# Understand the calculations in an RC circuit - voltage division to polar form

I'm missing some calculation steps in an AC RC-example, and I can't understand how to get to the next step. If someone could show me, I'm sure to understand the rest.

Example: A standard RC-circuit.

----R------
|
Vin    C      Vout
|
-----------


Vin=1 sin(wt) -> Vin = 1<0deg (polar form)

1. Vout is given by voltage division

Vout=Zc/(R+Zc)=...=(1/(1+jwRC))(Vin)

Now, apply R,C and w

R=1600ohm

w=2PIf

f=100

The example then shows:

1. f=100Hz gives Vout=(0.99-j0.10)1<0=0.995<-5.7deg

Given 1), applying numbers would give 1/(1+j0,100530965)

But then I'm stuck.

How do I solve the equatation to get to 2) the above from the voltage division? I will use it to calculate further different frequncies, i.e for 1000Hz and so on, but need to understand what is happening in between 1) and 2). I understand basic complex numbers, but reading right now upon converting to polar form. I understand -j(1/wc)=1/jwc. but have a problem getting to (0,99-j0.10)Vin from 1/(1+j0,100530965)Vin

• I don't know what you're expecting from us. You already have the equation you need and the values you need, and you plugged them values into the equation. Just complete the equation. Right now your question is the equivalent of "I am given A = 1, B = 2. The equation is A + B = C. What do I do?" The only real problem you could encounter there is if you did understand the operations algebraic manipulations. But you never stated any specific problem like not understanding complex numbers, converting from a complex fractional to polar form, or anything else like that. Oct 3 '21 at 6:11
• If that is your problem, show us your work up until where you got stuck. Because as it stands, you are basically just asking us to do the entire problem for you. Oct 3 '21 at 6:12
• I understand basic complex numbers, but reading right now upon converting to polar form. I understand -j(1/wc)=1/jwc. but have a problem getting to (0,99-j0.10)Vin. from 1/(1+j0,100530965)Vin. Updated Q Oct 3 '21 at 6:58
• It's late here and its a pain to type out in mathjax when you're typing fractional multiplications all over the place. But multiply by the conjugate divided by itself to get the j out of the denominator. Oct 3 '21 at 7:01
• @G36 But that's a negative phase shift. Just to be clear.
– jonk
Oct 3 '21 at 7:14

DK has told you the right viewpoint. You write that:

I understand basic complex numbers

But it appears you don't. One of the first things you are taught is how to put a fraction into standard form of $$\a+b i\$$ when presented with something of the form $$\\frac{a+b i}{c+d i}\$$. It's probably late chapter 1 or maybe chapter 2. Not sure. But it is very early. You multiply by $$\1=\frac{c-d i}{c-d i}\$$, which is 1 made up by a fraction composed of the complex conjugate of the denominator in both the numerator and denominator. Remember?

So,

\begin{align*} \frac{V_{_\text{OUT}}}{V_{_\text{IN}}}&=\frac1{1+j \omega R C} \\\\ &=\frac1{1+j \omega R C}\cdot \frac{1-j \omega R C}{1-j \omega R C} \\\\ &=\frac{1-j \omega R C}{1+\left(\omega R C\right)^2} \\\\ &=\frac{1}{1+\left(\omega R C\right)^2}-j\:\frac{\omega R C}{1+\left(\omega R C\right)^2} \end{align*}

So, $$\a=\frac{1}{1+\left(\omega R C\right)^2}\$$ and $$\b=-\frac{\omega R C}{1+\left(\omega R C\right)^2}\$$. And now the complex number is back in standard Cartesian form.

If you plug in the values from your problem, I think you will see where your #(2) example result comes from.

• You are right, Altough I studied engineering maths 25 years ago, I have not used it at all (sadly) since then. And, So the knowledge gaps and the brain "rust" is all over, some basics forgotten. Really, I had for example forgot what a conjugate is.. Maybe I should have read "maths for engineers, summary" first before tackling problems like this. But then I'm afraid I never would get started with electronics. I learnt a lot from yours and DK answer- I would like to give both of your answers credit. Again thanks you both for putting me right and making me slightly wiser, Oct 3 '21 at 7:38

$$\\frac{1}{A-jB}=\frac{1}{A-jB}\times\frac{A+jB}{A+jB}=\frac{A+jB}{A^2+B^2}=\frac{A}{A^2+B^2}+\frac{jB}{A^2+B^2}=\frac{A}{A^2+B^2}+j\frac{B}{A^2+B^2}\$$

• Thanks a lot this helps understand this a lot, but how do I get the number 0.99 in the first place. Oct 3 '21 at 7:06
• I entered in the wrong numbers. But the steps are the same for getting the j out of the denominator so you have cartesian form. Ugh. Yeah. I'm def not up for typing all this out on top of how to get that 0.99. It's more typing than it's worth. It just comes from first dividing everything by 0.1005 because that's a pain to work with. Oct 3 '21 at 7:08
• I entered in the wrong numbers. But the steps are the same for getting the j out of the denominator so you have cartesian form. Ugh. Yeah. I'm def not up for typing all this out on top of how to get that 0.99. It's more typing than it's worth. It just comes from first dividing everything by $\frac{1}{0.100530965}$ to get a rounder number to work with. Oct 3 '21 at 7:13
• Will try, thanks a lot for putting me in the right directions. Will comment back when I have understood fully and verified, bbl. Oct 3 '21 at 7:15