# Question about Norton equivalent on AC Circuit

#### jboavida

Joined Jul 10, 2008
23
Hi All,

I have to solve a problem regardin a circuit of resistors, inductors and capacitors. The circuit is powered from a 110V 100Hz source.

I have to find the Norton equivalent. I already calculate RNorton (73.74 +j 0.03 Ohm). How do I calculate INorton on points AB?

Joaquim

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#### The Electrician

Joined Oct 9, 2007
2,848
Since RNorton is the same as RThevenin, you could calculate the voltage at A-B due to the 110V source, and then INorton will be Vab/RNorton.

Also, it appears that you have an error in your calculation of RNorton.

I get (73.733 +j 0.58675)

#### jboavida

Joined Jul 10, 2008
23
Hi,

The current that flows thru R1 is 21mA. That gives a voltage drop of 1.548 V. This gives the current flowing in to C1 L1 mesh that is 14ma. Current in L2 is Sqrt(21^2 - 14^2) right?

I need a final voltage drop on L2, but I get only the imaginary part j240.96

I'm lost now... How to proceed?

Thanks

#### The Electrician

Joined Oct 9, 2007
2,848
You have to keep your currents in complex format.

The current in R1 is (.34077-j21.3895) milliamps.

The voltage drop across R1 is therefore 73.74*(.34077-j21.3895) millivolts.

Then the voltage at the right end of R1 is (109.975+J1.577) volts, giving a current in the C1-L1 branch of (-.20465+j14.2694) milliamps.

If you subtract the current in the C1-L1 branch from the current in R1, you get a current in L2 of (.136121-j7.12012) milliamps.

From this you can get the voltage at A of (.011709-j.525)

I'll leave it to you to get INorton.

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#### jboavida

Joined Jul 10, 2008
23
"From this you can get the voltage at A of (.011709-j.525)"

I don't quit get this part. I know that your calculations are right, but in mine I get voltage drop on L2 108.108+i1.544, so VAB gives (109.975+J1.57)-(108.108+j1.54)=1.846+j0.03

Obviously I'm wrong. I think I have calculated the voltage drop across L2 wrong, but (.136121-j7.12012 milliamps) * j15444 (L2 reactance) gives me 108.108+j1.54 V.

What I'm missing here?

Thanks

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#### The Electrician

Joined Oct 9, 2007
2,848
Obviously I'm wrong. I think I have calculated the voltage drop across L2 wrong, but (.136121-j7.12012 milliamps) * j15444 (L2 reactance) gives me 108.108+j1.54 V.

What I'm missing here?

Thanks
If I do the last calculation backwards to see what you are multiplying j15444 by, I get (108.108+j1.54)/(0+j15444) = (.0000997151-j.007). This should have given the current in L2, which is (.000136121-j.00712012).

But when I multiply (.000136121-j.00712012) * j15444, I get (109.96+j2.10225). Apparently, your complex multiplication is going wrong somewhere.

Have you got a calculator that can do complex arithmetic? If you don't, then you have to be very careful to get all the parts of complex arithmetic right.

#### jboavida

Joined Jul 10, 2008
23
I think I got it. I forgot that L2 and R2 are a voltage divider. By the voltage divider formula I get the same result. So I get Inorton = 0.000102 -j0.007 A

Right?

One last question. I have calcuted previously the equivalent impedance to calculate the total current of the circuit and I get 81.93+j5146.59

Thats is diferent from RNorton (73.733 +j 0.58675).

Is that ok? They must be equal or not?

Thanks

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#### The Electrician

Joined Oct 9, 2007
2,848
Using the known RNorton and the voltage at A-B, I get (.000102134-j.00712104) for INorton.

You're going to have to explain to me just exactly you mean by the "total current of the circuit".

Also, what do you mean by "equivalent impedance"?

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#### jboavida

Joined Jul 10, 2008
23
When the teacher give the circuit we have to calculate the current supplied by the transformer to the circuit. Since we only have the secondary voltage and frequency, I have calculated the total impedance of the circuit (thevenin impedance) so I have the total current supplied by the transformer. That's I called equivalent impedance. Since Znortorn and Zthevenin are supposed to be the same...

That's why I'm not sure.

#### The Electrician

Joined Oct 9, 2007
2,848
The impedance where the 110 volts is applied would normally be called the input impedance, not the Thevenin impedance, since we are applying a voltage there. The value I get for the impedance there is (81.9118+j5141.41). Usually when we are asked for the Thevenin impedance at a particular node of a network, we are expecting to get a voltage out there, not apply a voltage going in.

The impedance of a network will typically be different when calculated at different parts (nodes) of the network.

You could have calculated the Thevenin impedance at the right end of R1, and it would be different than what is calculated at A-B.

Beware that calling it a "Thevenin impedance" at each node of the network doesn't mean that it will have the same value at those various nodes.