# Homework Help: I need help to create resonance circuit

1. Feb 3, 2012

### STINGERX

1. The problem statement, all variables and given/known data
i need to find values of capacitors coils and resistors in order to create resonance mixed circuit on the frequency of 100 hrz (as shown in the following screenshot)

2. Relevant equations
ω0=1/√(L*C)
ω0=2∏f0

3. The attempt at a solution

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2. Feb 3, 2012

### tiny-tim

Welcome to PF!

Hi STINGERX! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

3. Feb 3, 2012

### STINGERX

Re: Welcome to PF!

as you can see from the screenshot, i built this circuit on the simulation program Orcad Pcspice and i try to find the values of the parts C,L,R in order to recive a resonance frequency in 100hrz,
my requirements are to build the circuit like this and to make the voltage 100v ac

i only know how to find those values on simple cascade/parralar circuits but not in circuits such as this one.

please give me the values and if possible show me how to find them

thank you

4. Feb 3, 2012

### Staff: Mentor

Re: Welcome to PF!

How did you determine what values of L,C and R to use for your simulation? Does this circuit not behave as you need it to?

5. Feb 3, 2012

### vk6kro

Resistors have a relatively small effect on resonance frequency, so, as a start, you could try to get each pair of LC circuits resonant.

The other thing to note is that you can have any value of capacitor and still choose an inductor to resonate with it at any frequency.

So, to limit the choice a bit, you could pick a reactance (say 200 ohms) and calculate a capacitor and an inductor that both have this reactance at 100 Hz.

To do this, you need the formulae for reactance of capacitors and inductors.
You can find these in a suitable text book or here:
http://en.wikipedia.org/wiki/Electrical_reactance.

Note that you have the reactance and you are trying to calculate the component values. Also note that the capacitance produced from the formula above is in Farads.

6. Feb 3, 2012

### STINGERX

ok i calculated this values by determin that the coil will be L=10mH
and calculated the Capacitor value to resonance frequency of 100 hrz
like this:
http://img593.imageshack.us/img593/8049/photo030212173055hdr.jpg [Broken]

then i have only two values
L=10mH
c=253.302uF

i placed them twice in the simulation program Orcad Pcspice
(is this the right thing to do?)
http://img94.imageshack.us/img94/286/42599051.jpg [Broken]

then i recived this
http://img580.imageshack.us/img580/8971/63874598.jpg [Broken]

is there any way to recive this type of wave?
(maybe if i place a voltage probe in the coil and in the capacitor that shows me when the voltage is maximum in them?)
http://img823.imageshack.us/img823/5848/11132918.jpg [Broken]

Last edited by a moderator: May 5, 2017
7. Feb 4, 2012

### Staff: Mentor

You haven't indicated exactly what you are supposed to be doing. You have constructed two parallel resonant circuits, and connected them in series. To investigate currents in L or C (with almost no effect on the circuit) you can insert a small current-sensing resistor in series with that element. You only need one L and one C to form the resonant circuit. You could insert a 10 Ohm resistor between an element and ground, and monitor the voltage across that resistor to see what the current in that element is doing. You don't need any other resistors, unless the lab notes say to add some.

8. Feb 4, 2012

### STINGERX

and one more question, is it possible to get two resonance frequencys in one circuit such as this one?

9. Feb 4, 2012

### Staff: Mentor

Yes. But the result might not be as clear as you think it will be.

10. Feb 4, 2012

### vk6kro

You can observe the current by placing a voltage probe on the top end of R4. In this case, a dip in current indicates a resonance.

If there is only one dip, then the two circuits are resonant at the same frequency.

You can deliberately move one of the resonances if you like. Just make the lower capacitor 400 uF. You will then get a double hump and you can see which is the sharpest. The lower frequency one is the tuned circuit with 400 uF in it.

What did you learn about the effect of resistors on tuned circuits?
Your reactances are about 6 ohms and this very low value will affect your resonance sharpness or Q.