Sorry, correction:

Xl = jWL = j(2*∏*50*0.7) = 219.91j Ω

Previous post corrected!

 

Sorry, correction:

Xl = jWL = j(2*∏*50*0.7) = 219.91j Ω

Previous post corrected!

Ok, thats correct, but why have you ignored the 'j' imaginary impedance in your Current calculation! and used only the Real part.???

Edit:
It would help if you studied this link.
http://www.electronics-tutorials.ws/accircuits/series-circuit.html

 

I have not ignored the imaginary part.

I did 220|0º/50-1371.64j = 0.1603|87.91ºA

In this value I have the real part and the imaginary part.
If I convert this polar number into rectalgular i get 0.00584+0.1602j A.

 

OK, I need new spectacles...

I misread your equations, duh!

E

 

So, if have have ignored the imaginary part, should i get the same result?

Like:

R = 50Ω
Xc = 1/WC = 1591.55Ω
Xl = Wl = 219.91Ω

Ztotal = sqrt(50^2 + (219.91-1591.55)^2) = 1372.55Ω

I = V/Z = 220/1372.55 = 0.1603A

This is also correct, right?

One other question about this problem...
How do I calculate the Overvolt factor??? I think this is the name!!!

 

hi,
Thats also correct.

The over voltage factor occurs at resonance, the reactive voltage across the Ind and Cap can be much higher than the voltage source, so the voltage Rating of the components should be able to withstand that voltage.

E

 

hi,
Thats also correct.

The over voltage factor occurs at resonance, the reactive voltage across the Ind and Cap can be much higher than the voltage source, so the voltage Rating of the components should be able to withstand that voltage.

E

Ok, thanks..

Now another problem. Hope you still have patiente to help me out.


RLC circuit again.

R = 60Ω
L = 1H
C = 6.25μF
Vin = 240V at a variable frequency...

Questions:
1 - Calculate the ω so that the circuit is at Resonance point.
2 - For the ω calculated above, calculate current and Voltage Drop at each circuit element.
3 - To an ω 10% higher than the calculated before, find the current and the Voltage Drop at each circuit element.

The 2 first questions, I think I've managed to solve them.

1 -
ω = 400rad/s


2 -
Xc = -400jΩ
Xl = 400jΩ

Z = R + Xc + Xl = 60 + (-400j) + 400j = 60Ω
I = 240/60 = 4|0ºA

Vr = 60 * 4|0º = 240|0ºV
Vl = 400j*4|0º = 1600|90ºV
Vc = -400j*4|0º = 1600|-90ºV


3-
ω1 = 10%*ω = 440rad/s

Re-calculated all Xc, Xl, I, Vr, Vl and Vc but i'm skeptical.

Xc = -363.64jΩ
Xl = 440jΩ
Ztotal = 60-363.64j+440j = 60+76.36jΩ

I = 240|0º/60+76.36j = 2.47|-51.84ºA

Then:

Vr = 60*2.47|-51.84=148.2|-51.84º
Vl = 440j*2.47|-51.84=1086.8|38.16º
Vc = -363.64j*2.47|-51.84 = 898.19|218.16º

I'm skeptical because shouldn't the sum of these 3 voltage drops be equal to Vin=240V????

 

Ok, thanks..

3-
ω1 = 10%*ω = 440rad/s

Re-calculated all Xc, Xl, I, Vr, Vl and Vc but i'm skeptical.

Xc = -363.64jΩ
Xl = 440jΩ
Ztotal = 60-363.64j+440j = 60+76.36jΩ

I = 240|0º/60+76.36j = 2.47|-51.84ºA

Then:

Vr = 60*2.47|-51.84=148.2|-51.84º
Vl = 440j*2.47|-51.84=1086.8|38.16º
Vc = -363.64j*2.47|-51.84 = 898.19|218.16º

I'm skeptical because shouldn't the sum of these 3 voltage drops be equal to Vin=240V????

Hi,
If you add algebraically the 148.2V 1086V and the -898.19V you get 339Vpeak, so the RMS voltage is 339v * 0.707 = 239.7V

E

 

Ouch... Great news...

Thanks...

I'm now trying to solve problems of Resonant Frequency but for parallel setups!!

 

Sir Eric, there is something here I can't understand...

I have a circuit that has 2 branches.

The 1st branch has a resistor (RL) and an inductor (L).
The 2nd branch has a resistor(RC) and a capacitor (C).
Both branches are at a parallel setup.

L = 2mH
C = 80μF

Teacher asks us to calculate RL and RC so that the circuit is at Resonante Point for any frequency!

Regarding math, I can't figure out what this means... Can you give a hunch???


Edited;
My teacher says that if (RL)^2 = (RC)^2=L/C, then the circuit will be at Resonant Pint for any frequency... But I can't understand why...

 

Sir Eric, there is something here I can't understand...

I have a circuit that has 2 branches.

The 1st branch has a resistor (RL) and an inductor (L).
The 2nd branch has a resistor(RC) and a capacitor (C).
Both branches are at a parallel setup.

L = 2mH
C = 80μF

Teacher asks us to calculate RL and RC so that the circuit is at Resonante Point for any frequency!

Regarding math, I can't figure out what this means... Can you give a hunch???


Edited;
My teacher says that if (RL)^2 = (RC)^2=L/C, then the circuit will be at Resonant Pint for any frequency... But I can't understand why...

It's possible to define resonance for a parallel RLC circuit in 3 different ways:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html

You will get different frequencies for the 3 different definitions. To get the answer your teacher gave you, the 3rd definition must be used. The definition itself should give you a clue about what you need to calculate.

 

It's possible to define resonance for a parallel RLC circuit in 3 different ways:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html

You will get different frequencies for the 3 different definitions. To get the answer your teacher gave you, the 3rd definition must be used. The definition itself should give you a clue about what you need to calculate.

Thanks for replying.

It should, indeed give me a clue, but I cannot understand why (RL)^2 must be equal to (RC)^2. Is it a condition so that the circuit is a lossless circuit, meaning that it cannot have elements dissipating heat, and as inductance and capacitance cancel out themselves each other, this should be the only way to ensure that the circuit is lossless????

 

Well if the teacher says

\({R_L}^2={R_C}^2\)

Then provided neither term is negative (improbable) doesn't that simply reduce to

\({R_L}={R_C}\) .....?

Also, it is not a necessary condition that a passive parallel circuit at resonance should also be lossless. Otherwise no real circuit could ever be considered at resonance as there are always losses involved with real components carrying current.

 

Thanks for replying.

It should, indeed give me a clue, but I cannot understand why (RL)^2 must be equal to (RC)^2. Is it a condition so that the circuit is a lossless circuit, meaning that it cannot have elements dissipating heat, and as inductance and capacitance cancel out themselves each other, this should be the only way to ensure that the circuit is lossless????

morning Psy,

Consider the Phase relationship of the Voltage and Current related to a Power Factor of unity.

Post your calc's and answers and we can check them out.

E

 

morning Psy,

Consider the Phase relationship of the Voltage and Current related to a Power Factor of unity.

Post your calc's and answers and we can check them out.

E

Hi Sir Eric...

Thanks for replying.

Yesterday I had an exam... Not from this class but from another one.
I think I'm going to ask my teacher to help me on this one...

In the meantime I need help to you to teach me how to check the Resonant Frequency at LTSpice for parallel circuits... Can you help?

 

I the meantime I've tried to check it with LTSpice but I don't know If it's the right way to check for the RF at a parallel setup.

I've calculated the RF with the following formula:

which gave 722.15Hz.

I'm plotting the Ic and Il and I'm checking where do they cross each other.
But I really don't know if this means anything because I can't make it a match with LTSpice results!

 

Since the frequency of resonance is deemed (by the equation you used) to be when the source voltage and current are exactly in phase then this is the condition one would look to confirm. That is, at 722.15 Hz (correct), the source voltage and current are measured to be exactly in phase.

This condition is also interpreted as the frequency at which the parallel circuit is purely resistive in nature.

It's possibly worth noting that the maximum parallel impedance (an alternative measure of resonance) is obtained at 959.4 Hz.

 

It's possibly worth noting that the maximum parallel impedance (an alternative measure of resonance) is obtained at 959.4 Hz.

Did you get this frequency from spice?

I gave a formula here:

http://forum.allaboutcircuits.com/showpost.php?p=675743&postcount=39

The formula gives 959.177 Hz

 

Yes from a spice based simulation.

 

Since the frequency of resonance is deemed (by the equation you used) to be when the source voltage and current are exactly in phase then this is the condition one would look to confirm. That is, at 722.15 Hz (correct), the source voltage and current are measured to be exactly in phase.

This condition is also interpreted as the frequency at which the parallel circuit is purely resistive in nature.

It's possibly worth noting that the maximum parallel impedance (an alternative measure of resonance) is obtained at 959.4 Hz.

Well, the condition we were said by the teacher to look for was when the imaginary part of the circuit impedance was equal to zero.
So, that is what we do and what I have done!

Did you get this frequency from spice?

I gave a formula here:

http://forum.allaboutcircuits.com/showpost.php?p=675743&postcount=39

The formula gives 959.177 Hz

You have a lot of formulas there.

I have used the one I posted in the small image.

How, then, can I make LTSpice to match my result?
And, what formula you used to get the 959.177Hz????

The plot I'm getting from LTSpice is this: