# Calculating the angle on impendance - RL circuit

Thread Starter

#### V1ck5

Joined Sep 1, 2023
2
Hi everyone, I'm new and happy to be here with you.
The subject of this particular excercise is
a) to calculate the θ angle of the impendance given, which is Ż= 40^θ and
b) To calculate the frequency (f) in H.
It's an RL circuit, and the given values are L= 0.04 H and R = 20Ω.

The formulas that come to my mind are the Ż = √(R² + X²L) and L= XL/2πf, but I obviously can't use them and I'm stuck. I'd appreciate any help!

#### Papabravo

Joined Feb 24, 2006
20,626
Hi everyone, I'm new and happy to be here with you.
The subject of this particular excercise is
a) to calculate the θ angle of the impendance given, which is Ż= 40^θ and
b) To calculate the frequency (f) in H.
It's an RL circuit, and the given values are L= 0.04 H and R = 20Ω.

The formulas that come to my mind are the Ż = √(R² + X²L) and L= XL/2πf, but I obviously can't use them and I'm stuck. I'd appreciate any help!
For any impedance Z, represented by a complex number in cartesian form with real and imaginary parts, it can be converted to polar form with two relations, one of which you already know. The one you already know is:
$$M\;=\;\sqrt{R^2+X_{L}^2}\;\;\;\text{; Magnitude of the complex number}$$
$$\theta\;=\;tan^{-1}\cfrac{X_L}{R}\;\;\;;\text{Angle of the complex number}$$

Expressing a complex impedance as:
$$R\;+\;jX_L$$
is entirely equivalent to expressing the number in polar form as:
$$M\angle\theta$$

#### WBahn

Joined Mar 31, 2012
29,534
Hi everyone, I'm new and happy to be here with you.
The subject of this particular excercise is
a) to calculate the θ angle of the impendance given, which is Ż= 40^θ and
b) To calculate the frequency (f) in H.
It's an RL circuit, and the given values are L= 0.04 H and R = 20Ω.

The formulas that come to my mind are the Ż = √(R² + X²L) and L= XL/2πf, but I obviously can't use them and I'm stuck. I'd appreciate any help!
I don't understand your part (a). You say that you need to calculate the angle of the impedance given that the angle of the impedance is 40° (I'm assuming that's what you meant by Ż= 40^θ). Or did you mean that the magnitude of the impedance is 40 Ω?

Is it a series RL circuit or a parallel RL circuit?

#### MrAl

Joined Jun 17, 2014
10,931
Hi everyone, I'm new and happy to be here with you.
The subject of this particular excercise is
a) to calculate the θ angle of the impendance given, which is Ż= 40^θ and
b) To calculate the frequency (f) in H.
It's an RL circuit, and the given values are L= 0.04 H and R = 20Ω.

The formulas that come to my mind are the Ż = √(R² + X²L) and L= XL/2πf, but I obviously can't use them and I'm stuck. I'd appreciate any help!
Hello there and welcome to the forum,

I have a couple questions too.

First, as WBahn asked, when you typed 40^theta did you really mean Z=40 Ohms?
Second, what is the reason you put a dot over the "Z"?
Third, did you mean "f in Hz" rather than "f in H", because H is not used for frequency while Hz or Hertz is.

The phase angle is best calculated by using the two argument inverse tangent function:
Ph=atan2(imaginary_part, real_part)
To calculate that though first you need to calculate the frequency, and that can work if we assume that you meant that Z=40 Ohms rather than Z=40^theta.

Maybe most important of all, you never said if this is a series or parallel circuit. The calculations are different for each of them. I can assume you meant a series circuit, but you may not have actually meant that.

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Thread Starter

#### V1ck5

Joined Sep 1, 2023
2
Papabravo, wBahn and mrAI thank you for your feedback and I apologize for the delayed reply. I asked my professor about the 40^θ and he said it is a simplified version of 40•e^iθ (go figure...). It is an RL in series circuit.
mrAI, the dot over the Z is how our professor symbolizes it on board. I've seen in the Uni books too. So I used it the same way. Yes, my bad, I meant Hz for Herz and I keep messing it up with H which stands for Henry I've pretty much solved it minus few details, thanks to your help. Thank you for your time, I truly appreciate it! If anything doesn't make sense I'll come back.

#### MrAl

Joined Jun 17, 2014
10,931
Papabravo, wBahn and mrAI thank you for your feedback and I apologize for the delayed reply. I asked my professor about the 40^θ and he said it is a simplified version of 40•e^iθ (go figure...). It is an RL in series circuit.
mrAI, the dot over the Z is how our professor symbolizes it on board. I've seen in the Uni books too. So I used it the same way. Yes, my bad, I meant Hz for Herz and I keep messing it up with H which stands for Henry I've pretty much solved it minus few details, thanks to your help. Thank you for your time, I truly appreciate it! If anything doesn't make sense I'll come back.
Hi,

You're welcome. I see now he used the dot over the Z to represent the complex impedance, which would mean the magnitude is 40, so obtaining the solution is pretty much the same way as shown before.

Congrats on finding the answer. Maybe you want to post it here so others can check it for you, unless you already had a way to do that.

One more little note...
I had to correct post #4 because I wrote the solution to the phase angle with real and imaginary parts reversed. I have corrected it:
Ph=atan2(imaginary_part, real_part)

There is another function used sometimes that is named simply "angle" and it would be written:
Ph=angle(real_part, imaginary_part)
which requires the reversal of the two arguments, but then again I suppose this could be the other way too depending on who writes the function and how they want to use it.

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