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In summary, the conversation is about calculating the resonance frequency and value of L in a two-branch parallel circuit. The formula for calculating w0 is different from that of RLC series circuits. The value of L is found by setting the impedance of the inductive element (ZL) equal to the capacitive element (ZC). The value of L is 2 millihenry and the impedance at resonance is j10 ohms. The imaginary part of the admittance is frequency dependent and can be calculated using the given rules.

- #1

toxique

- 19

- 0

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.

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- #2

SGT

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

- #3

toxique

- 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!

- #4

young e.

- 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

- #5

young e.

- 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 MILIHENRYZL= j10 ohms at an angular frequency of 5000 **

...at resonance

Last edited:

- #6

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.

- #7

toxique

- 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.

- #8

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 + 0.080000iy1 + 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?

- #9

toxique

- 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.

- #10

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.

Last edited by a moderator:

## Related to Formula for calculating w0 (w at resonance)

## 1. What is the formula for calculating w0 at resonance?

The formula for calculating w0 at resonance is w0 = 1/sqrt(LC), where w0 represents the resonant frequency, L is the inductance, and C is the capacitance.

## 2. How do I know when a circuit is at resonance?

A circuit is considered to be at resonance when the inductive reactance (XL) is equal to the capacitive reactance (XC), and the impedance of the circuit is at its minimum value. This can also be determined by finding the point where the phase angle between the voltage and current is 0 degrees.

## 3. Can the formula for w0 be applied to any type of circuit?

Yes, the formula for w0 at resonance can be applied to any type of circuit, as long as it contains an inductor and a capacitor.

## 4. How does changing the values of L and C affect w0?

Changing the values of L and C will affect w0 at resonance. An increase in inductance or capacitance will result in a decrease in w0, and a decrease in inductance or capacitance will result in an increase in w0.

## 5. What is the significance of w0 in a circuit?

W0 is the resonant frequency of a circuit, which is the frequency at which the circuit will naturally oscillate. It is an important parameter to consider when designing circuits for specific applications, such as filters or oscillators.

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