Solving unkown impedance in circuit

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Discussion Overview

The discussion revolves around solving for an unknown impedance in a circuit involving resistors and a capacitor. Participants are analyzing the application of Kirchhoff's Current Law (KCL) and the relationships between voltage, current, and impedance in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant outlines a method to find the unknown impedance Z by calculating currents and voltages in the circuit, but expresses uncertainty about the correctness of their approach.
  • Another participant suggests checking the application of KCL, indicating a potential misunderstanding of current directions at a node.
  • There is a discussion about the angle of the calculated current Iz, with participants questioning its rectangular form and quadrant placement.
  • One participant acknowledges a mistake in quadrant identification related to the angle of Iz, admitting fatigue may have affected their calculations.
  • Despite the identified issues, the participant maintains that their complex part of the answer remains unchanged.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the calculations or the application of KCL. Multiple viewpoints regarding the interpretation of angles and signs in the calculations are presented, indicating ongoing uncertainty.

Contextual Notes

There are unresolved issues regarding the sign convention used for currents and the identification of quadrants for complex numbers, which may affect the overall solution. The discussion reflects a reliance on specific assumptions about circuit behavior and mathematical representations.

x86
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Homework Statement


Selection_030.png


Homework Equations


V=IZ

The Attempt at a Solution


So my plan is as follows:
Find the current through the 1 ohm resistor, multiplying this by the combined impedance of the capacitor & 1 ohm resistor. Now I know the voltage across the unkown impedance Z. Subtract this voltage from the 12 V source, now I know the voltage through the 9 ohm resistor. Use KCL to find the current through Z, and use V=IZ to find the unknown impedance.

Current through 1 ohm resistor: 4(45d)
Voltage through 1 ohm resistor & capacitor: 4(45d)(1-j1)=5.66(0d)
Voltage through 9 ohm resistor: 12(0d)-5.66(0d) = 6.34(0d)
Current through 9 ohm resistor: 6.34(0d)/9 = 0.7044(0d)

By KCL:
-Ic + Iz + Io = 0
Iz = Ic - Io = 0.7044(0d) - 4(45d) = 3.54(53d)

V/Iz = 5.66(0d)/3.54(53d) = 1.6(-53d) = 0.963-1.278j

But this is wrong, and I've been trying to figure this out for a good hour now, but keep getting the same answer. So I conclude that there's something wrong with my approach.
 
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I think you want to check your KCL. There is one current flowing into the node and two flowing out. Which is which?
 
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gneill said:
I think you want to check your KCL. There is one current flowing into the node and two flowing out. Which is which?

Yes. My sign convention is (-) for in, (+) for out. Iz is the current across Z, Io is the current through the resistor/capacitor, Ic is the current through the 9 ohm resistor.
 
The angle of your Iz looks suspicious. What was your rectangular version?
 
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gneill said:
The angle of your Iz looks suspicious. What was your rectangular version?

0.7044 - 2sqrt(2) - 2sqrt(2)j
-2.124-2.823j
3.537(53d)
 
x86 said:
0.7044 - 2sqrt(2) - 2sqrt(2)j
-2.124-2.823j
3.537(53d)
Yeah, both terms are negative. In what quadrant should the resulting angle lie?
 
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gneill said:
Yeah, both terms are negative. In what quadrant should the resulting angle lie?

The quadrant where x<0 and y<0, so this means that the answer is 3.537(-53d)

Even so, this doesn't change the real answer, it is still 0.963

But the complex part is correct. Perhaps I'll redo this problem tomorrow when I'm more awake. I really need to focus on the little things more!

Thanks
 
x86 said:
The quadrant where x<0 and y<0, so this means that the answer is 3.537(-53d)
-53° is in the 4th quadrant where x > 0...
 
gneill said:
-53° is in the 4th quadrant where x > 0...

Yes you are right. Sorry about that, I was up very late last night and havent' slept well lately.
 

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