Complex Numbers and AC Electronics.

[MushManG]

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

a) The Impedance of the circuit in both $R + jX$ and $Z\angle\theta$ 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 $V_{s} + V_{c}$ is equal to the voltage applied to the circuit.

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

My answer to a)

$Z_{total} = Z_{C} + Z_{R}$

$Z_{C} = \frac{-j}{\omega C}$

$C = 10\times 10^{-9}$, $\omega = 2\pi f = 2\pi(5000) = 31415.9$

$Z_{C} = \frac{-j}{31415.9 \times 10 \times 10^{-9}}$

$Z_{C} = -j(\frac{1}{13.14 \times 10^{-4}})\Omega$

$Z_{C} = -j3184.7\Omega$

$Z_{R} = 4700\Omega$

$Z_{total} = 4700 -j3184.7$ Which is my answer in R + jX format.

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

Using Pythagoras:

$Z^{2} = 4700^{2} + 3184^{2}$

$Z = \sqrt{4700^{2} + 3184^{2}}$

$Z = 5677.35\Omega$

Using Trigonometry:

$\theta = tan^{-1}\frac{3184.7}{4700}$

$\theta = 34.12\degree$

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

$Z = 5677.35\angle-34.12\degree$

Let's see who can spot all of Mush's mistakes on that one then :)?

 

[Ouabache]

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.

 

[piercas]

My response:

Firstly, I would like to clarify that this is a forum for scientific discussions and not for checking answers to specific questions. However, since you have already provided your answer, I will provide feedback on it.

To start with, your calculation of Z_C is incorrect. The correct formula for calculating impedance for a capacitor is Z_C = \frac{1}{j\omega C}. This will give you an impedance of 31.83-j318.47 \Omega. Additionally, your calculation for Z_R is correct, but you have not added the two impedances correctly. The total impedance should be Z_{total} = 4700-j318.47 \Omega.

For part b), the magnitude of current can be calculated using Ohm's law, I = \frac{V}{Z_{total}}. This gives a value of approximately 0.0035 A.

For part c), the phase difference between the applied voltage and the current can be calculated using the formula \theta = tan^{-1}\frac{318.47}{4700} which gives a value of approximately 3.88 degrees.

For part d), the voltage across the resistor and capacitor can be calculated using Ohm's law, V_R = IZ_R and V_C = IZ_C. This gives values of approximately 16.45 V and 0.11 V, respectively.

For part e), we can see that V_s + V_c = 20 V, which is equal to the voltage applied to the circuit.

Lastly, for part f), if the phase difference is actually 31 degrees, then we can calculate the output resistance using the formula \theta = tan^{-1}\frac{318.47}{R_s}. Solving for R_s gives a value of approximately 1004.35 \Omega. This indicates that the output resistance of the voltage source is not negligible and should be taken into account in future calculations.