How do I calculate the total impedance of two parallel impedances? - Page 1
Forum section: electronics / beginners

author: cvriv
author_rank: Frequent Contributor
timestamp: October 11, 2016, 04:56:36 am
content: The parallel circuit has 3 branches total. On one branch there's a signal source and a resistor. The second branch has a capacitor and a resistor. The third branch has a capactior and two resistors.I know how to calculate the impedance of each branch along with their phase angles, but how do I calculate the the impedance of those two parallel impedances?I checked my book and it doesn't have anything on how to do that. It only shows how to do simple calculations of total impedance on a series-parallel RLC circuit.I also checked online and still Im only finding simple series-parallel RLC type circuits.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2016, 06:26:35 am
content: Admittances in parallel add:Ytot = Y1 + Y2 + ... + YnAdmittance is simply reciprocal impedance (Y = G + jB = 1/Z). Thus, you get the "reciprocal of the sum of reciprocals" that's the conventional rule for parallel circuits.Or for two branches,Ztot = Z1*Z2 / (Z1 + Z2)which you will see follows from the above.Tim

author: sainter
author_rank: Regular Contributor
timestamp: October 11, 2016, 06:28:27 am
content: Without seeing your assignment I would say: when i doubt, kirchhoff laws.https://www.youtube.com/watch?annotation_id=annotation_3932286205&feature=iv&src_vid=WBfAEeEzDlg&v=8f-2yXiYmRI

author: cvriv
author_rank: Frequent Contributor
timestamp: October 11, 2016, 07:55:22 am
content: So it doesn't matter that the phase angles for each impedance value are different? Thanks.

author: cvriv
author_rank: Frequent Contributor
timestamp: October 11, 2016, 08:07:19 am
content: Here's a picture of the curcuit. If i product-over-sum the impedance values, which have different phase angles, what would be the new phase angle?When i simulated the circuit and measured the impedance of the two branches with capacitors, the values were just about the same. The value was about 10kOhm, if i remember correctly.

author: dmills
author_rank: Super Contributor
timestamp: October 11, 2016, 08:48:15 am
content: Easiest to do the work in the complex plane, remembering that multiplying by Z*/Z* then expanding the denominator can be used to end up with a real denominator.This is in my view one of the places where working in Cartesian space is easier then polar, start by converting everything to complex admittance and then you can just sum them before converting back to impedance.Regards, Dan.

author: greenhorn
author_rank: Newbie
timestamp: October 11, 2016, 09:50:07 am
content: Quote from: dmills on October 11, 2016, 08:48:15 amYes, you can do product-over-the-sum, but with complex impedances. Then the argument of the resulting complex impedance is the phase angle you are looking for.Btw, as the two impedances are almost the same, adding them in parallel should yield about half the impedance, 25kO, with angle about the same.And btw, the phase angles on your drawing are correct values, but in radians not degrees.

author: cvriv
author_rank: Frequent Contributor
timestamp: October 11, 2016, 09:53:07 am
content: This is as far as I got with this. I don't know how to combine the two impedance values with different phase angles. My book says nothing about this, which just blows my mind.I don't fully understand what your telling me to do. Im familiar with the reciprocal equivalents of R, Xc, Zt..... G, Bc, and Y, but don't know what to do with that.... The phase angles of Y would still be different and and the relationship between the G's and the Bc's are still in parallel.

author: cvriv
author_rank: Frequent Contributor
timestamp: October 11, 2016, 09:57:39 am
content: Yea i realized the angles were in radians. Im tired. The angles are about -78 and -72 degrees. I know the impedance would be about half. I still need to figure out how to get the phase angle though so i can convert to rectangular values. Also, i have to sim this again because i swear that when i measured the resistance across each branch, the impedance was a lot less than 24k.

author: greenhorn
author_rank: Newbie
timestamp: October 11, 2016, 10:12:17 am
content: It is best to learn to work with complex numbers. I dont know of nice way to solve your problem without them.If you use complex impedances, thenZ1 = 10 - i*48.3Z2 = 15.6 - i*48.3(in kOhms)i is imaginary unit, i^2=-1.Then you can use product-over-the-sum to find the resulting impedance.

author: nfmax
author_rank: Super Contributor
timestamp: October 11, 2016, 10:49:49 am
content: You could always use a decent scientific calculator that handles complex arithmetic. I tend not to bother with the product-over-sum formula and just use the 1/x key.

author: JPortici
author_rank: Super Contributor
timestamp: October 11, 2016, 11:52:05 am
content: they are complex numbers.for impedances in parallel i follow the general ruleZp = (Z1^-1 + Z2^-1 + ... + Zn^-1)^-1Because i am lazy and my calculator can perform operation with complex numbers in any form they are written. I can mix algebric, polar and exponetial, she won't care.if i have to do that by hand,sum/subtracion of complex number is easier when they are in the "a + ib" form, real sums with real. imaginary sums with imaginary.multiplication/division is easier when they are in the "mod<phi" form, modules are multiplied/divided between them. phases are summed when multiplying and subtracted when dividing.

author: dmills
author_rank: Super Contributor
timestamp: October 11, 2016, 01:36:01 pm
content: The OP seems to have a problem with cartesian form, polar is the wrong way to do this.Let Z1 be R1 + C1Z1 = 10 - 48.3j (- because it is a capacitive reactance).Then Y1 = 1/Z1 = 1 /(10-48.3j), units in ms because input was in k ohms.Multiply numerator and denominator by complex conjugate:Y1 = (10 + 48.3j)/((10 - 48.3j)(10+48.3j))Multiply out the denominator 10 * 10 + 48.3j * 10 -48.3j * 10 - (48.3j)^2But j^2 = -1, so expansion is 10^2 + 48.3^2(Electronics types use j rather then i because i was already used for current).Y1 = (10 + 48.3j)/(10^2 + 48.3^2)Y2 will have essentially the same form.Someone check my working, it has been a LONG time.I leave taking the reciprocal and calculating the tangent to the student.

author: greenhorn
author_rank: Newbie
timestamp: October 11, 2016, 03:52:24 pm
content: QuoteLooks OK.Let's go all the way to the end.So, since parallel admittances simply add:Y = Y1 + Y2Z = 1 / Y is your impedance of the two parallel branches.Z is a complex number a +bj.Absolut value of Z is sqrt(a^2 + b^2).b/a is the tangent of phase angle.

author: BobsURuncle
author_rank: Regular Contributor
timestamp: October 11, 2016, 04:18:00 pm
content: QuoteYou do it the same way you would do it with all resistors except you need to use complex arithmetic. 1/Zp = 1/Z1 + 1/Z2.... +1/Zn; or with only two branches you can use Zp = (Z1*Z2)/ (Z1 + Z2).Generally it is easy to multiply and divide using polar coordinates and to add and subtract if the phase angles are the same. It is easier to add and subtract using cartesian coordinates if the phase angles are different. There are multiple paths you can use to get the final impedance (Zp) but in most approaches some conversion between polar and cartesian coordinates is necessary. If you don't have a calculator to determine angles from tangents then you pretty much have to stay in cartesian form which I find a little messy.In the attached approach I use the formula (Z1*Z2)/ (Z1 + Z2). First converting Z1 and Z2 to polar form for multiplication in the numerator and use the cartesian form for addition in the denominator. Then convert the denominator to polar to divide into the numerator for the final answer:25K/_-75deg.If you are not completely comfortable with complex number math you should take this opportunity to quickly bone up on it because it is the bread and butter tool of electronics and you will quickly fall behind if you have not absorbed it.

author: cvriv
author_rank: Frequent Contributor
timestamp: October 11, 2016, 06:00:43 pm
content: Thanks everyone. So I calculated 25kOhm and reading through this thread and speaking to my professor as well. I dont fully understand this. I'll have to look into this more later. I haven't seen anything about this in my book which sucks. I have two other books that I have to look through to see if this is covered.So i calculated Z total as 29.2kOhm and with that I calculated the voltage across C1 as 573.5mV. I know I cant measure reactance in multisim, but I did measure the voltage across C1 and the value was 593mV. Why are the values off? It's only a 3.43% difference but we're talking about a calculation being compared to a sim. Why is it off so much?

author: BobsURuncle
author_rank: Regular Contributor
timestamp: October 11, 2016, 08:33:35 pm
content: QuoteSince you don't show your work, I can't speak for the accuracy of your calculation or how well you set up the model in MS. In any case you are solving with a simplified abstraction of the circuit to get a specific time indepenent answer which involves just a few calculations. Multisim, when solving circuits with reactive elements, is using numerical methods (such as Taylor series approximations) to solve differential equations to determine the state of the circuit at hundreds of intervals of time. You can see that the accumulation of error could become significant. At the sacrifice of speed those errors can be minimized by choosing appropriate parameters for the run, e.g. smaller time interval steps.With typical capacitor tolerances of 20% or worse, 3% error is likely to be a lot less than the deviation from nominal of a physical circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2016, 10:14:01 pm
content: Quote from: BobsURuncle on October 11, 2016, 08:33:35 pmWell, if you've done an AC analysis, it's just doing plain static nodal analysis, just as you'd do it on paper. But with a matrix inversion on the CPU, so it's super fast.If you're piddling around with impedances in a transient simulation, euch...Tim

author: BobsURuncle
author_rank: Regular Contributor
timestamp: October 12, 2016, 12:49:54 am
content: QuoteThat won't provide an AC plot vs time, switch on ramp ups or other transient responses.http://www.ni.com/white-paper/5808/en/#toc2Quote from: National Instruments

author: IanB
author_rank: Super Contributor
timestamp: October 12, 2016, 01:03:33 am
content: Quote from: cvriv on October 11, 2016, 06:00:43 pmYou are writing down the impedances as if they are resistances. Do you have an understanding of complex numbers and how to do complex number arithmetic? If you do not have that in your mathematical foundation then everything written in this thread will make no sense to you.

author: Ratch
author_rank: Regular Contributor
timestamp: October 13, 2016, 03:42:57 am
content: Quote from: cvriv on October 11, 2016, 04:56:36 amWow! You sure are making a simple problem a hard slog. First of all, if the impedances are given in rectangular coordinates, why convert to polar? I do hope you can do complex arithmetic. If not, you are going nowhere fast. The hardest part of this problem is doing the product divided by the sum, using complex numbers representing the impedances of the circuit components. After that, it is a snap. Look at the attachment I enclosed.Ratch

author: danadak
author_rank: Super Contributor
timestamp: October 13, 2016, 10:52:05 am
content: Simple analysis with Laplace, just HS algebra.https://web.stanford.edu/~boyd/ee102/laplace_ckts.pdfJust google "laplace circuit analysis", moe than you want to know.Regards, Dana.

author: Ratch
author_rank: Regular Contributor
timestamp: October 13, 2016, 02:57:15 pm
content: The OP did not ask for the circuit response. He asked for the parallel impedance of a two branch circuit.Ratch

author: danadak
author_rank: Super Contributor
timestamp: October 14, 2016, 01:02:25 am
content: Starting at page 20 of the attachment an impedance S parameter analysis.Regards, Dana.

author: Ratch
author_rank: Regular Contributor
timestamp: October 14, 2016, 01:13:25 am
content: Yes, but then you have to substitute s=j omega to switch to the time domain. That is just as much computing work as staying in the time domain in the first place.Ratch

author: danadak
author_rank: Super Contributor
timestamp: October 14, 2016, 10:31:47 am
content: Poster is looking for Z and Phase angle, no need to go to time domain.Or am i missing something ?Regards, Dana.

author: helius
author_rank: Super Contributor
timestamp: October 14, 2016, 10:57:05 am
content: The concept of impedance is only necessary when there is an AC component: just paralleling the complex impedances won't tell you what happens in the AC case, you need the time domain to model LC tank circuits and the like.

author: danadak
author_rank: Super Contributor
timestamp: October 14, 2016, 07:55:14 pm
content: s parameters will give you Q, BW, Z, V, I........

author: Ratch
author_rank: Regular Contributor
timestamp: October 14, 2016, 08:06:28 pm
content: Quote from: danadak on October 14, 2016, 07:55:14 pmFor sinusoidal steady state like this problem is, LaPlace offers no advantages over time-domain analysis. All the above calculations you listed are available just as easily with time-domain.Ratch