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.

 

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.

 

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.

 

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.

 

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?

 

View attachment 191239

View attachment 191240

View attachment 191241

Thanks.

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.

 

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 !

 

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.

 

I forgot it, thank you !

What do you get for the 3 node voltages with that error fixed?

 

V3=0.7167∠-174.89°
V4=5.5582∠133.605°
and
V2=10.2855∠37.3431°

 

 

I get the same result. How many total hours do you think you spent on this problem?

 

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.

 

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.

 

Yes, the instructor told me using it this way.

 

Yes, the instructor told me using it this way.

Ok great.

Now did anyone try to simulate this circuit yet?