Q factor prob

Discussion in 'Homework Help' started by JimmyB, Jun 26, 2010.

Jun 1, 2010
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hi all,

another confusing one

A coil of inductance 120mH & resistance 150 ohms is connected in parallel with a variable cap across a 20 V, 4Khz supply. determine for the condition when the supply current is a minimum:

(a) the capacitance of the capacitor
(b) the dynamic resistance
(c) the supply current
(d) the Q-factor
(e) the bandwidth
(f) the upper and lower -3dB frequencies
(g) the value of the circuit impendance at the -3dB frequency

I can do all except (d)
My text book says in parallel Q=R/X but govener says its wrong
he says its Q=WrL/R but my text book says thts for series.... is it something to do with the dynamic impedance??

any help would be good!!

Apr 5, 2008
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3. t_n_k AAC Fanatic!

Mar 6, 2009
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You are confusing the inductor series resistance (R_inductor) with the equivalent parallel resistance (Rp_equiv) at resonance.

Q=XL/(R_inductor)=(Rp_equiv)/XL

where XL=ωL at resonance

Jun 1, 2010
38
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Is Rp equivalant got something to do with admitance?

wht is Rp?

Last edited: Jun 26, 2010
5. t_n_k AAC Fanatic!

Mar 6, 2009
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Not really. It's the equivalent resistance you would calculate if you treated the inductor and capacitor as ideal and lumped the losses due to their series resistance values into one single parallel resistive element.

For the sake of an exercise consider that only the inductor has a series resistance [RL] of any significance and that the capacitor series resistance is negligible. This would correspond to the conditions in your question - since no resistance in series with the capacitor is stated.

In this case at resonance the equivalent resistance in parallel with the ideal inductor and capacitor would be

$R_p=R_L$1+\frac{X_L^2}{R_L^2}$$

If you calculate the Q values using the two options I stated in my previous post, you should get the same Q value whether you use the actual inductor series resistance version or the calculated parallel equivalent resistance version of the Q formula.

6. t_n_k AAC Fanatic!

Mar 6, 2009
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Here's another point to ponder.

If you drew the parallel equivalent circuit with the value of Rp as calculated, would the equivalent parallel ideal L & C values be exactly the same as those given in the original circuit?

7. Ron H AAC Fanatic!

Apr 14, 2005
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I don't understand why we are talking about equivalent parallel resistance here. The resistor is in series. The OP has the equation for series resistance Q.

8. t_n_k AAC Fanatic!

Mar 6, 2009
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I think it went this way because the OP asked this in post #1 ....

"can do all except (d)
My text book says in parallel Q=R/X but govener says its wrong
he says its Q=WrL/R but my text book says thts for series.... is it something to do with the dynamic impedance??"

They appeared to be concerned as to why there were (apparently) two formulas for Q which seemed incompatible to them.

Last edited: Jun 27, 2010

Jun 1, 2010
38
0
Thanks, i think i got it,
so if the cap had a series resistance aswell, u find the parallel resistanace between the RL and Rc then u divide by X.
if a there is Rc but no RL u use 1/WrCR

Last edited: Jun 27, 2010
10. t_n_k AAC Fanatic!

Mar 6, 2009
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Not exactly sure what you mean but finding the equivalent resistance at resonance in the case where both L and C have resistance, isn't a trivial exercise. It's more difficult to analyse than the series resonance RLC case having the same values.

Since you just want the Q then I'd use the formula you gave in another post

$Q=\frac{Q_L Q_C}{Q_L+Q_C}$

which works for both series and parallel resonance when both L & C have resistance. [I've never actually derived this for the parallel case.]

Presumably you are already aware that the presence of resistance in either L or C will shift the parallel resonant frequency away from that of the simple series resonance RLC case using the same values.

i.e.

$\omega_r=\frac{1}{\sqrt{LC}}$

is only true for the parallel resonance case when there is no resistance in either L or C