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### Network # Series resistance in parallel resonance circuits

T

#### TweedleDee

Jan 1, 1970
0
I have a lab exercise using a AC Voltage source with series resistance
hooked up to a parallel capacitor and inductor with winding resistance.
Now in my book, all the examples use a current source with parallel
restance as opposed to the voltage source with series resistance. I
can find the resonant frequency and transform the inductor with winding
resistance to an inductor with parallel resistance. But when I go to
find the total impedance of the circuit by adding up the two resistors
(which I assume are in series) I don't get the correct answer.

Transforming the voltage source with series resistance to a current
source doesn't seem to work either. It would seem to change the Q of
the circuit. Here is a feable attempt at a picture...

----R--------------
| | L
V C |
| | R
------------------

My goal is to find the total impedance, Q, and then the Bandwidth.
Can anyone shed some light on this? Thanks.

Alan

J

#### Joe McElvenney

Jan 1, 1970
0
Hi,

At phase resonance the impedance of a parallel tuned circuit is
equal to L/CR which is a pure resistance. 'Phase Resonance'
incidentally is the point where the current in both arms are equal
and out of phase. For practical purposes this occurs for Q-values
of 10 or greater.

Cheers - Joe

S

#### skeptic

Jan 1, 1970
0
I know you said you can find the resonant frequency and transform the
inductor with winding resistance to an inductor with parallel
resistance but I'd like to start with a review of that anyway. You
didn't provide any values so let's say Xl = 100 and the series winding
resistance is 1. To transform into parallel values:
Rp = (Rs^2 + Xls^2)/Rs = 10001
Xlp = (Rs^2 + Xls^2)/Xls = 100.01

To convert back again...
Rs = (Rp^2 + Xlp^2)/Rp
Xls = (Rp^2 + Xlp^2)/Xlp

To be parallel resonant Xcp must be equal to -Xlp or -100.01

The other R in the circuit which I'll call R1s doesn't affect either
the resonant frequency or the Q of the tuned circuit. R1p = 1/R1s.

The equivalent circuit for a current source would look like this. Note
all series components become shunt components and vice versa.

____________
| | |
| | Cp
^ I R1p |_____
| | | |
| | Lp R2p
|_____|______|_____|

T

#### TweedleDee

Jan 1, 1970
0
Thank you for the replies. Although I'm a little confused by the
comment that the other R does not affect the Q of the tuned circuit?
That other R definitely effects the Bandwidth of the circuit as shown
in the lab we did. By tuned circuit do you mean just the tank circuit
portion?

J

#### Joe McElvenney

Jan 1, 1970
0
Hi again,

Thank you for the replies. Although I'm a little confused by the
comment that the other R does not affect the Q of the tuned circuit?
That other R definitely effects the Bandwidth of the circuit as shown
in the lab we did. By tuned circuit do you mean just the tank circuit
portion?

Although I have given you one expression for the dynamic
impedance at resonance, you could also use Zd = (omega x L)^2 /
R(winding).

Further, remembering that an ideal voltage source has zero
resistance, where does that place the series input resistance with
respect to the tuned circuit? How does that then affect the Q?

And of course, bandwidth = fo/Q.

Cheers - Joe

S

#### skeptic

Jan 1, 1970
0
My apologies.

You're right, it does affect the bandwidth but not the resonant
frequency of the circuit.

T

#### TweedleDee

Jan 1, 1970
0
Thanks guys. I think I'm clear on this now for the most part. My
mistake was putting both resistors in series with each other as pointed
out by Joe. They should be in parallel with each other with regard to
the calculations for either a voltage or a current source.

I'm glad I got this figured out before our test since this exact
circuit was on it. Alan

R

#### Rich Grise

Jan 1, 1970
0
Thank you for the replies. Although I'm a little confused by the
comment that the other R does not affect the Q of the tuned circuit?
That other R definitely effects the Bandwidth of the circuit as shown
in the lab we did. By tuned circuit do you mean just the tank circuit
portion?

Please distinguish which R you're talking about. Let's assign, say, R1
to the resistor from the signal source to the "resonant circuit", which
consists of R2 in series with L1, this series string in parallel with
C1, and that whole thing in series with R1.

R1 doesn't affect Q at all. There's some equation for figuring out Q
from the component values, but I've long since forgotten even where
to look up such a thing.

I remember in tech school, the first time we did this experiment with
real inductors and capacitors, and the instructor didn't tell us that
the inductor in the bench setup had about 133R series resistance. Cheers!
Rich

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