Help with Norton eq. circuit

[Patrik]

Hi

Well thing is I'm supposed to do a norton eq. of a circuit. The circuit is looking like this

ImageShack - Image Hosting :: circuitm.jpg

VAdc is a Voltage depending current source,which depends on the joint V, in the picture I got they are shown as DC sources, but if we have DC sources won't that mean the capacitor is a deadend hence everything beyond the C1 capactiro can be discarded?

I do understand how to do this with only resistors, but what about when capacitors come in play?

Should I tackle this with the generalized Ohms theorem?

I've been stuck on this for quite some time so if anybody can help out or give pointers it's much appriciated

//Patrik

 

[Patrik]

IS there no one in here who has a clue?

 

[ccurtis]

Why are impedances given for C1 and C2 if, indeed, the sources are DC current sources?

 

[Patrik]

well that's what I want to find out, there are basiclly 2 possiblitys as I see it

1. they act as DC and are vut off, hence, the circuit is only the part furthest to the left.

2. Norton can be used on single frequencys which would be described as DC and then we can simply look at them as resistors and use the generic Ohms formula(V*G=I)

 

[ccurtis]

The question becomes: which possibility is it? It could be either, and the Norton equivalent is different for the two possibilities.

I agree that if the sources are purely DC, the steady state condition separates the circuit into two halves. The left half would be used to calculate V so that the magnitude of the current source on the right half is known (V Adc). The Norton equivalent circuit would then be current source VAdc in parallel with a 50 ohm resistor (20mS=0.02mho=50 ohms).

 

[user_88]

This looks like it is a superposition problem, with C1 as the load.
That is, you are required to determine the resultant complex current vector going through C1.

Basically, you null, or omit, one of the current sources, and find the C1 load current current value, then you null the other current source, and find the other C1 current value. Then you add the two current vectors.

This just doesn't look like a straight Norton or Thevenin equivalent problem.
It appears, as you have suggested, that the current sources are just complex vectors, with a zero value imaginary part.

Maybe after you find the current going through C1, you can convert to a Thevenin or Norton equivalent problem.

Also note that the problem diagram provides conductances and admittances for the parallel elements. C1 is also given as an admittance, even though it is in series.

I would ask for a review of the problem definition... wherever you got it from.

 

[ccurtis]

well that's what I want to find out, there are basiclly 2 possiblitys as I see it

There is a third possibility. It could be an instantaneous snapshot after the time zero when the sources are first applied but before the steady state is reached.

 

[Patrik]

the problem definition is "Find a Norton equalient for the circuit below".

this is a problem I have in school on an old exam of a course I failed and are now trying to pass, my teacher hasn't responded to mail so I'm now asking on forums for help.

resultant complex current vector would be the same current as in the joint V?

admitance is just another way of writing 1/R right? It's probably just to confuse us students a bit.

why is the unit given depending on if it's in series or parallell?

option 3 looks like a possibility thanks for that

I did calculations with kirschhoffs laws to get V to -10nA with phase of -90degress.

 

[user_88]

... Re-evaluation .....
If you are to find the Norton equivalent, then maybe the load terminals are the two lines drawn at the right of the circuit .... with the load not actually given.

Therefore, for Norton equivalence, you find Rn by looking into the load terminal, and 'open circuit' all the current sources. There are two in this case. Then find the resulting admittance/impedance of this circuit. This will require a little complex number manipulation, but should be simple with a calculator.

Next, short the output terminals with a straight line .... to find the short circuit current, looking into the circuit from the load viewpoint. Doing this will actually short out the VAdc current source, and you can find the resulting total admittance ... just the sum of the leftmost resistor admittance, inductor admittance, and then add the center capacitive admittance. Take the reciprocal of this summed admittance quantity, obtaining a total impedance, and multiply by the leftmost current source. This resulting quantity is the complex voltage at V. The one remaining step is to take this complex voltage at V quantity, and divide it by the impedance of the center capacitor... or multiply by the center capacitor admittance. By doing this, you will find the short circuit, or Norton current.... That is, the current that will flow through a short circuit across the load terminals, and also the center capacitor element.

It seems that it is very convenient to use the Norton equivalent, in that the variable current source, VAdc, is shorted out during the calculation of the Norton short circuit current at the load terminals.

You have calculated the short circuit current ... at the load terminals .... the same as that in the j2ms center capacitor.
You have found the Norton Resistance, Rn.
Therefore you have found the Norton equivalent of the original circuit.

Does this seem to make sense? Surely, the underlying principles were provided in your reference materials....

I get:
Rn = (49.1-j4.97) Ω .... or in terms of admittance: (20.2+j1.94) ms
In= (0.587+j3.91) mA

.... will try to check this result later.

 

[Patrik]

Hi and thank you very much for your answers

first i'd like to apologize, the load is not given and are to be connected to the terminals just as you said.

Next, short the output terminals with a straight line .... to find the short circuit current, looking into the circuit from the load viewpoint. Doing this will actually short out the VAdc current source, and you can find the resulting total admittance

so the resolving circuit for Isc is

the source
the 20ms resistor //5jms capacitor //2jms capacitor

just the sum of the leftmost resistor admittance, inductor admittance, and then add the center capacitive admittance. Take the reciprocal of this summed admittance quantity, obtaining a total impedance, and multiply by the leftmost current source. This resulting quantity is the complex voltage at V.

if I understod you right, and then calculating by means of Ohms law, gives us

Y = 20m+5m*i+2m*i=(20+7i)ms

Z=1/Y=1/G+iB=G-iB/(G^2+B^2)=R+iX=56-20i

U=Z*I=4m*(57-20i)=0,228-0,08*i V

The one remaining step is to take this complex voltage at V quantity, and divide it by the impedance of the center capacitor... or multiply by the center capacitor admittance. By doing this, you will find the short circuit, or Norton current.... That is, the current that will flow through a short circuit across the load terminals, and also the center capacitor element.

and the current through 2jms is

U/Z=I=>(0,228-0,08*i)*2m*i=(0,16+0,456i)mA

so the question then as we get different results, were did I go wrong

best regards/
Patrik

 

[user_88]

Sorry ....I may have made a mistake in my calculation.... I will have to check the numbers again.....

 

[ccurtis]

I am used to working with ohms, so R1=R2=50 ohms, C2=-i200 ohms, C1=-i500 ohms. I used the impedance calculator at:Impedance

Calculate the Norton Current (In): Shorting the output terminals to obtain In (the current through the shorted output terminals), we have current source I1 in parallel with R1//C2//C1, for I1 in parallel with impedance 44.55-i15.59. Then V=.004 × (44.55-i15.59)=0.1782-i0.06236. Norton current, In, is then V÷-i500= 0.00012472+i0.0003564 A or .125+i.356 mA

Calculate Norton Resistance (Rn): Since I2 is a dependent current source, Rn is determined by placing a 1A current source at the output terminals, finding the voltage across that 1A current source, then dividing that voltage by the 1A current source. I think .

 

[user_88]

I re-calculated, and got about what ccurtis got for the node voltage, V:
V=0.1769-j0.061

However, if the Norton Current is the short circuit current taken through the load terminals, at the right side of the circuit, then the Norton Current is the voltage, V, divided by the C1 impedance, or -j500 ohms:
In=0.122 + 0.354 ma

Rn is found by open circuiting the two current sources, and looking into the load terminals....
This gives Rn = [(R1||ZC2) + ZC1] || R2

or
Rn = 49.11 - j4.72 Ω

I'm not absolutely sure about the I2 current source. ... If it is a special case or not. .... Haven't found any similar examples yet.

It would seem that according to the Norton rules, by short circuiting the load terminals, the effects of I2, the voltage dependent current source, would be negated, or have no effect on the short circuit current calculation.

 

[ccurtis]

I saw an error and got the same In as you, User88. I think the dependence of I2 on V requires substituting a 1A current source at the output terminals, then solving V to get the voltage across the 1A current source (the output terminals).

 

[Patrik]

okey my error was in the translation to impedans from admitance, as if I calculated with adminttance I got the same as you for V.

as I understood it from my reference material,

we should "zero all independent sources then connect a 1A current source between the load terminals and determine Vterminal then Rn=Vterminal/In"

so if we do this then it would mean that the source is "shortened" out hence the R1 and C2 components will be neglected.

But now V is directly conencte dto the grounding point, does this mean(conviently) that the dependent source is 0, hence, we have C1//R2//1ADC

this gives me a Voltage of (49.5+4.95i)V and hence gives the resulting Vterminal/Isc=(3.4-1.6i)ms

are these calculations correct?

We are getting there and want to thank you all for trying to help me out

 

[ccurtis]

as I understood it from my reference material, we should "zero all independent sources then connect a 1A current source between the load terminals and determine Vterminal then Rn=Vterminal/In"

From the english wiki article on Norton Theorem:

However, when there are dependent sources, the more general method must be used...

Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. The quotient of this voltage divided by the 1 A current is the Norton impedance RNo. This method must be used if the circuit contains dependent sources, but it can be used in all cases even when there are no dependent sources.​

I see nothing in the method above that allows shorting independent sources while the 1A source is in place. The magnitude of I2 depends on I1, so it does not seem reasonable to short I1 to find Nr. It makes sense that both sources must be functioning before calculations are made. My take, anyway.

 

[user_88]

One relevant reference which I have located, Hayt and Kemmerly, Engineering Circuit Analysis, states that with a dependent source, the best procedure is to determine Rth, or equivalently, Rn, by using a ratio or Voc to Isc, referring to the output terminal of the circuit.

.... have not yet resolved this idea to get a meaningful result.

The unique location in the circuit of the dependent source, I2, is such that it is short-circuited when the load terminals are shorted.

 

[ccurtis]

Since the Thevenin and Norton resistances are interchangable, it seems to me that what is left to do is calculate the open circuit voltage at the output terminals (Voc), just as you see the circuit with no opens or shorts applied, then divide that by the Norton short circuit current we already calculated (In) to get the Norton resistance. Lots of math, but doable.

 

[ccurtis]

After much juggling, using the reasoning in my post above, I get Voc=200 mV (nice round and real number).

So, Nr = Voc/In=.2 Volts ÷ .000125+.000356 Amps=176-500i ohms

 

[Patrik]

okey I wil ltry and reach the same results, seems we solved it, if I have more questions I'll update the thread but for now I once again thank you for the assistance