# Tricky circuit analysis

Discussion in 'Homework Help' started by JoyAm, Sep 20, 2015.

1. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
Hello everyone, i had a really hard time with this one because i cant find the angles. I know the angle of the source voltage as zero and the angle of the empedance of the line to be 86.64. I can also say that the angle of the coil current is -90 of the angle of the coil voltage.But after that i am suck.The currents seem to form an orthogonal triangle but i dont know how to use that fact ( if its even helpful)

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Too many words.
Too few digits.

3. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
What do you mean ?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Source voltage?
Impedance?
Circuit layout?

5. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
did you check the picture i have in thumbnail ? the only impedance i know is the one of the line

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
I don't see picture. I don't see thumbnail.

7. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
Idk why i can see it myself try this link ( its a screenshot) http://prntscr.com/8iip3r

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Are A1, A2, A3 current sources?

9. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
I thought it is so because it has cos(Omega*t+0)

10. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
You are right, sorry, I was a bit dense.
Should it not be 240*1.41? Since the magnitude has the square root of 2?

I only have two questions.
Are A1, A2, A3 independent current sources?
Do you have value for L or XL (I assume it is inductor)?

11. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
No they are amp meters or whatever they are called

12. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
Yes it is an inductor but i do not have a value ( all i know is that this element is an inductor)

13. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Ok. If A3 is 3 A, then Iload is 3 A.
I know it sounds too simple. But... given the information you provided it is that simple.

14. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
What about square root of 2 in the amplitude of Vn? That would make it 339.41 volts @ 0°.

15. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Overall, the solution looks like this:
-You know A1.
-Find the voltage drop across the impedance that is series with Vn (I will call it Z1).
-ZL and Load are in parallel, that means that voltage across ZL and Load is the same. So take Vn substract voltage across Z1. That will give you voltage across ZL and Load since it is the same. This will give you values in complex form.
-Now do Inverse Fourier Transform to get value in time domain.

16. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
i divide it by sqrt(2) because i have the rms value on the phasor and yes i agree with what you say Iload will be 3 but what will be the angle ?(because if you dont take angles into consideration and do a kvl you get something like 4=5+3 )

17. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Are you telling me that current meters in the circuit give you magnitudes only?

If the load is purely resistive (a resistor), there will not be a phase angle to the current.
In previous post I showed how to get Vload. You can take it and either add 90° or subtract 90° to get the phase angle of the Iload. To add or to subtract depends on the load. Is it resistive? Is it iductive? Is it capacitive?

18. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
How can i find the voltage drop on z1 since i dont know the angle of the curent I1?

19. ### JoyAm Thread Starter Member

Aug 21, 2014
112
4
Yes, magnitude only and i am not given information about the load so i guess it can be anything

20. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,397
497
Then you have too many unknowns. I don't think it can be solved.