power rating

the remaining diagrams

Hi again,

In real life we may or may not know the Q as in this exercise, buy with exercises like this we often find we are told to solve for one thing knowing the other thing, and it doesnt always mean that it is really possible in real life. I guess it is possible that someone else measured the required parameters and then came to the conclusion about what Q was and then conveyed that information rather than the other information you really wanted. So in real life we might find the information given to us in a lot of different ways for different reasons, and one of our jobs is to convert that information into something we can easily use that more closely fits what we are trying to accomplish.

As JoeJester pointed out, we might also change the phase to get the graph to stabilize faster, or, we could just solve for the initial current in the inductor. For this circuit if we use the initial current of -266.666 amps we'll find the inductor current stabilizes almost immediately using E*sin(wt) as the source. Not that level of current is when we use the peak value to show us the RMS value in the graph, and the sign may change depending on what simulator you use and possibly how the inductor graphic is oriented within the schematic diagram in the simulator software schematic view.

we could just solve for the initial current in the inductor

Click to expand...

Thanks for mentioning that method. But I think it has two inconveniences,
1) It is tedious. Every time we change a element value, for example, L or R, we need to re-evaluate the initial current.
2) We don't know how long a circuit will get to steady state.
Of course, it's just my own opinions.

"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.

Click to expand...

This is what you told me when I asked about the power in series RL circuit.

I belive that in a series RL or RC circuit with AC source, as long as the circuit is an isolated complete one, current just flowing from the voltage source, through the resistor, inductor and back to the voltage source repeatedly, nowhere else, then whenever the inductor is returning its energy, that energy is ultimately dissipated by the resistor in this single loop. So if this could be proved and if 'inductor in my house' were supplying energy 'to my neighbors', then I imagine that the connection relationship between electric utility company and houses would be something like (1), not (2) in the picture below. But that has a problem: every house in the 'series circuit' has the same voltage. Maybe it's not just a simple series or palallel question? Hmmm....

I've been thinking about this silly question. Do you have any ideas?

Hi there Heidi,

1) Well maybe a little, but we could look into this. Maybe it is just -E/wL as an approximation. Or, waiting for the simulation to reach steady state, just read off the current level from the graph exactly at the point where the line voltage goes through zero and use that (or similar).
2) Well if we solve for the time response we get the required time constant, so if you want to take 5 time constants you could do that. But since the series resistance is so low in comparison to the parallel resistance,we could use L/Rs as an approximation to the time constant.

For your question about 'supplying energy to the neighbors'...
Usually this isnt going to happen anyway because your own house is going to be consuming energy in other devices so it will never reach the neighbors. But if that isnt the case, then you have to consider what we might call "micro voltages". A micro voltage could be a small increase in voltage that is not noticeable not necessarily a real 1uv or so. That small increase is enough to LOWER the energy coming from the energy company line to the neighbors, and thus you are supplying a small but still present energy to the neighbors. In other words both your lines might be exactly 120v, but then when your inductor starts to fire into the line YOUR voltage goes up to 120.1v while the neighbors only goes up to 120.05v. The electric company line is still 120, so it's obvious your are supplying at least a little energy to the neighbor.
This can happen because of the real life line impedance, which is never zero. That depends on a large number of factors such as where are you relative to the transformer and where the neighbor is relative to the you and the transformer, but we could simulate it with just resistances for simplicity as that would still clarify the situation.
If your resistance is 0.1 ohm and the neighbors is 0.1 ohm and the utility company is 0.1 and you are all connected to the same node, if you inject some energy into your line then the node voltage MUST rise at least a little bit. That in turn means the energy company supplies a tiny bit less energy, which means the neighbor gets it
If this still isnt clear we could do a little simulation.

Hello MrAl

Please don't be mad if I didn't use words or phrases properly.
A thought came to my mind after reviewing my last post that I might have used the word 'tedious' improperly. I thought something is tedious means it takes a long time doing it until just now I looked up the word in my dictionary.
Here is how I get the inductor initial curent
voltage source: vS(t)=600sin(wt)=600cos(wt-90°), phasor form Vs=600(-90°)
R1=1u Ω
R2=3 Ω
L=7.162m H
w=100∏=314 rad/s, wL=2.2489
first change the voltage source to its Norton current source
iS(t)=vS(t)/R1=600*10^6 cos(wt-90°), phasor form Is=6*10^8(-90°)
equivalent resistance R of parallel R1, R2
R=R1*R2/(R1+R2)=10^-6
then the inductor current in phasor form
IL=Is*R/(R+jwL)=266.8(-180°)
finally inductor curent iL(t)=266.8cos(wt-180°)
iL(0)=-266.8 amps RMS

And whenever I change any one of R1, R2 or L, I have to repeat that process over again. This is why I said it is "tedious", it "takes a long time" to get the initial current.

Apparantly I need to spend more time learning English and be more careful in choosing words to avoid misunderstanding.

About the initial current, until I read

Maybe it is just -E/wL as an approximation.

Click to expand...

suddenly I realize that maybe I should do some analysis before I actually do the calculation!

Since R1 is quite small, 1u, so the source voltage basicly is all acroee R2 and L, therefore, inductor current = 600/wL = 600/2.2489 = 266.8 approximately. As for the phasor angle, since 600(-90°) is across the inductor, and inductor curent lags voltage by 90 degrees, so it's -180 degrees.

Is the cause to 'micro voltage' the same as the source termianl voltage that appears when the source has a series internal resistance that you once explained to me?

If this still isnt clear we could do a little simulation.

Click to expand...

Well, before that, I think I need to understand a little bit, for example, my neighbors and I are connected in series or what? I have completely no ideas about this.
Could you please maybe draw a picture showing conceptually how an electric company supplies electricity to houses? Is it similar to the (1) in my last drawing?

Thank you!

Hello Heidi, or should i say hello neighbor,



If you look at the diagram, we are now neighbors

As you can see, if we both are loading the line (remove our batteries) the Electric Company supplies our RL loads with energy. But if you start to generate energy your battery starts to take effect and since it is more like a current source it raises the node voltage at N a tiny bit higher, and so that causes current flow through R3 and so i get more energy because of that because my line voltage goes up slightly (when i am not generating energy my battery is not in the circuit).
If on the other hand i started to generate energy then your line voltage would rise slightly and thus your RL would get more power.

You probably also want to look into the principle of "Superposition" as that simplifies the view of a circuit with several sources.

Dear my neighbor,

Thank you very much for your detailed and easy-understanding explanation. Now I have much much better understanding about the mysterious reactive power. Thanks again!

Hello Heidi,

You're welcome, and more good luck with your studies.