# Basic AC circuit analysis

1. Apr 15, 2014

### dcrisci

Okay so I have been working on a problem for practice for an exam and this one question is really pissing me off. I can't find what I am doing wrong and I've redone it about five times.

The question is asking for the admittance of the circuit and the answer is given as 2.29 S at an angle of -42.2 degrees (don't know how to make the phasor symbol)

Picture of the circuit

Picture of my solution

I realize my answer is not the same but I have no clue where I am going wrong with this. I began from the right side of the circuit doing the following:
1. Calculate the impedance of the inductor and resistor in series
2. Calculate the admittance of part 1 and the second inductor since they are in parallel
3. Convert part 2 into impedance
4. Calculate impedance of part 3 added with the capacitor and resistor
5. Calculate the admittance of part 4 added with the parallel resistor
6. Convert part 5 to impedance, add this impedance with the final inductor
7. Convert back to admittance to find out the answer is wrong

2. Apr 15, 2014

### Staff: Mentor

Can you show your calculated results for the steps?

You know that you can do the whole thing with admittance values, right? No particular need to convert back and forth from admittance to impedance. Admittances in series add like impedances in parallel.

3. Apr 15, 2014

### dcrisci

My calculated values should be in the one picture, is it not showing for anyone other than me? I figured I could have done it with only admittance I was only following an example that was very similar

4. Apr 15, 2014

### Staff: Mentor

My apologies, I somehow managed to ignore the second attachment. I'll have a look.

5. Apr 15, 2014

### Staff: Mentor

Okay, right off the bat when you calculate your Z1, you're adding two admittances as though they were impedances. The values given on the schematic for the components in question are in Siemens, so they are admittances and must be added as such. How do admittances in series add?

6. Apr 16, 2014

### dcrisci

sorry I got caught up with other studying. Now that I am back to circuits..

Okay so I don't know how I didn't think of that the whole time (the units being Siemens), Let me try this again and get back to you.

7. Apr 16, 2014

### dcrisci

I am still having a problem with it. I can't seem to get it right and don't know what I am doing wrong.

This time around I was trying to only use admittances so when adding the parallel components I added them as (1/G1) + (1/G2) and then I thought this would give me units of impedance so I took the inverse of this answer.

Also sorry for the messy work I have some chicken scratch writing.

8. Apr 16, 2014

### Staff: Mentor

That's the right idea, but it's not reflected in your calculation of Y1 where it seems you've just added the admittances of the 2 S resistor and -j2 S inductor. They should be combined as you've described above.

You'll probably find it convenient to leave the admittances in rectangular form through most of the calculations, except perhaps where polar form (magnitude/angle) might make performing a division easier.

Believe me, I've seen a lot worse

9. Apr 16, 2014

### dcrisci

I dont understand why it would be like that for the 2 S resistor and -j2 S inductor, aren't the two in series so the admittances should just combine as a normal sum?

Edit: would I technically need to convert the two into their relative impedances to add them as if they were series?

Last edited: Apr 16, 2014
10. Apr 16, 2014

### Staff: Mentor

No No No No No! Conductances and admittances add via a normal sum when they are in parallel. When they are in series they add like resistors do in parallel, namely:
$$Y_{series} = \frac{1}{\frac{1}{Y1} + \frac{1}{Y2}+ ... + \frac{1}{Yn}}$$
You spelled it out in a previous post, so I thought you had that well in hand.

11. Apr 16, 2014

### dcrisci

Okay so if I'm using that formula I seem to get the same answer when adding the first two admittances. I just felt like the 2 - j2 S was all wrong and don't like seeing it here as well

12. Apr 16, 2014

### Staff: Mentor

The series "addition" of 2 S and -j2 S should yield (1 - j) S. You can use the shortcut formula for two components:
$$Y = \frac{Y1 \cdot Y2}{Y1 + Y2}$$
$$Y = \frac{(2) (-j2)}{(2) + (-2j)}$$

and carry out the algebra to reduce it.

EDIT: Oops. Fixed the sign on the "sum". The -j2 S would make the imaginary component of the "sum" negative too. So 2S and -j2 S in series yields (1 - j) S.

Last edited: Apr 16, 2014
13. Apr 16, 2014

### dcrisci

Yeah I just got so confused trying to use this. My professor barely covered using complex numbers so I don't really know what I am doing. I just tried googling it and understand it is something to do with rationalizing it but I am just way too lost.

14. Apr 16, 2014

### Staff: Mentor

You'll find that the basics of complex number arithmetic are essential for dealing with the math of AC circuits. It's something you'll need to spend some time on mastering.

There are online complex number calculators that you can find with a web search. They can help you to check your math while you're practicing.

15. Apr 16, 2014

### dcrisci

Wooooo got it but I had to switch between impedances and admittances. I'm sure this way is longer than how you were trying to get me to approach it but nonetheless I finally got the correct answer. I think my problem with initially trying it this way was that I was forgetting that 1/jA = -j(1/A)

Here is how I went about it all

I converted the admittances to impedances in the beginning as well.

16. Apr 16, 2014

### Staff: Mentor

Yup. That's certainly one way to go about it. The result looks fine. Lot's of brain sweat involved though

17. Apr 16, 2014

### dcrisci

You sir have no idea! haha
I have been going at this one for a while now so it feels great to get it right finally.