Calculating real power of a circuit

[nForce]

How do I calculate real power of this circuit (green):



So do I write down the impedance, and use 1/2*(U^2/)Z?

[pigrew]

Only the resistors will have real power, not the inductors or capacitors.

You'll have to calculate the voltage across or current through the final resistor, and then use the appropriate formula.

The first step I'd do is to substitute the source and two resistors with a Thévenin equivalent.

[Benta]

There are several ways to attack this one.
Personally, I'd calculate the total impedance [seen from u_g(t)] using R, 1/sC and sL for the impedances (LaPlace notation). Substituting jw for s, you can calculate real and imaginary current (and thus power).
I'm certain there are simpler ways, others will chime in, but this is usually foolproof for me.

PS: how do you insert symbols? Using w instead of greek omega is sub-optimal.

[nForce]

Sorry benta, but pigrew said that only R has real power, why bother then calculating impedance for L and C?

[Benta]

Sorry benta, but pigrew said that only R has real power, why bother then calculating impedance for L and C?

Because otherwise you won't know the currents through and voltages across your resistors. I'll give you my solution when you've tried it yourself.

[dmills]

There is a bit of a trick hiding in here.

You are given that omega = 1/sqrt(LC) which is nice, think about what you know about series resonance in LC networks, it makes this real easy to solve.

Regards, Dan.

[Benta]

There is a bit of a trick hiding in here.

You are given that omega = 1/sqrt(LC) which is nice, think about what you know about series resonance in LC networks, it makes this real easy to solve.

Regards, Dan.

Aaah! Well spotted, Sir.
I missed that little point, as the circuit analysis is not trivial at all. The impedance seen from the source works out as:

[ s²2CLR + s3CR² + 2R ]  /  [ s²CL + s2CR + 1 ]

Not nice at all.

[Vtile]

If you remember the generator topic of yours..

[Benta]

If you remember the generator topic of yours..

Cryptic.

[Vtile]

If you remember the generator topic of yours..

Cryptic.

He/she made a topic of "reactive power in AC generators" @general section around at xmas. This topic relates to concepts discussed there.

[Benta]

'Nuff said. If you have a thing about referring to previous topics somewhere else that nobody understands, the Finnish winter is apparently too boring.

[dmills]

[ s²2CLR + s3CR² + 2R ]  /  [ s²CL + s2CR + 1 ]

I hate it when they do that in an exam, 40 minutes or so of fucking about with algebraic manipulation (and no marks to show for it), because you missed one little critical detail.

I make the impedance seen by the generator 3R/2 at the specified frequency without any need for Laplace.

Regards, Dan.

[Vtile]

'Nuff said. If you have a thing about referring to previous topics somewhere else that nobody understands, the Finnish winter is apparently too boring.

Sometimes.
Unfortunately I'm writing from phone so I can't easily link it here.

[Benta]

I hate it when they do that in an exam

Yep - and I fell right into it    - fail.

[IanB]

PS: how do you insert symbols? Using w instead of greek omega is sub-optimal.

You can use MathJax. So for example:

Code: [Select]

\$u_g(t)\$, \$R\$, \$1/sC\$ and \$j\omega\$
will produce:

\$u_g(t)\$, \$R\$, \$1/sC\$ and \$j\omega\$

[rstofer]

PS: how do you insert symbols? Using w instead of greek omega is sub-optimal.

You can use MathJax. So for example:

Code: [Select]

\$u_g(t)\$, \$R\$, \$1/sC\$ and \$j\omega\$
will produce:

\$u_g(t)\$, \$R\$, \$1/sC\$ and \$j\omega\$

Thanks for that!  I'll have to cut and paste that code into a file and save it somewhere.
I found this:

https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

[bson]

There's also a handy MathJax demo .  Just type there until you like the result, then insert the latex into a post.

[tatus1969]

you don`t need *any* complex number math to solve this one. Start with L, C, and w=1/sqrt(L*C). You can solve this one by looking at the picture only.

[nForce]

You are given that omega = 1/sqrt(LC) which is nice, think about what you know about series resonance in LC networks, it makes this real easy to solve.

Yes, that green circuit is original series LC circuit with resonant frequency \$\omega= \frac{1}{\sqrt{LC}}\$

But how do i use that for calculating real power?

[dmills]

I read that as telling you to calculate the reactive power at frequency w=1/Sqrt(LC), which is trivial if you think about what impedance a series LC network has at resonance.

Xc + Xl = 0 at resonance, so the green box degenerates into what?

The rest is just trivial arithmetic.

Regards, Dan. 

[nForce]

I read that as telling you to calculate the reactive power at frequency w=1/Sqrt(LC), which is trivial if you think about what impedance a series LC network has at resonance.

Xc + Xl = 0 at resonance, so the green box degenerates into what?

The rest is just trivial arithmetic.

Regards, Dan. 

Oh, I get it. In series resonance the impedance of L and C (both of them) is 0. So only the resistor remains. Because of that I can calculate with P = (U^2) / Z

[nForce]

I have another question which is related to this.

Why do we look at impedance in series RLC circuit and admitance in parallel RLC circuit?

[dmills]

Impedances sum in series, admittances sum in parallel, it just makes the maths less annoying.

Regards, Dan.