Series/Parallel Cap question

DamienH

Hello,

I've just started an electronics course and I've already found this site invaluable in my understanding of electronics in general. I've been assigned a series of questions to complete as part of my year and I'm just wondering if someone could give me a little help.

Diagram

These are the diagrams I have been given. I'm asked to "Write the parametric equations for the above network and calculate the values of A, B, C and D parameters, at a frequency of 1kHz, "


First step I'm taking is to add the impedances together so that I can get some kind of value for current. Obviously the resistors are no problem but I'm having trouble with the cap. I'm just having trouble trying to see if it's in parallel or series with the other resistors i.e. Is it in parallel with the second resistor ( 1K ) I'm thinking series but I'd just like to be sure before I do the calculations.

I'm thinking that if I can get the correct impedance, I can then let vin = 100v and hopefully, with a bit of tinkering around with formulas, I can get either a or b?? Tbh the a,b,c and d are throwing me a bit.

I'll probably have one or two more q's , hence the big thread

Thanks for any help

Michael Collins

With the circuit as it is, the capacitor is neither in series or parallel with either of the two resistors. This is known as a 'STAR' or 'Y' or connection.

I think the best thing to do is to help you through getting one parameter and then hopefully you'll be able to do the rest.

So first things first. We have to work out the impedance of the capacitor (I'm assuming you know how to do this?). The answer is -j5000/pi.

So let's concentrate on the A parameter. We can see from the parametric equations that

Vin = A.Vout + B. Iout

so to find A, we let

Iout = 0

Then the second term goes to zero and we see that

A = Vin/Vout

But what does letting Iout = 0 mean in terms of the circuit? Well, it means that there is no current flowing through the 1 kΩ resistor - hence we can ignore it. So Vout is now the voltage across the capacitor.

We can now calculate Vout in terms of Vin (and R) using simple voltage division. (Again I'll assume you know how to do this).

So we get, after some manipulation

Vout = Vin. 5000 / (j.pi.R + 5000) [Equation 1]

But we know that A = Vin/Vout, so substituting Vout for its value in terms of Vin according to equation 1, we get

A = Vin / [Vin. 5000 / (j.pi.R + 5000)]
= 1 / [5000 / (j.pi.R + 5000)]
= (j.pi.R + 5000) / 5000
= (j.pi.R / 5000) +1

And that's it. All you have to do is apply a similar procedure for the other three parameters.

When you try this you'll find that you must set Vout = 0 (for the B and D parameters), just this means a short circuit is applied at the output.

Hopefully this helps a bit, let me know if anything isn't clear.

Post your answer and i'll let you know if it's right!

DamienH

Excellent!!! Thanks for all the help I really, really appreciate it!! I'm watching the england match now but I'll try it afterwards. Thanks a lot

DamienH

Hey,

Just trying this out now. I've been looking for a while and I can't see where you got this line.


Vout = Vin. 5000 / (j.pi.R + 5000) [Equation 1]

I get the feeling it's staring me in the face but I just can't see it. I can follow everything you do after and before that line.

The value I got for the impedance of the cap is -j1591.55 which looks to be correct.

The answer I got for A=2.13 , which judging from you calculations seems about right.

I am unsure what u mean exactly about a short circuit occurring when Vout = 0. I realise this means that no voltage flows through vr2. Am I right in thinking that If Vout = 0 then Iout= the current across the capacitor (Ic) ? I am slightly confused in that I thought Itotal=Ir1=Ir2=Ic and if Vout=0 then Itotal=0 which I no can't be correct.

Sorry for all the q's but I haven't done this kind of thing before ( not two port networks ) as you might of guessed

B=Vin/Iout and D = Iin/Iout, these are what I can get of B and D, but I know that once I can see where you got equation1 from I can at least begin to start working these out further, as you did for A.

Thanks a lot for any help

Newb

DamienH wrote:

Hey,

Just trying this out now. I've been looking for a while and I can't see where you got this line.


Vout = Vin. 5000 / (j.pi.R + 5000) [Equation 1]

It's Kirchoff's voltage law

Michael Collins

DamienH wrote:

...I can't see where you got this line.

Vout = Vin. 5000 / (j.pi.R + 5000) [Equation 1]

It's just from the voltage division formula - if you have two impedances Z1 and Z2 in series connected to a voltage V like this:

______
|     |
|     Z1
|     |
V     |
|     |
|     Z2
|     |
|_____|

Then the voltage across Z2 is simply:

V. Z2/(Z1+Z2)

I'm sure you know this anyway. That's all I did, with Z1 = R and Z2 = -j5000/pi, followed by some manipulation to get the denominator real.

The answer I got for A=2.13 , which judging from you calculations seems about right.

How did you get an actual value for A? The resistor on the left only has the symbolic value R right? In that case only the C and D parameters will come out as actual numbers.

I am unsure what u mean exactly about a short circuit occurring when Vout = 0

Well if Vout is zero, it means that, effectively you can join the two output wires together...since the voltage between them is zero, so the 1 kΩ resistor is now in parallel with the capacitor. If you don't get it immediately just think about it for a bit, it'll make sense after a while...

I realise this means that no voltage flows through vr2

Voltage does not "flow"! Voltage can be "across" an element or "between" two points, but it doesn't "flow through" anything. A CURRENT will flow through the 1 kΩ resistor and so there will be a voltage across it.

Iout= the current across the capacitor (Ic) ? I am slightly confused in that I thought Itotal=Ir1=Ir2=Ic and if Vout=0 then Itotal=0 which I no can't be correct.

No such thing as a current across something, only through. Also this formula

Itotal=Ir1=Ir2=Ic

is wrong. The current through each device is not necessarily the same. Maybe you're getting confused with the sum of currents flowing into a node?

To be honest with you two ports are quite diffiuclt at the start, even more so if you arn't good a solving basic circuits first. Just keep at it, try anything you can think of, that's the best way to learn.

Let us know how you get on with this