Complex Numbers and AC Electronics.

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SUMMARY

The discussion focuses on calculating the impedance of a series circuit comprising a resistor (R = 4.7 kΩ) and a capacitor (C = 10 nF) connected to an AC supply of 20V at 5 kHz. The impedance is calculated in both R + jX form as Z_total = 4700 - j3184.7 Ω and in polar form as Z = 5677.35∠-34.12°. The current flowing through the circuit and the phase difference between the voltage and current are also explored, with a noted phase difference of 31° indicating a non-zero output resistance (R_s) of the voltage source. The calculations and methods used are confirmed to be conceptually correct, despite potential minor errors.

PREREQUISITES
  • Understanding of complex impedance in AC circuits
  • Familiarity with phasor representation and Argand diagrams
  • Knowledge of basic circuit components: resistors and capacitors
  • Proficiency in trigonometric functions and their application in electrical engineering
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  • Study the calculation of impedance in series R-C circuits
  • Learn about the effects of output resistance on circuit behavior
  • Explore the use of phasors in analyzing AC circuits
  • Investigate the relationship between phase difference and circuit parameters
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Electrical engineering students, circuit designers, and professionals working with AC electronics who need to understand complex impedance and its implications in circuit analysis.

MushManG
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Okay. I'm just looking for someone to check my answer to the following question!

I'm not sure this is the correct forum for such a question, if not, feel free to have it moved :)

QUESTION:

An a.c. supply of amplitude [itex]20V[/itex] and a frequency of [itex]5 kHz[/itex] is connected across a Resistor [itex]R = 4.7 k\Omega[/itex] and a capacitor [itex]C = 10nF = 10 \times 10^{-9}F[/itex] connected in series. If the output resistance of the voltage source [itex]R_s[/itex] is assumed to initially be [itex]0\Omega[/itex], find:

a) The Impedance of the circuit in both [itex]R + jX[/itex] and [itex]Z\angle\theta[/itex] form.

b) The magnitude of the current flowing in the circuit.

c) The phase difference between the applied voltage and the current.

d) The voltage across the resistor and the voltage across the capacitor.

e) Show that [itex]V_{s} + V_{c}[/itex] is equal to the voltage applied to the circuit.

f) It is suspected that [itex]R_{s}[/itex], the output resistance of the voltage course, is not actually [itex]0\Omega[/itex]. When the circuit is investigated experimentally, it is fond that the phase difference between the applied voltage and the current is actually [itex]31\degree[/itex]. From this observation, calculate R_{s}.
By the way this is an Engineering problem, so we're using [itex]j[/itex] for the complex numbers jargon :).

My answer to a)

[itex]Z_{total} = Z_{C} + Z_{R}[/itex]

[itex]Z_{C} = \frac{-j}{\omega C}[/itex]

[itex]C = 10\times 10^{-9}[/itex], [itex]\omega = 2\pi f = 2\pi(5000) = 31415.9[/itex]

[itex]Z_{C} = \frac{-j}{31415.9 \times 10 \times 10^{-9}}[/itex]

[itex]Z_{C} = -j(\frac{1}{13.14 \times 10^{-4}})\Omega[/itex]

[itex]Z_{C} = -j3184.7\Omega[/itex]

[itex]Z_{R} = 4700\Omega[/itex]

[itex]Z_{total} = 4700 -j3184.7[/itex] Which is my answer in R + jX format.

To get it in [itex]Z\angle\theta[/itex] format I construct an Argand diagram, and deduce from it that:

Using Pythagoras:

[itex]Z^{2} = 4700^{2} + 3184^{2}[/itex]

[itex]Z = \sqrt{4700^{2} + 3184^{2}}[/itex]

[itex]Z = 5677.35\Omega[/itex]

Using Trigonometry:

[itex]\theta = tan^{-1}\frac{3184.7}{4700}[/itex]

[itex]\theta = 34.12\degree[/itex]

But I see from the Argand Diagram that this is a negative degree.

[itex]Z = 5677.35\angle-34.12\degree[/itex]

Let's see who can spot all of Mush's mistakes on that one then :)?
 
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This is the correct forum for your problem.
A little hint: If you had a specific question with your homework, you would get more replies.
Without checking your math, for what you've shown, conceptually you are on the right track.
 

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