Formula for calculating w0 (w at resonance)

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19
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Main Question or Discussion Point

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
W = 5000 radian

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

Answers and Replies

SGT
  1. Calculate the impedance [tex]Z_1(\omega)[/tex] of branch 1.
  2. Calculate the impedance [tex]Z_2(\omega)[/tex] of branch 2.
  3. Calculate the impedance [tex]Z(\omega) = X (\omega)+ iY(\omega)[/tex] of the two parallel branches.
  4. The resonant frequency [tex]\omega_0[/tex] is that which makes null the imaginary term [tex]Y(\omega)[/tex]
 
19
0
Thanks SGT for your prompt reply.

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!
 
64
0
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:

W= 5000 radians

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)

given, W=5000 rad, C=-j10

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
 
64
0
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:

W= 5000 radians

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

given, W=5000 rad, C=-j10

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

answers:

L= 2 MILIHENRY

ZL= j10 ohms at an angular frequency of 5000

.....at resonance
:smile:
 
Last edited:
SGT
toxique said:
Thanks SGT for your prompt reply.

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!
[tex]Y(\omega)[/tex] is the reactive component.
Using the set of rules I suggested you, we can conclude that [tex]\omega_0 = \frac{1}{\sqrt{LC}}[/tex] exactly like in a series or parallel circuit.
 
19
0
Thanx a lot guys for your replys.

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.

Best regarss.
 
SGT
toxique said:
Thanx a lot guys for your replys.

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.

Best regarss.
The imaginary part of the admittance is frequency dependent. What frequency have you used to calculate it?
 
19
0
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.
 
SGT
Forget my previous post. [tex]\omega_0[/tex] is not [tex]\frac{1}{\sqrt{LC}}[/tex]. 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:
[tex]Y_1(\omega)=\frac{1}{R_1 + j\omega L}[/tex]
The imaginary part of this admittance must be equal to the negative of the admittance of branch 1.
[tex]imag(Y_1) = - 0.08[/tex]
From this you get a second degree equation in L.
 
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