# Need help with this circuit problem

1. Jul 22, 2013

### MissP.25_5

i1(t) = √2sin(2t+2π/3) [A]
i2(t) = √2sin(2t+π/6) [A]

1. Find R and L.
2. Given that the phase difference between e and i is arg(e/i)=θ. Find tan θ.

I am not sure how to do this. I know that first we need to find e. Is it easier to use phasors to solve this or solve it in time function form? Did I get the e correct?

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Last edited: Jul 22, 2013
2. Jul 22, 2013

### rude man

Too late tonite, someone else will probably chip in. Or me, 10 hrs later ...

BUT:
you can find e from i1(t) since R = 3 and C = 1/4F are given in the left branch (inner loop).
Then solve the outer loop for R and L, knowing e and i2(t).

Yes you should use phasors for e, i1 and i2, and complex forms of reactance. Using time-domain is messy but doable of course. Wouldn't try it myself.

3. Jul 22, 2013

### MissP.25_5

I will try to use phasors, then. But did get I the e(t) correct?

4. Jul 22, 2013

### rude man

I don't know right off. Wouldn't express e that way.

Why not use phasors? E = phasor of e(t). Then

I1 = ej2π/3
E = I1*Z
E, I1 and Z all complex. ω = 2.

5. Jul 22, 2013

### MissP.25_5

Can't we just take the magnitude of I1 and I2, without taking into account the phase? So that would be 1 amp for both I1 and I2. And here's what I have done so far, I got stuck in the end. It seems tedious too, I am sure there must be an easier way to do this but I just can't figure out.

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Last edited: Jul 22, 2013
6. Jul 22, 2013

### Staff: Mentor

Nope. The phases are critical. The voltage source e is going to have phase, as will the total current, I.

When you have a phasor of the form $A e^{j \theta}$, you can write that as a complex number in polar or rectangular form. In polar form, A ∠ θ. You should be able to convert between polar and rectangular form as needed.

So, begin by determining the voltage source phasor E in complex form.

7. Jul 22, 2013

### MissP.25_5

I did use phasors in the form of $A e^{j \theta}$ but still got the same result as the attached solution. What do you think of my working there?

8. Jul 22, 2013

### Staff: Mentor

You dropped the angle and worked with magnitudes. That doesn't work; when complex numbers are multiplied or divided, the imaginary terms interact with the real terms.

Convert your current to rectangular form and then multiply by the rectangular form of the impedance.
The conversion is made easy because the angles involved have well known, exact values of sine and cosine.

9. Jul 22, 2013

### MissP.25_5

Rectangle form meaning x+yj form? And after I got the E, what should I do next?

10. Jul 22, 2013

### Staff: Mentor

Yup.
Well, you'll have the voltage E across and the current i2 through the unknown impedance...

11. Jul 22, 2013

### MissP.25_5

So does that mean solving simultaneous equations? I mean, when we have E, we know that E is the same in each parallel and it's also the voltage for the whole circuit. The unknown impedance is R+j2L,right? That gives one equation E=(R+j2L)/(I2). And for the second equation, I'd find the total impedance of the circuit and get an equation E=Zs/(I1+I2). But this seems to be a really long work and takes time! Is there an easier way to get it?

12. Jul 22, 2013

### Staff: Mentor

Nothing nasty at all! Use Ohm's Law to find the complex value of the impedance of that branch. Equate terms.

13. Jul 22, 2013

### MissP.25_5

Is this what you mean? I equate the real terms and imaginary terms, but the values seem unlikely, especially R, cuz I got it negative.

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14. Jul 22, 2013

### MissP.25_5

Ok, I re-did it and got L=3 but I got R=2-3√3. Did I get them right? R is negative, though.

15. Jul 22, 2013

### Staff: Mentor

Yeah, something's amiss with the real part of E; Overall the term should end up with a positive value. Better check that calculation.

16. Jul 22, 2013

### Staff: Mentor

I'm seeing both terms of the impedance as simple integer values.

17. Jul 22, 2013

### Staff: Mentor

What is your final complex value for E?

18. Jul 22, 2013

### MissP.25_5

Finally!!Is this correct? L=3/2 and R=2.

19. Jul 22, 2013

### Staff: Mentor

Yes, that looks good. Huzzah!

20. Jul 22, 2013

### MissP.25_5

Thanks! I still have to do the second question, finding tanθ.