# Tricky circuit analysis

#### JoyAm

Joined Aug 21, 2014
126
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)

#### shteii01

Joined Feb 19, 2010
4,644
Too many words.
Too few digits.

#### JoyAm

Joined Aug 21, 2014
126

#### shteii01

Joined Feb 19, 2010
4,644

#### JoyAm

Joined Aug 21, 2014
126
Source voltage?
Impedance?
Circuit layout?
did you check the picture i have in thumbnail ? the only impedance i know is the one of the line

#### shteii01

Joined Feb 19, 2010
4,644
did you check the picture i have in thumbnail ? the only impedance i know is the one of the line
I don't see picture. I don't see thumbnail.

#### JoyAm

Joined Aug 21, 2014
126

#### shteii01

Joined Feb 19, 2010
4,644
Are A1, A2, A3 current sources?

#### JoyAm

Joined Aug 21, 2014
126
Run me through the math of how you get Vn=240@0°
I thought it is so because it has cos(Omega*t+0)

#### shteii01

Joined Feb 19, 2010
4,644
I thought it is so because it has cos(Omega*t+0)
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)?

#### JoyAm

Joined Aug 21, 2014
126
Are A1, A2, A3 current sources?
No they are amp meters or whatever they are called

#### JoyAm

Joined Aug 21, 2014
126
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)?
Yes it is an inductor but i do not have a value ( all i know is that this element is an inductor)

#### shteii01

Joined Feb 19, 2010
4,644
No they are amp meters or whatever they are called
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.

#### shteii01

Joined Feb 19, 2010
4,644
Yes it is an inductor but i do not have a value ( all i know is that this element is an inductor)
What about square root of 2 in the amplitude of Vn? That would make it 339.41 volts @ 0°.

#### shteii01

Joined Feb 19, 2010
4,644
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.

#### JoyAm

Joined Aug 21, 2014
126
What about square root of 2 in the amplitude of Vn? That would make it 339.41 volts @ 0°.
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 )

#### shteii01

Joined Feb 19, 2010
4,644
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 )
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?

#### JoyAm

Joined Aug 21, 2014
126
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.
How can i find the voltage drop on z1 since i dont know the angle of the curent I1?

#### JoyAm

Joined Aug 21, 2014
126
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?
Yes, magnitude only and i am not given information about the load so i guess it can be anything

#### shteii01

Joined Feb 19, 2010
4,644
Yes, magnitude only and i am not given information about the load so i guess it can be anything
Then you have too many unknowns. I don't think it can be solved.