Formula for calculating w0 (w at resonance)

1. Sep 6, 2005

toxique

Hi dudes. sory if it is not the correct area for this question.

A question. What is the formula for calculaing w0 (w at resonance) in a two-branch parallel circuit consisting of:
A voltage source
branch 1: R1 + L (in series)
branch 2: R2 + C (in series)

I thought it was the same as RLC series, but it actually is pretty different.

BTW I got a parallel resonance exercise but cannot figure out how to solve it.
Same circuit as above, but:
r1 = 2 ohm
r2 = 5 ohm
C = -10j ohm

Find out the value of L (both in ohms and Henry)
Any help on this subject is really appreciated.

2. Sep 6, 2005

SGT

1. Calculate the impedance $$Z_1(\omega)$$ of branch 1.
2. Calculate the impedance $$Z_2(\omega)$$ of branch 2.
3. Calculate the impedance $$Z(\omega) = X (\omega)+ iY(\omega)$$ of the two parallel branches.
4. The resonant frequency $$\omega_0$$ is that which makes null the imaginary term $$Y(\omega)$$

3. Sep 6, 2005

toxique

By Y(w) you mean the reactive component or admitance?

somewhere i've read that for this kind of circuits:

w0 = [ 1 / sqrt(LC) ] * [ sqrt(r^2 - L/C) ] / [ sqrt(r^2 + L/C) ] or something similar. This is what i do not understand.

sorry for not posting in Latex!

4. Sep 7, 2005

young e.

at resonance the circuit will act as pure resistive circuits, meaning the power factor is zero or the value of capacitive and inductive impedance is the same. thus:

given:

then:

ZL=WL...............impdedance (Z) of inductive element
ZC= 1/WC..........impedance (Z) of capacitive element

since ZL=ZC (at resonance)

then:

WL= 1/WC or

L= 1/(W^2*C)

C= -j10= -j/(WC)

C= 20x10 exp (-6) or 20 microfarad

therefore:

L= 1/ {[5000^2]*[20 x 10 exp (-6)] }

L= 2 x 10 exp (-3) or 2 mili Henry

or, ZL= WL= [5000]*[2 x 10 exp (-3)]

ZL= 10 ohms,,,,thats it

5. Sep 7, 2005

young e.

at resonance the circuit will act as pure resistive circuits, meaning the power factor is zero or the value of capacitive and inductive impedance is the same. thus:

given:

then:

ZL=WL...............impdedance (Z) of inductive element
ZC= 1/WC..........impedance (Z) of capacitive element

condition: ZL=ZC (at resonance)

then:

WL= 1/WC or L= 1/ (W^2*C)...this is the key formula

C= -j10= -j/(WC)

C= 20x10 exp (-6) or 20 microfarad

therefore:

L= 1/ {[5000^2]*[20 x 10 exp (-6)] }

L= 2 x 10 exp (-3) or 2 mili Henry

or, ZL= jWL= [5000]*[2 x 10 exp (-3)]

ZL= j10 ohms,,,,thats it

L= 2 MILIHENRY

ZL= j10 ohms at an angular frequency of 5000

.....at resonance

Last edited: Sep 7, 2005
6. Sep 7, 2005

SGT

$$Y(\omega)$$ is the reactive component.
Using the set of rules I suggested you, we can conclude that $$\omega_0 = \frac{1}{\sqrt{LC}}$$ exactly like in a series or parallel circuit.

7. Sep 7, 2005

toxique

About the solution by "young e.", a question:

From the results you obtained, it means that Ztotal = 15.43 + 4.2857i.

should not the reactive component on Ztotal be equal to 0, thus leaving only a resistive part <> 0 ? Even working it out as admittances, Y1L and Y2C are different.

y1 = 0.019231 - 0.096154i
y2 = 0.040000 + 0.080000i

y1 + y2 = 0.059231 - 0.016154i

Whats wrong here? I do not get it. AFIK, parallel resonance has a rather ****ty behaviour compared to series resonance.

8. Sep 7, 2005

SGT

The imaginary part of the admittance is frequency dependent. What frequency have you used to calculate it?

9. Sep 8, 2005

toxique

Hi SGT.

I have used W = 5000 (which is 2*pi*796.178) as stated in the excercise. It supposed to be W0 = 5000

Best Regardz.

10. Sep 8, 2005

SGT

Forget my previous post. $$\omega_0$$ is not $$\frac{1}{\sqrt{LC}}$$. Use the rules I suggested you to obtain the resonant frequency.
If you want only to calculate the inductance, the admittance of branch 1 is:
$$Y_1(\omega)=\frac{1}{R_1 + j\omega L}$$
The imaginary part of this admittance must be equal to the negative of the admittance of branch 1.
$$imag(Y_1) = - 0.08$$
From this you get a second degree equation in L.

Last edited by a moderator: Sep 9, 2005