# Complex Impedance

1. Mar 30, 2012

### BuddhaBelly34

1. The problem statement, all variables and given/known data
Time-domain expression for i(t), based on measures above:
Time-domain expression for i(t), based on measured R and L:

Vp-p on 16Ω resistor: 7.25mV "measured above"
Ip-p for resistor: 0.676mA "measure above"
Ip for resistor: .338mA "measured above"

R: 14.8Ω L: 10.1mH "measured R and L"

Source voltage is 5Vp-p 0 off-set at 1kHz
2. Relevant equations
? The issue is that I need to make the equations I suppose.

3. The attempt at a solution
This is for the measured R and L values:

Ip = Vp/(R + jωL) = 2.5/(14.8 + j2000∏*10.1) = 2.5/(14.8 + j63.5) = 37/622 - j(159/622) = 59E-3 - j256E-3

i(t) = R(cosωt - β)

R = √(x2 + y2) = √((59E-3)2 + ((256E-3)2) = 263E-3

β = tan-1(-256/59) = -77°

i(t) = 0.263cos(2000∏t + 77°)

Some specific issues:
1. How in the world can i(t) = Rcos(ωt - β)? Wouldn't that make voltage lag current which shouldn't happen when an inductor is in the circuit?

2. When I do the problem with
Vp-p on 16Ω resistor: 7.25mV "measured above"
Ip-p for resistor: 0.676mA "measure above"
Ip for resistor: .338mA "measured above"
should I be using Vp = 7.25E-3/2 or Vp = Vsource p-p/2? The instructions in the lab manual and the slides from the professor that wrote the manual where not clear.

Last edited: Mar 30, 2012
2. Mar 30, 2012

### Staff: Mentor

Is there a particular circuit configuration that we should know about?

3. Mar 30, 2012

### BuddhaBelly34

Lol I'm not sure, but here it is:

PS, this is an intro-lab if that wasn't obvious enough. lol

Last edited: Mar 30, 2012
4. Mar 30, 2012

### Staff: Mentor

You don't want to mix p-p and p quantities. Choose one "type" for your computations and stick to it.

Something misfired in your complex number calcs above; The result looks off.

Also, if I go by the 5Vpp @1KHz source and the given component values, the magnitude of the current that I calculate would be about 77 mA p-p, or 38 mA peak. So something is fishy about your "measured above" values or the component values.
What's a "current voltage"?
Well, it depends upon what it is you're trying to calculate. Make a clear statement of what it is you're trying to determine. If it's the magnitude of the current flowing through the resistor, then Ip = Vp/R, or Ipp = Vpp/R, where both Vp and Vpp are measured across R.

5. Mar 30, 2012

### BuddhaBelly34

Edited my post, I didn't mean "current voltage" I was typing "voltage lag current" but added a current in front of voltage.

As for picking "p-p" or "p" it was just a type-o going to fix that too.

"Also, if I go by the 5Vpp @1KHz source and the given component values, the magnitude of the current that I calculate would be about 77 mA p-p, or 38 mA peak. So something is fishy about your "measured above" values or the component values."
This is something that I, unfortunately, cannot fix. This is my professor's first time teaching this class, so there is more self teaching than anything and I probably made some mistakes with the oscilloscope (which is basically push buttons and hope you get it right :( ).

"Something misfired in your complex number calcs above; The result looks off. " Like rounding errors? Or something in the algebra? I was rounding as I was going.

6. Mar 30, 2012

### BuddhaBelly34

"Well, it depends upon what it is you're trying to calculate. Make a clear statement of what it is you're trying to determine. If it's the magnitude of the current flowing through the resistor, then Ip = Vp/R, or Ipp = Vpp/R, where both Vp and Vpp are measured across R."

I'm assuming it's Ip because the lab manual says "... i(t) = Ipcos(wt-b)".

7. Mar 30, 2012

### Staff: Mentor

$$\frac{2.5\;V}{14.8 + 63.46j \; \Omega} \cdot \frac{14.8 - 63.46j}{14.8 - 63.46j} → \frac{37 - 158.65j}{4246}\;A = 8.71 + 37.36j \; mA$$

Last edited: Mar 30, 2012
8. Mar 30, 2012

### BuddhaBelly34

edit since there is no delete.

Last edited: Mar 30, 2012
9. Mar 30, 2012

### BuddhaBelly34

okay I found my mistake. this guy in my class keeps changing my ti-83's notation to eng from normal and i miss the E's half the time. I get what you get now (now being paying attention to the Es lol)

So if the measured values look wrong should I go in to the lab and remeasure?

Okie doke, phase angle stays the same but now I have R = 0.0387. :P

Last edited: Mar 30, 2012
10. Mar 30, 2012

### Staff: Mentor

That might not be a bad idea. You might want to determine at least the order of magnitudes to expect for the values. Also confirm the part values and supply frequency. If you have an Ohmmeter handy, check to see if the inductor has a non-negligible resistance.

When using an oscilloscope, always check to see if any scope probe you're using has a scaling factor associated with it (quite often they divide by ten).

For the current that you measured, what type of instrument did you use? Was it an ammeter? Does it provide values in peak or RMS? (typically they present RMS values assuming a sinusoidal signal).

Supposing that R = 14.8 Ω, L = 10.1 mH, and f = 1 KHz, then what do you think you should see for the voltage across the resistor?

Last edited: Mar 30, 2012
11. Mar 30, 2012

### BuddhaBelly34

I thought that I was supposed to calculate the current? http://utdallas.edu/~dodge/EE1202/lab5.pdf Top of page 7.

Not sure what I should be seeing across the resistor. I don't understand AC circuits that well. Since it is in series the voltage across the resistor and the inductor must equal the source voltage right? So VL + VR = VS. I think VL = IXL where XL = wL.

Last edited: Mar 30, 2012
12. Mar 30, 2012

### Staff: Mentor

Ah. From the caption on the figure there: "The 'current' trace is really resistor voltage, but it is in phase with the current."

So the actual (p-p) current would be given by that curve's p-p voltage divided by the resistance.

13. Mar 30, 2012

### BuddhaBelly34

Yeah that is why I made the edit. Sorry, too much going on in my head right now. I had that problem first last night then the professor corrected me. Assuming my lab partner took the right reading for voltage on the resistor I have V = 7.25mV.

Ip = Vp-p/2R = 244.9E-6 A

14. Mar 30, 2012

### Staff: Mentor

You can treat the impedance of the inductor just as you would a resistance as far as the usual circuit equations are concerned. The only difference is that you'll be dealing with complex values in the calculations.

Your two components (resistor, inductor) are going to behave like a voltage divider. So if E is the input voltage and R and ZL the impedances of the parts, what voltage should you expect across the R?

15. Mar 30, 2012

### BuddhaBelly34

I'm not sure. Sorry for being ignorant, but there was no "intro to EE" class before this lab course. So I'm learning as I go.

16. Mar 30, 2012

### Staff: Mentor

If you can calculate the (complex) current, then you can determine the (complex) voltages for the individual components via Ohm's law. Their magnitudes should match those of the traces you see on the scope.