An AC-circuit

1. Nov 23, 2008

Niles

1. The problem statement, all variables and given/known data
Hi all.

Please take a look at the attachted circuit.

I have found the following expression for the complex current in the resistor (the hat indicates that it is complex):

$$\widehat_{I (t)} = \frac{{\omega ^2 LC {\widehat{U_0} }}}{{\omega ^2 LRC + i\omega L - R}}\exp ( - i\omega t),$$
where U_0 is the amplitude of U(t), and the hat indicates that it is complex.

Now I wish to find the real current, and I want to write it as:
$$I(t) = I_0\cos(\omega t + \phi).$$

But how do I do this? I have spent like 2 hours trying, but I don't know how to rewrite the complex amplitude of the current to have a phase.

Regards
Niles.

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2. Nov 23, 2008

tiny-tim

Hi Niles!

(have an omega: ω and a phi: φ )

The standard way would be to write the left bit as A + iB (multiply top-and-bottom by the complex conjugate of the bottom), and then

Re(A + iB)eiωt = Re(A + iB)(cosωt + isinωt)

= Acosωt - Bsinωt,

which maybe you know how to rewrite as Ccos(ωt + φ) ?

3. Nov 23, 2008

Niles

Ok, so you are telling me to multiply the first term on the RHS with the complex conjugate and then take the square root. Thus I get the amplitude of the current, and now we have to multiply that with exp(iφ).

But this gives me: ... * exp(-i(ωt-φ))? Don't we want a plus instead?

4. Nov 23, 2008

tiny-tim

Why are you taking the square root?

(And you did multiply top and bottom, didn't you? To get a complex over a real?)

Anyway, you don't multiply the amplitude with exp(iφ) …

you can't find the amplitude (or the phase) until you've got the whole function …

in other words, until after you multiply.

5. Nov 23, 2008

Niles

Ok, so what I do is:

1) Multiply the first term on the RHS with the complex conjugate of the denominator.

2) I find the phase, and write the first term as: z = |z|*exp(iφ).

3) Find the real part, and all ...

4) Done? If yes, then I will go to bed.

EDIT: Can you confirm me in that -0.1 * 2*Pi radians is equal to 2*Pi - 0.1 * 2*Pi radians?

6. Nov 23, 2008

tiny-tim

I'm not convinced that you've got it.

You're given (K + iL)/(M + iN) times eiωt

that's (K + iL)(M - iN)/(M + iN)(M - iN) times eiωt

= (K + iL)(M - iN)/(M2 + N2) times eiωt

the bottom is real, so the whole of the left can now easily be put in the form A + iB …

go on from there.
Yes, adding multiples of 2π to the phase makes no difference … e2πi = 1

… so if your result comes out as more than 2π, just subtract 2π (or 4π or … )

7. Nov 23, 2008

Niles

And then I write z = A + iB as z = |z|e, right?

8. Nov 23, 2008

tiny-tim

sorry!

ooh, I misunderstood what you wrote earlier …

even though it was clear
Yes, that's right … then the real current will be the real part of |z|*exp(-iωt + iφ), = |z|*cos(-ωt + φ)

To answer a previous point, cos(x) = cos(-x), so the minus sign doesn't matter (though it will mess up the sign of φ).

Ignore some of my previous comments.

9. Nov 24, 2008

Niles

Re: sorry!

Nono, they were all good. Thanks for helping me - I see it much clearer now!

Sincerely,
Niles.

10. Nov 24, 2008

Niles

I have found the phase φ to be -0.63, which is equivalent of 5.65. Does this mean that I am allowed to write:

I(t) = Re [|z|*exp(-iωt - 0.63i)] = Re [|z|*exp(-i(ωt + 0.63)] = |z|*cos(ωt + 0.63),

where Re [..] denotes the real part of the argument "...".

Last edited: Nov 24, 2008
11. Nov 24, 2008

tiny-tim

Yup!

12. Nov 24, 2008

Niles

Great! Thanks!

Just one last thing: Is it correct that sometimes the expressions for the current are very nasty? I mean, they are sometimes quite long :uhh:

13. Nov 24, 2008

tiny-tim

oooh … dunno!

i never did current 'n' stuff …

i just know about the maths bit!

14. Nov 24, 2008

Niles

Ahh, I see.

But thanks anyway!

15. Nov 24, 2008

Nissen, Søren Rune

Hah, yes.

Consider a serial RCL circuit, that is, a voltage source connected to a condensator, a loop and a resistor. Let's say the Resistor is 200 ohm, the loop is 1 millihenry and the condensator is 20 nanofahrad. If you apply a simple voltage, not even timevariable, of 10 volt to this system, the current will look like this:

i(t)=0.05*e^(-100000*t)*sin(200000*t)

(this assumes i(0)=0 and i'(0)=10000)

If, instead, you apply an AC source, the voltage function (again assuming i(0)=0 but i'(0)=2*10^9) of which being v(t)=10*sin(20000*t), the resulting current function looks like this:

i(t) = (1/20)*e^(-100000*t)*sin(200000*t)+B*sin(200000*t)+A*cos(200000*t).

Explaining why this is so takes about two full pages.

Last edited by a moderator: Nov 24, 2008
16. Nov 25, 2008

Nissen, Søren Rune

Oh, I forgot to substitute the variables. A=1/85 and B=4/85

17. Nov 25, 2008

Niles

I am familiar with Kirchhoff's laws (I guess that's what you are referring to?), and I used them to find my expression.

But I guess I should stop fearing long expressions

18. Nov 25, 2008

Nissen, Søren Rune

Nah, it's

L*i''(t)+R*i'(t)+1/C*i=V'(0)

where

L=1 milliHenry, R=200 ohm and C=20 nanofahrad.
Applying 10 volt (DC) to this system gets you

0.001*i'' + 200*i' + 5*10^7*i = 0.

Solving this equation:

0.001*y^2 + 200*y + 50000000 = 0

gives you two imaginary numbers,

y1=100000 + j*200000
y2=100000 - j*200000

which means that

i(t) = e^(-100000*t)*( A*sin(200000*t) + B*cos(200000*t) ).

(This is the general solution to the "ordinary differential equation" shown above.)

To find A and B, you need to know some starting conditions. We know i(0)=0 because the loop has "infinite resistance*" when you apply the initial potential.

Knowing that i(0)=0 means you can rewrite, so:

i(t) = e^(-100000*t)*( A*sin(200000*t) + B*cos(200000*t) )
=>
i(0) = e^(0)*(A*sin(0) + B*cos(0))
=>
0 = A*sin(0)+B*cos(0)
=>
0 = B*1
=>
B=0

so you can simplify the initial expression for the current to

i(t) = e^(-100000*t)*A*sin(200000*t)

from elsewhere (and I honestly can't remember) we know that i'(0)=10000, so A can be isolated and determined. I don't want to write out di/dt here because it is a damn long function, but suffice to say, it turns out A = 0.05 so

i(t) = 0.05*sin(200000*t)*e^(-100000*t)

All in all, current determination becomes complicated once you have a lot of loops and capacitors, but not so difficult that you should feel discouraged - there are a lot of shortcuts and neat tools at your disposal if you ever start in on the subject in a serious manner.

*not true