# AC circuit: 2 Dependent sources (Confirmation with software)

#### michgkou

Joined Oct 3, 2019
38
Hello,

I was found the node's voltages of the circuit and I want a confirmation. Can someone verify using a software ?

This is my try: https://docdro.id/AontV5p

The circuit:

Thanks.

#### The Electrician

Joined Oct 9, 2007
2,832
On the second page of your calculations you have:

I6 + I1 = 2 Vo + I5 ==>

(10-V3)/2 +.........

The expression for I6 is not correct; it should be: (V4+10-V3)/2

Show us your results after you make the correction.

#### MrAl

Joined Jun 17, 2014
8,150
Hello,

I was found the node's voltages of the circuit and I want a confirmation. Can someone verify using a software ?

This is my try: https://docdro.id/AontV5p

The circuit:
View attachment 191155

Thanks.
Hi,

The general rule for this site is to upload all of your images to the forum itself so that this site does not have to depend on external sites to maintain images. If the site you upload to goes away for any reason then this thread becomes useless to others in the future.

#### michgkou

Joined Oct 3, 2019
38
On the second page of your calculations you have:

I6 + I1 = 2 Vo + I5 ==>

(10-V3)/2 +.........

The expression for I6 is not correct; it should be: (V4+10-V3)/2

Show us your results after you make the correction.

Thanks.

#### The Electrician

Joined Oct 9, 2007
2,832
In your first working (in the image attached to post #1) you had an expression for I5 of (V3-V4)/-j2

In your latest working you have an expression for I5 of V3/-j2. This doesn't seem right; can you explain why you changed it?

#### MrAl

Joined Jun 17, 2014
8,150
Hi again,

Question, why did you make the cosine source 0 degrees and the sine source 90 degrees. Shouldnt it be the other way around? Normally a sine source is taken to be at 0 degrees and a cosine at 90 degrees because the sine source passes though 0 volts (or amps) at 0 degrees.

#### michgkou

Joined Oct 3, 2019
38
In your first working (in the image attached to post #1) you had an expression for I5 of (V3-V4)/-j2

In your latest working you have an expression for I5 of V3/-j2. This doesn't seem right; can you explain why you changed it?
I forgot it, thank you !

#### michgkou

Joined Oct 3, 2019
38
Hi again,

Question, why did you make the cosine source 0 degrees and the sine source 90 degrees. Shouldnt it be the other way around? Normally a sine source is taken to be at 0 degrees and a cosine at 90 degrees because the sine source passes though 0 volts (or amps) at 0 degrees.
sine is minus 90.

#### The Electrician

Joined Oct 9, 2007
2,832
I forgot it, thank you !
What do you get for the 3 node voltages with that error fixed?

#### michgkou

Joined Oct 3, 2019
38
V3=0.7167∠-174.89°
V4=5.5582∠133.605°
and
V2=10.2855∠37.3431°

#### michgkou

Joined Oct 3, 2019
38

#### The Electrician

Joined Oct 9, 2007
2,832
I get the same result. How many total hours do you think you spent on this problem?

#### michgkou

Joined Oct 3, 2019
38
I get the same result. How many total hours do you think you spent on this problem?

View attachment 191327
Probably more than they should have. I have to be more careful.

Thanks.

#### MrAl

Joined Jun 17, 2014
8,150
sine is minus 90.
Hello again,

Well thanks but that is not an explanation of why you made the cosine source 0 degrees and the sine source minus 90 degrees. The cosine source peak starts at 0 degrees not the zero crossing.
A sine source is normally placed at 0 degrees. If you change that to -90 then all your phase shifts will be different by 90 degrees.

So the question is what is making you place the sine source at -90 degrees? What told you to do that? We already know you did that, but what caused you to believe you should make it minus 90 degrees? Did some instructor tell you to consider the cosine source to the the reference source or what?

For another example, if you had a source of sin(w*t) and another source of sin(w*t+30 degrees) would you arbitrarily make the 30 degree source the reference phase? No, not unless there was something else in the system that suggested you should do that or the instructor told you to do that or some other good reason.

Last edited:

#### michgkou

Joined Oct 3, 2019
38
Yes, the instructor told me using it this way.

#### MrAl

Joined Jun 17, 2014
8,150
Yes, the instructor told me using it this way.
Ok great.

Now did anyone try to simulate this circuit yet?