and Ratch.

I would like to focus on the period (between 10ms ~ 14ms in Voltages&Current.jpg below) where the instantaneous voltage across the source V_S(t) is negative (the Green curve in Voltages&current.jpg below), the voltage across inductor V_L(t) (Red) is negative while | V_L(t) | is greater than | V_S(t) |, and the current (Blue) through the entire circuit i(t) is positive and decreasing, naming

during 10ms < t < 14ms, V_S(t)<0, V_L(t)<0, V_L(t)<V_S(t)

In THIS case, it seems to me that the power delivered/returned by the inductor has tow parts, one part is dissipated by the resistor, that is , [V_L(t) - V_S(t)] * i(t) is dissipated by the resistor, but (1)where does the other part, V_S(t)*i(t), go, in what form(s)? Assume that the source voltage is NOT an ideal voltage, but is the voltage from an electric company, (2)what influence does this part of reactive power have? (3)Is it causing a waste?

Thank you!

 

[Ratchit]

Heidi,

I would like to focus on the period (between 10ms ~ 14ms in Voltages&Current.jpg below) where the instantaneous voltage across the source V_S(t) is negative (the Green curve in Voltages&current.jpg below), the voltage across inductor V_L(t) (Red) is negative while | V_L(t) | is greater than | V_S(t) |, and the current (Blue) through the entire circuit i(t) is positive and decreasing, naming

during 10ms < t < 14ms, V_S(t)<0, V_L(t)<0, V_L(t)<V_S(t)

All three curves you submitted have different periods. How can that be? Where is the curve representing power? Why don't your curves look like the ones I submitted?

In THIS case, it seems to me that the power delivered/returned by the inductor has tow parts, one part is dissipated by the resistor, that is , [V_L(t) - V_S(t)] * i(t) is dissipated by the resistor, but (1)where does the other part, V_S(t)*i(t), go, in what form(s)?

All the energy in the inductor is stored and returned to the circuit in equal amounts. None of it is dissipated in the resistor because the phases of the voltages of the inductor and resistor are 90° apart.

Assume that the source voltage is NOT an ideal voltage, but is the voltage from an electric company, (2)what influence does this part of reactive power have?

It means that the power factor is lower, and the energy company won't supply as much energy to the customer, thereby losing revenue. If they increase the voltage to make up for the lower energy, they will suffer higher IR losses.

Ratch

 

[Heidi]

All three curves you submitted have different periods. How can that be? Where is the curve representing power? Why don't your curves look like the ones I submitted?

Sorry, the curves were in initial transient state, I should have included the initial condition for the inductor, that is, the initial current = -0.57805A. The instantaneous power curves are also attached, P_S(t) represents power supplied by source, P_L(t) for power absorbed by inductor, and P_R(t) for power dissipated by resistor.

All the energy in the inductor is stored and returned to the circuit in equal amounts.

I really cannot understand the description that all the energy in the inductor is stored and returned to the circuit in equal amounts. In this particular RL series circuit with voltage source, ignoring the resistance of the wire, 'returned to the circuit' means returnning to what element? None of the energy stored in inductor is delivered to the resistor?
The reason why I have this question is that when V_L(t)<0, V_S(t)<0 and
V_L(t)<V_S(t) (please refer to thr RL circuit attached in the next post), then the voltage across the resistor V_R(t) is
V_R(t) = |V_L(t)| - |V_S(t)|, so the voltage across the resistor at those moments satisfying the above conditions is coming from the 'inductor acting as a source', so is the power dissipated in the resistor. Do I have the right conclusion?

None of it is dissipated in the resistor because the phases of the voltages of the inductor and resistor are 90° apart

If I didn't get any calculation errors,
V_L(t)= 1.90578 COS(2∏ft - 72.3432degrees),
V_R(t)= 0.60663 COS(2∏ft - 162.3432degrees).
They are not out of phase by 90 degrees.

 

[Heidi]

Sorry, forgot the graphs

 

[MrAl]

Hello again,


Since the current is 90 degrees out of phase with the voltage in the inductor, that means the resistor current is 90 degrees out of phase with the voltage in the inductor, so that means the voltage across the resistor is 90 degrees out of phase with the inductor voltage.

But when the line voltage goes to zero (twice per cycle) the inductor still contains energy, and the resistor is dissipating power. Therefore the resistor gets it power from the inductor alone during that brief period, and after which it would get it's power from the inductor and the line.

I think it might help to do two things here:
1. Plot the energy stored in the inductor (not i(v)*vL(L) which is different). The energy is (1/2)*L*i^2 and is always positive.
2. Start with the initial inductor current of zero as before, but this time look at the waveform after 5 time constants, which would be 50ms, but to sync with the line period start looking at 60ms. A good time to view is between 60ms and 80ms. This ensures that all transient phenomena have died out.
3. Plot the resistor power.
4. Try to look at times when something goes through zero. When something goes through zero that eliminates that as a contribution to the circuit because a zero voltage can not deliver nor absorb power for example.

If the line is supplied with energy and you dont have anything else plugged in, then you are now supplying the neighbors with energy
This is in fact a key point with solar powered line tied inverters that are used to convert solar energy into electrical energy and pump it into the line. There we do it on purpose, but with the inductor resistor circuit it's just natural. So the line itself does not absorb the bulk of the energy but another load somewhere else would absorb that energy. There is some energy absorbed in the line itself however because the line wires are not perfect conductors so the conductors themselves dissipate energy as heat in the process of delivering the energy to another load.
You can look at this as the two battery circuit you mentioned, but instead of connecting them directly in parallel connect them with a small resistor in series with the positive terminals. When one battery is 3.0v and the other 3.0v no current flows between the batteries and thus no current through the resistor. But when one is 3v and the other 1.5v then there is a potential difference of 1.5v across the resistor so current flows from the 3v batt to the 1.5v batt, thus attempting to charge the 1.5v batt. But if the 1.5v batt also had a load resistor across it, the load resistor would absorb some of that energy so the battery would not charge as much. Note also that if the 1.5v batt had a large series resistance inside of it, the terminal voltage would go up to almost 3 volts. If this happened in the line/inductor/resistor circuit the line voltage could shoot up to a much higher value ruining other equipment plugged into the line.

 

[Heidi]

Since the current is 90 degrees out of phase with the voltage in the inductor, that means the resistor current is 90 degrees out of phase with the voltage in the inductor, so that means the voltage across the resistor is 90 degrees out of phase with the inductor voltage.

Ooops! You and Ratch are right. I was wrong.

But when the line voltage goes to zero (twice per cycle) the inductor still contains energy, and the resistor is dissipating power. Therefore the resistor gets it power from the inductor alone during that brief period, and after which it would get it's power from the inductor and the line.

In the RL series circuit with AC source case, I guess this would hold:
P_R(t) = P_S(t) - P_L(t), at any instant, namely,
power dissipated by resistor = (power supplied by source) - (power absorbed by inductor), at any instant, ignoring the resistance in wires.

I can understand what you said about the situation when the line voltage (meaning the source voltage?) is at zero - the resistor gets it power from the inductor alone during that brief period. What is bothering me is when the line voltage and the inductor voltage both go below zero, meanwhile the current value is positive. Would you please tell me what exactly happens during this period?
According to the above equation (if it is not incorrect), since the power dissipated by the resistor is always positive, if the line voltage went below zero, the inductor voltage would go further below zero. I guess this situation is, like you said, similar to the one that a small resistor is connected in series with the positive terminals of two batteries. Is this case also similar to the one where two people are pushing each other? In real life, when the electrons in wires are pushed by two different opposite electric forces, will it cause heat or something in the wires? In the point of view of the line voltage, is this causing a waste and is undesirable? Hmmm, confusing...

I think it might help to do two things here:
1. Plot the energy stored in the inductor (not i(v)*vL(L) which is different). The energy is (1/2)*L*i^2 and is always positive.
2. Start with the initial inductor current of zero as before, but this time look at the waveform after 5 time constants, which would be 50ms, but to sync with the line period start looking at 60ms. A good time to view is between 60ms and 80ms. This ensures that all transient phenomena have died out.
3. Plot the resistor power.
4. Try to look at times when something goes through zero. When something goes through zero that eliminates that as a contribution to the circuit because a zero voltage can not deliver nor absorb power for example.

I guess the point here is that you want me to analyze the circuit in its steady AC state.(?)
I do so by finding out the current at t=0 from the source voltage phasor being devided by the equivalent impediance of the circuit, and using the initial current as the inductor's initial condition in PSpice. Is this method also correct?

If the line is supplied with energy and you dont have anything else plugged in, then you are not supplying the neighbors with energy

Please forgive me for my poor English. I'm not sure I get the point

This is in fact a key point with solar powered line tied inverters that are used to convert solar energy into electrical energy and pump it into the line. There we do it on purpose, but with the inductor resistor circuit it's just natural.

Sorry, I cannot understand this paragraph completely. Could you please tell me more, in terms of easier English? Thank you!

 

[MrAl]

Hello again,

Sorry that was a big typo on my part, and i've corrected it in my last post. That line should have read:
"and you are now supplying energy to the neighbors"
and that happens because you have stored energy in your inductor and you are not using all the energy in your resistor.

BTW i think of this circuit as having the inductor connected to the line voltage with one leg of the resistor grounded. That way i can see the resistor voltage easily and know it is the same as the current so i have both quantities right off. If you prefer the other way i can change it.

The inductor voltage falls through zero before the line voltage falls through zero. As the inductor voltage falls through zero, the inductor energy starts to decrease, but at the moment it goes through zero the energy does not change so the power in the resistor comes from the line. This is easy to think about because if the inductor has no voltage it is a short circuit at that point in time yet the line is still slightly positive, so the resistor gets power from the line voltage.
Soon afterward the line voltage falls though zero, with the inductor voltage still negative. As that happens, again we have the inductor supply energy to the resistor and this is evident by the decrease in energy in the inductor. As the line goes negative then both inductor voltage and line voltage are negative, but at first the inductor voltage is more negative than the line so a positive voltage appears across the resistor, and that means it's getting energy from the inductor and the line is also getting energy from the inductor. Sometime later the inductor voltage and line voltage become equal and both still negative, and that that point there is no more energy anywhere for the resistor so the resistor power goes to zero (line and inductor voltage cancel each other out), and also the inductor energy goes to zero. Once they become non equal again this time the line voltage is more negative than the inductor voltage, and so the line starts to supply the inductor with energy again, and of course the resistor gets some of that energy too so it starts to dissipate power again.

The paragraph with the solar powered inverter was talking about a device that is capable of converting solar power into electrical energy, then driving the line with that energy. It's like a AC generator.

It's ok if you want to solve for the initial inductor current as long as you are sure you get the right value. You know you have the right value when you see all the waves exactly the same, including the very first wave cycle.

If you want to check your equation for I(t), you can use this:
I(t)=Epk*(sin(t*w)*R-w*cos(t*w)*L)/(R^2+w^2*L^2)

where Epk is the peak of the sin source we are calling the 'line' voltage. w=2*pi*f.
The sine source is of course:
V(t)=Epk*sin(w*t)

The average power between any two times T1 and T2 is:
Pavg=(1/(T2-T1))*Epk*(2*w*L*cos(w*T2)^2-R*sin(2*w*T2)+2*w*R*T2+R*sin(2*w*T1)-2*w*L*cos(w*T1)^2-2*w*R*T1)/(4*w*(R^2+w^2*L^2))

The average power for the entire wave is with T1 equal to some time t and T2=T1+1/f with f the frequency. If you have solved for the initial inductor current then you can use T1=0 and T2=1/f no problem.

I thought that might help because your original questions were about the average power.

 

[Heidi]

Thank you very much for your detailed explanations and your time, MrAl, they really help. I have attached a graph showing the relationships between the line voltage, inductor voltage and the total energy stored in inductor for other people who are also interested in this topic.

I have one more question about the statement in your last post.

Note also that if the 1.5v batt had a large series resistance inside of it, the terminal voltage would go up to almost 3 volts.

I can understand why it is so, but I can't understand the following description

If this happened in the line/inductor/resistor circuit the line voltage could shoot up to a much higher value ruining other equipment plugged into the line.

In what manner a large resistor was inserted into the line/inductor/rsistor circuit would result in a much higher line voltage? Isn't the amplitude of the line voltage fixed?

Thanks again!

 

[MrAl]

Hello again,

The line voltage is only as fixed as it can be through at least some impedance. That means you have the ability to raise the line voltage as least a little bit if you have a generator connected to the line, and an inductor could act as a generator for a short time period. Usually this wont be much because the line impedance is usually low and the amount of energy from the internal source would be somewhat low too, but it's all just a network and a generator connected to a network can raise the voltage at the node it connects to.
Maybe i should not have brought this up, but since we were talking about the inductor supplying energy to the line for some short time periods i thought it might be interesting. And the two batteries are similar in that one battery supplies energy and the other receives energy.

 

[Heidi]

Yes, it is interesting, but it just shows that I still have a lot lot more things to learn. I appreciate your help.

 

[MrAl]

Hi again,

Well not too hard to understand really, because if you have a voltage source in series with a resistor and you power that other side of the resistor with a voltage that is higher than the first source then the current will flow the other way, from the second source through the resistor into the first source. It's a matter of determining the voltage across the resistor and it's polarity. That's not hard to do either because the difference in voltage is just the difference between the two source voltages:
vR=Vs1-Vs2

Knowing the resistance is R we then know the current:
iR=vR/R

and that is signed, and the sign tells us which way it is flowing. So if Vs1 is greater than Vs2 then we get a positive current, and if Vs2 is greater than Vs1 we get a negative current.

And the main point was that if the source Vs1 had it's own internal resistance R1 then the voltage drop across R1 would make the Vs1 source look higher if it was being powered by Vs2 (because Vs2 is higher than Vs1). If you do a quick analysis you'll see how simple this really is. In the diagram R1 is part of the Vs1 real source, so if Vs2 powers Vs1 then the voltage at the terminal marked 'v' goes up, and because we are really measuring the first source voltage at node v it appears that the first source voltage has risen. Any real life source has series resistance so when it gets powered by another higher source it's terminal voltage goes up, at least by a little.

                v
     +---R1-----+----R-----+
     |                     |
    Vs1                   Vs2
     |                     |
     +---------------------+

Here's a snapshot of the average power from the line and in the resistor as time progresses...

 

[Heidi]

Now I understand. Thank you very much. I learn a lot from you.

 

[MrAl]

Hi,

You're welcome Heidi and i am happy to hear that i could have helped, and good luck with your future studies.