# Resonance and impedance matching

Discussion in 'General Electronics Chat' started by The Electrician, Jan 12, 2016.

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1. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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RRITESH, can you calculate the voltage appearing at node 3 in this circuit?

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2. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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OK, As the circuit is at resonance mean both Xc and Xl are out of 180 phase cancel each other.
the total current wil be E=I*R
I=10/5=2Ams
so, Exl=2*30=60V i think

Jun 29, 2010
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right?

4. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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Yes, albeit, strictly speaking, 'phase' is not properly a property of reactance itself.... Anyway - good work!

@RRITESH KAKKAR
Inasmuch as this thread represents @The Electrician 's approach to instructing you, I hesitate to 'interfere' beyond 'the superficial' -- Please understand that trial of truly diverse technique is in your best interest

Best regards
HP

Last edited: Jan 13, 2016
5. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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That is exactly correct. Notice that the voltage source Vs is only supplying 10 volts, but you have 60 volts appearing across the inductor. This is what a series resonant circuit can do for us. It is an important phenomenon that we can use for impedance matching.

Now, can you calculate the Thevenin equivalent impedance appearing at node 3? That will be the impedance at node 3 when the voltage source is replaced with a short circuit.

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6. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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The total impedance by shoring the Voltage source and open node 3
will give us:
Z=√(R²+Xc²)
Z=30.41ohm, right?

7. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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No. You have R1 and C1 in series, with that combination in parallel with L1, like this:

What is the impedance seen at the A-B terminals? You will have a complex impedance, with a real and an imaginary part.

8. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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OK,
At 30.41ohm // with L1 30ohm
Z=1/(30.41²+30²)
Z=42.7ohm may be.

9. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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(5-30j) // (0+30j)
(5-30j)*(0+30j)/ (5-30j)+(0+30j)
(5-30j)*(0+30j)/(5)
...........
..............

10. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Ok. I see your problem here, RRITESH. You are having difficulty working with complex impedances.

I'm assuming you know how represent complex impedances using the imaginary constant j which is SQRT(-1). If you don't you should tell me so we can brush up on that technique.

The impedance of the series combination of R1 and C1 is given by (5 - j 30) and the impedance of L1 is (0 + j 30), so you need to calculate (5 - j 30) || (0 + j 30)

The impedance of the parallel combination is $\frac {(5 - j 30)*(0 + j 30)}{(5 - j 30)+(0 + j 30)}$, using the conventional product over the sum formula. Can you do this complex arithmetic?

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11. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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OK. But carry the multiplication in the numerator further, so you end up with an expression in the form A + j B.

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Jun 29, 2010
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How to do?

13. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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(5-30j) // (0+30j)
(5-30j)*(0+30j)/ (5-30j)+(0+30j)
(5-30j)*(0+30j)/(5)

0+150j+0-900j²/5
as j²=-1
150j+900/5
30j+180
180+30j i think

14. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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It's simple complex arithmetic. I think you must have forgotten how to do this.

You do it the same way you would do it with real arithmetic. Do you remember how to do this:

(a + b) * (c + d) = a*(c + d) + b*(c + d) = a*c +a*d + b*c + b*d

Proceed in the same way with complex arithmetic:

Code (Text):
1. (5 - j30) * (0 + j30) =
2. 5*(0 + j30) - j30*(0 + j30) =
3. 5*0 + 5*j30 - j30*0 - j30*j30 =
4. 0   +  j150 -  0    -  j*j*900 =
5. In this next step it's very important to remember that j*j = -1
6. 0   +  j150 -  0    - (-1)*900 =
7.       900 + j150
And finally, divide by 5 which was in the denominator:

We get 180 + j30

This is the Thevenin impedance seen at node 3 of the original circuit.

Do you follow everything so far?

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15. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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I have also done the same thing above.

Ok now what to do?

16. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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I see that you worked it all out while I was preparing my response!

Yes, you have the correct result.

Now, imagine that we wished to get rid of the imaginary part of that impedance. Since the imaginary part is j30, which is the reactance of an inductor, we see that if we placed a capacitor whose reactance is -j30 in series and to the right of node 3, after the capacitor we would see an impedance of 180 ohms, which is a pure resistance of 180 ohms. We would have a circuit like this (ignore the R2 designator):

Do you understand why the impedance at node 4 would be 180 ohms?

17. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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Now i got it that because the reactive power and impedance has no use.
Why we say it is imaginary??

18. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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Yes i understand importance of it.

19. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Well, actually reactive power has a use. It performs the impedance transformation we want to happen in this case, but in the end we want to get rid of it.

Complex arithmetic first came into use because the mathematicians of a long time ago were trying to find a solution of this equation:

X^2 = -1

Since they knew that multiplying two negative numbers gave a positive result, they didn't understand how there could be a number X such that X*X = -1. It seemed that no matter what X was, X*X couldn't be negative. Thus the concept of SQRT(-1) was born, but it seemed mysterious and they called it "imaginary", with imaginary meaning "not real".

Now we use the symbol j for SQRT(-1).

20. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
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ok, now what is next step?