Tonight we will jump right into solving for the total impedance, total current, and the currents through and the voltage across each individual components. In doing this we will use complex numbers and as you will find is not all that complex to do. The circuit is the same as the circuit I talked about in the Series RLC resonance example, except I added one complication. The Capacitor has a leakage resistance in this example. I labelled it Rl (small L) for Resistance leakage. We now have a Series-Parallel RLC circuit and I know of no other way to handle it, except using complex numbers. Note: This “Leakage Resistance: is unrealistically low. The value was chosen to make this problem interesting not to be realistic.

A couple of announcements before we dive into the meat of this post. First, I completed much of the thermistor board this weekend. I keep track of where I am at by changing the line color to black as I complete it. As you can see there is very little color left on the board. Second, I took a very much needed day off on Sunday and did some outdoor work and enjoyed the sunshine, so this post will probably come out on Monday, and possibly Tuesday. The final announcement is I seem to be under some sort of spam attack. I know of no way those of you who are subscribed will be getting bogus e-mails, but should that happen, please alert me as soon as possible. If you have become subscribed to this blog and did not intend to be please drop me an e-mail with the word UNSUBSCRIBE in the subject and I will remove your name from the list. If that happens, you also need to change your e-mail password, because your e-mail has be broken into by the bad actors. (Not through this site, but they are the ones that subscribed you.)

OK, back to the job at hand. The first thing I want to point out is sinusoidal analysis can only operate on one frequency at a time. In this problem I set the frequency a little below the resonance frequency of 100 Hz. It will provide us with a more interesting problem.

The first step will be to convert all the of components into an impedance represented by a complex value. The impedance of a resistor only contains a real part and Zr= (R+0j). Therefore: ZR = (1000 + 0j) and ZRl = (10000 +0j). It is not necessary to put the ( ) around each value, but I find that makes “housekeeping” much easier. The biggest problem with doing complex number math is just keeping all the values straight and anything you can to to help this will make your chances of success that much greater.

Now we will deal with the reactive components. These only have an imaginary component. ZL = XL * j where j is the symbol for √-1. XL = 2 * Π * f * L = 2 * 3.14 * 90 * 1.59. ZL = ( 0 + 898.67j).

ZC = XC*-j = -j *[ 1/( 2 * Π * f * C)] = -j *[ 1/(2 * 3.14 * 90 * 0.00000159)] = (0 – 1112.76j).

Our goal now is to combine all of these impedance values to end up with a total impedance of the circuit. Lets start with the easy one first. We will combine ZR with ZL. The rules for combining impedance are exactly the same as combining resistors. The total value for series resistors is the sum of the resistances. The total of ZR and ZL is ZR + ZL and this is (1000 + 0j) + (0 + 898.67j). To add two complex numbers simply add the real parts and add the imaginary parts. We will call this new impedance ZRL and it equals (1000 + 898.67j).

Now comes the more complicated part calculating the parallel part. Two resistors in parallel = (R1 * R2) / (R1 + R2). Again the same rule will apply for our two impedance values in parallel. (From now on we will take the normal short cut of writing things in parallel by using || to indicate parallel.) ZC||Zrl = (ZC * ZRl) / (ZC + ZRl). First, we need to learn how to multiply two complex numbers to calculate the numerator.

To multiply two complex numbers we have to multiply each part of the 1st number by each part of the 2nd number. In this case it will not be a major problem ZRl is all real, and ZC is imginary. ZC * ZRl = (0 – 1112.76j) * (10000 +0j) = 0 – 11127600j + 0 – 0*j^{2}. This becomes in this case (0 – 11127600j), but soon we will not be so lucky.

We have the numerator figured out so it is time to do the denominator. This is addition and using the method I described earlier the answer will be: (10000 – 1112.76j).

Now comes the difficult part, we have to divide a complex number by another complex number. There is no simple way to do that except by using a math trick. The first step is to multiply both the numerator and the denominator by the “complex conjugate” of the denominator. The complex conjugate is simply the same complex number but the imaginary part is negated. In this case the complex conjugate is (10000 + 1112.76j).

About now you are probably ready to throw up your hands and say “this ain’t worth it”. Please hang in there, Soon I will show you how to do it with a computer program and things will be much much better.

We will multiply the denominator by its complex conjugate first. (10000 -1112.76j) X (10000 + 1112.76j) gives 100,000,000 – 11127600j + 11127600j – 1238234.8176 * j^{2}. The two middle terms cancel each other and the j^{2} of the last term becomes -1 so the answer become 100,000,000 + 1,238,234.8176 or 101,238,234.8176.

Now multiplying the numerator by the complex conjugate gives: (0 – 11,127,600 j) * (10000 + 1112.76j) = 0 – 111,276,000,000 j + 0 j – 12,382,348,176* j^{2}. Again we make j squared = -1 and the total for the numerator becomes. (12,382,348,176 – 111,276,000,000 j).

The final answer for the parallel impedance is arrived at by dividing the numerator by the denominator which is now just a real number and becomes: Zpar = (122.309 – 1099.15j).

Because I have shown in complete detail the math operations the rest of the steps will just describe the operations and the answer for each.

The total impedance Ztot = ZRL + Zpar which equals: (1122.309 – 200.48j)

Vin needs to be written in complex form and since it is the reference phase it has phase angle of zero, so it only has a real term Vin = (1+0j)

The total current in the circuit is Vin/Ztot and after doing the math this becomes:

Itot = (0.000863 + 0.0001542j).

At this point we are probably interested in the absolute current and the phase angle.

The Absolute current = square root of (real part squared +imaginary part squared) = 0.000877 A

The phase angle is the arc-tangent of the (imaginary part / real part) (opposite/ adj).

This calculates to be +10.128 degrees. (leading because Xc > Xl.)

My guess is that I have smoke coming out of both ears by now… It is a lot to absorb and the math does get messy. However, it is worth doing just to understand the picture of what is going on. Soon I will be showing how it is done using a computer program. Things get much easier very quickly.

If you really want to make this your own, use the print/pdf button at the bottom of this page and print this post out and get out a calculator and pencil and paper and attempt to do this problem yourself. It is really the only way to learn it.

We could continue on and calculate the voltage across ZR and ZL using Ohms law and the total current which must flow through each of those two components. Next we would add those two voltages and then subtract the sum from Vin to get the Voltage across ZC and ZRl. (We just used Kirchhoff’s Voltage Law.) We could then use that Voltage and Ohms Law to calculate the current through each component. The concepts just became very simple but the math is no fun. (It will be once with get the computer to help us.)

if you have any questions please e-mail me and I will try to answer them. This has been a difficult subject and although I tried very hard to make it clear I may have not met my goal.

Gary

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