# Phasor Current

#### cheeseburger

Joined Jan 10, 2007
4
Hey im new to these forums and would like to ask a question.. could anyone explain to me how to evaluate phasor current in an AC circuit? I have a circuit in which the current needs to be calculated but im not sure how to do this as the resistors are complex numbers ( 2 resistors labelled 20+j20 in series) with a voltage source of 100V.
Any help would be muchos appriciated
Thanks

#### Dave

Joined Nov 17, 2003
6,969
Your question is about being able to analyse Complex Numbers. The best place to start is Volume 2 - Chapter 2 - Complex Numbers. This section will explain more thoroughly and concisely than I could do. I recommend you read through and get back to us if there are specific things you do not understand.

The core aspect you need to take on board is that complex numbers can be plotted on an Argand Diagram. Don't worry too much about the name, all you need to note is that an Argand Diagram has a real axis (the x-axis on normal graphs) and an imaginary axis (the y-axis on normal graphs). For an arbitary complex number, say 20 + j20, you would start at the origin (0,0) and go along 20 on the real (x) axis and up 20 on the imaginary (y) axis. You would end up at point (20,20). The line drawn from (0,0) to (20,20) is the phasor. This is the important component from the perspective of AC circuit analysis.

To characterise the phasor we have two things:

1) The modulus - the length (or magnitude) of the phasor line, calculated:

r = √(Re^2 + Im^2)

2) The argument, θ - the angle the phasor makes with the real (x) axis, calculated:

θ = arctan(Im/Re)

Where Im is the magnitude of the Imaginary component, which is 20 in the above example. And Re is the magnitude of the Real component, which is also 20 in the above example.

The modulus and argument constitute the polar form of the complex number.

Ok, I hope you've followed that! As for your query about the circuit arrangement you have, the first thing I want to make you aware of is that in AC circuits, theoretical resistors can only have a real component, whereas theoretical inductors and capacitors can only have imaginary components (the reality is often different, but for the purposes of your exercise this is the assumption you can take). Therefore, the "resistor" of 20 + j20 is in fact a resistor and inductor in series (resistor of 20Ω and inductor with reactance of 20Ω - don't concern yourself with the details of reactance at this point). This is known as an impedance (combined real and imaginary components indicating the presence of resistors and inductors or capacitors) denoted Z. Therefore, the two "resistors" in series of 20 + j20 are 2 resistors and 2 inductors all in series. The best thing to do is lump the resistors together and lump the inductors together.

Resistors in series:

Rt = R1 + R2 = 20 + 20 = 40

Inductors in series:

Lt = L1 + L2 = +j20 + j20 = +j40

Following me so far? So now we have lumped the resistance and inductance, we can express them combined as the impedance: which is 40 + j40. So your circuit now comprises the 100V source with a single complex impedance.

Now I have shown you how to get your circuit into a suitable form, following this tutorial here on AAC should allow you to derive the current quite simply. Have a go and feel free to post up your attempts to allow somone here to check for you.

Good luck.

Dave

#### cheeseburger

Joined Jan 10, 2007
4
Is that simply the answer then? 40+j40 ohms? Its a lot easier than i anticipated. Is there no way of calculating current in the circuit? Maybe i didnt understand your feedback correctly.
Thanks

#### cheeseburger

Joined Jan 10, 2007
4
In the first question on that tutorial, how actually is the 10mH inductor converted into 3.7699ohms? Dont really understand...

#### Dave

Joined Nov 17, 2003
6,969
Is that simply the answer then? 40+j40 ohms? Its a lot easier than i anticipated. Is there no way of calculating current in the circuit? Maybe i didnt understand your feedback correctly.
Thanks
Sadly, that is wrong!

In the first question on that tutorial, how actually is the 10mH inductor converted into 3.7699ohms? Dont really understand...
That comes from the equation for calculating the inductive reactance:

XL = 2πfL

In the example:

f = 60Hz
L = 10mH = 10e-3 H

Therefore:

XL = 2*π*60*10e-3 = 3.7699111848 Ω

For the purposes of your exercise, you don't need to do this conversion since your impedance is already in complex form.

Dave