# Circuit Analysis

Discussion in 'General Electronics Chat' started by kaosad, May 8, 2008.

Apr 14, 2008
19
0
Hi,

I have a circuit which I want to analyse using phasor at its natural frequency.

The circuit consists of a parallel inductor-capacitor $(L || C_1)$ in series with a parallel resistor-capacitor $(R || C_2)$ which is connected to an AC voltage source. So we have parallel inductor-capacitor impedence, call it $Z_1$, and parallel resistor-capacitor impendence, call it $Z_2$.

Therefore their impedences are:
$Z_1 = \frac{1}{\frac{1}{j \omega L} - \frac{\omega C_1}{j}} = j\frac{\omega L}{1 - \omega^2 L C_1}$
and,
$Z_2 = \frac{1}{\frac{1}{R} - \frac{\omega C_2}{j}} = \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2}.$

Since these impendences are in series the total impedence of the circuit is:
$Z_1 + Z_2= j\frac{\omega L}{1 - \omega^2 L C_1} + \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = \frac{\frac{1}{R}}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} + j \left[ \frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} \right].$

The natural frequency of this circuit occurs when the reactance of the series is zero so,
$\frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = 0.$
Rearranging this we get,
$\omega_o = \sqrt{\frac{C_2 -\frac{L}{R^2}}{LC_2^2 + L C_1 C_2}}.$

Let the voltage source be denoted by $V_s$. At natural frequency, $\omega_o$, the voltage across the impedence $Z_1$ is,
$V_1 = \frac{Z_1}{Z_1 + Z_2} \times V_s = j \frac{\frac{\omega_o L}{1 - \omega_o^2 L C_1}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \times V_s$
and the voltage across the impedence $Z_2$ is,
$V_2 = \frac{Z_2}{Z_1 + Z_2} \times V_s = \left[ 1 - j \left( \frac{ \frac{\omega_o C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \right) \right] \times V_s = \left[ 1 - j \omega_o C_2 R \right] \times V_s$

Okay, now let $V_s = 12, \ L = 0.1mH, \ C_1 = 100nF, \ C_2 = 80nF, \ R = 100 \Omega$. Using the formulae above I got,
$\omega_o = 220479.275$Hz, $V_1 = 12 \times (j 1.04364 \times 10^{3})$, and $V_2 = 12 \times (1 - j 1.04364 \times 10^{3})$.

These mean $|V_1| = 12.52368 \times 10^3$ and $|V_2| = 12.52368 \times 10^3$. But my simulation using SPICE, I got both $|V_1|$ and $|V_2|$ roughly around 12 volts only! Can someone tell me where I am wrong?

Last edited: May 8, 2008
2. ### ametso Member

Apr 30, 2008
13
0
Wow... you frighten me...

I'm not an expert in mathematic but I just wonder your starting equations. How did you got them?

I suppose that the reactance equation for parallel L||C should be

Z = (jXL)(-jXC) / (jXL) + (-jXC)

And after few trick (to get the "j" away from dominator) the equation comes to form

Z = -j(XL)(XC) / (XL - XC)

And set the

XL = wL
XC = 1 / wC

the imbedance for (L||C) equation comes to form

Z = -j(wL)(1/wC) / (wL - 1/wC)

As I say. I'm not an expert in mathematics - so I can be wrong...

Finally a question for you. How the heck you did genarate those equations? I did not found any way how to generate mathematical fonts on this site.

3. ### DC_Kid Distinguished Member

Feb 25, 2008
643
9
quote his post and you shall see.

4. ### Caveman Senior Member

Apr 15, 2008
471
1
Your last V2 equation is incorrect. It's actually
V2 = 1 - jw*C2*R

Apr 14, 2008
19
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Whoops! You are right. I have now changed to the correct expression. I made some mistake in my calculation as well. Okay, I got quite reasonable values now, but still not the same as simulation. My new calculation shows $|V_1| \approx |V_2| \approx 21.1654$

Thanks ametso, your equations can be simplified to what I have written.

Apr 14, 2008
19
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Funny, I rerun the simulation and got $|V_1| \approx 21.4169$ and $|V_2| \approx 23.4816$.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,347
347
All your algebra appears to be correct down to and including the expression for wo.

But you have incorrectly calculated the numeric value for wo using the component values you gave. It seems like you may have taken L as 1 uH, rather than 1 mH.

When I evaluate wo, I get 7453.09 Hz. When I then plug that value into your expressions for V1 and V2, I get:

V1 = j 715.497

V2 = 12 - j 715.497

The expressions you gave for V1 and V2 give the correct voltages for w = 7453.09, but not for other values of w.

8. ### Dave Retired Moderator

Nov 17, 2003
6,960
150
In the "Reply to Thread" window (where you write your replies to posts) you will see the sigma (Ʃ) icon on the formatting toolbar. Select this icon and you will see the LaTeX drop-down where you can select any of the LaTeX symbols, equation formatting and a whole host of other mathematical typesetting options.

Dave

Apr 14, 2008
19
0
The inductance is 0.1mH and not 1mH. By the way how did you get that frequency? Mine is 35090.366Hz (Sorry, in my original post I wrote 220479.275Hz, but that should be in radian per second). I got:
V1 = j 21.16543 and V2 = 12 - j 21.16543, and therefore |V1| = 21.16543 and |V2| = 24.3305.

Last edited: May 8, 2008
10. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,347
347
The 1 mH in my post was a typo. I meant .1 mH and that's what I used to calculate wo, V1 and V2.

However, the 7453.09 number is radians/sec, not Hz. That was from me just copying what you had and not paying attention! :-(

In your response to Caveman, you mentioned editing the expression for V2, but you forgot to tell us that you originally had C1 = 100 uF and C2 = 80 uF which in the meantime you changed to 100 nF and 80 nF. When I put in those values, I get the same thing you are getting now for wo, and V1 and V2.

However, my comment about your expressions for V1 and V2 giving the correct value for wo but not for any other value of w is, I believe, still correct. You may not care about that since you get the right answers for the problem at hand.

Apr 14, 2008
19
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Oh yes, I did change the capacitance values because I then noted my typos. Sorry for not making a note. Anyway, thanks for re-verifying all my calculations--really appreaciate it!

12. ### ametso Member

Apr 30, 2008
13
0
Yes you are right... 12 years of professional electrical designing behind (and one un-slept night) of me and still a blind to some basic transformations... But as I said - I'm not the pro in mathematics ;o)