Complex Numbers and AC electronics.

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 :smile:

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 :smile:

.

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 :smile:

?

Reply 1

laeti

When you write:

\theta = \arctan \frac{3184.7}{4700}

surely the numerator should be negative?

Either that, or use a CAST diagram to figure out which quadrant the complex number should be in.

Reply 2

Mush

OP

17

laeti

When you write:

\theta = \arctan \frac{3184.7}{4700}

surely the numerator should be negative?

Either that, or use a CAST diagram to figure out which quadrant the complex number should be in.

Ah true enough.

How does it look, apart from that ?

Reply 3

Wangers

16

Maybe try the Mathematics fora?

Reply 4

Mush

OP

17

I thought physics would have been the closest thing to electronics...

Reply 5

Wangers

16

Mush

I thought physics would have been the closest thing to electronics...

True, but alot of this would be covered as part of the appied side of a decent maths course, besides they'll be more able to check the maths behind the derivations :smile:

Reply 6

teachercol

9

All this used to be A level Physics back in the day - including the derivations .

Reply 7

EdwardCurrent

10

laeti

When you write:

\theta = \arctan \frac{3184.7}{4700}

surely the numerator should be negative?

Either that, or use a CAST diagram to figure out which quadrant the complex number should be in.

When you're converting a complex number into polar form you're typically only interested in the principle argument. In this case its easier and more common to consider the complex number as a point on the Argand diagram, and take the acute angle theta to be

Unparseable latex formula:

$\arctan \left|\frac{\Im(z)}{\Re(z)}\right|$

, finding the principle argument is then trivial. So the OP's method was fine.

OP: Your method is sound. But, you only posted your solution to the first part? Not sure why you bothered typing out the whole question?

Reply 8

Mush

OP

17

A_Fool's_Paradise

When you're converting a complex number into polar form you're typically only interested in the principle argument. In this case its easier and more common to consider the complex number as a point on the Argand diagram, and take the acute angle theta to be

Unparseable latex formula:

$\arctan \left|\frac{\Im(z)}{\Re(z)}\right|$

, finding the principle argument is then trivial. So the OP's method was fine.

OP: Your method is sound. But, you only posted your solution to the first part? Not sure why you bothered typing out the whole question?