# Parallel RLC Resonance bandwidth

Good day ,I need help in understanding a practice problem that my professor gave us.

A parallel resonant circuit has Q = 20 and is resonant at ωO = 10,000 rad/s. If Zin = 5kΩ at ω = ωO what is the width of the frequency band about resonance for which |Zin| ≥ 3kΩ?

I know that in a parallel RLC circuit , the quality factor Q is given by the equation Q=ω/BW and that the question seems to ask about the bandwidth . What I don't understand is the relevance of Zin and |Zin| in the question. I know that in resonance,the impedance is purely resistive but I can't understand what |Zin|≥ 3kΩ means.

Any help would be appreciated.

• The impedance can be capacitive or inductive, hence the need for the abs(). Jul 11, 2018 at 9:04

Any help would be appreciated.

You will need to calculate L and C to get the required answer. Only with values for L and C can you calculate the magnitude of the impedance when not at resonance. At resonance L and C cancel each other and the input impedance is purely 5 kΩ hence this is the value of the parallel resistor.

So, knowing that Q = 20 and ωO = 10,000 rad/s (at resonance), you can calculate the inductive reactance because Q = R/ωL for a parallel tuned RLC. From the inductive reactance you get L. From L and ωO you get C: -

$$ω_O = \dfrac{1}{\sqrt{LC}}$$

Now you know all the component values for R, L and C.

Next, an impedance magnitude of 3 kohm means that the net reactance is: -

$$\sqrt{5000 Ω^2 - 3000 Ω^2}= 4000Ω$$

In other words a 5 kohm resistor in parallel with a 4 kohm reactance is a magnitude impedance of 3 kohm.

Can you take it from here?

• I'm sorry I'm still confused on how to approach the question "what is the width of the frequency band about resonance for which |Zin| ≥ 3kΩ?" Am I correct that the question is asking for the bandwidth? Jul 11, 2018 at 10:20
• Does that also mean the inductive reactance and capacitive reactance that are in parallel is equal to 4000Ω? Jul 11, 2018 at 10:35
• 1) yes, the question is asking for a bandwidth 2) yes, XC || XL = 4000 ohms. Jul 11, 2018 at 10:43
• I got 10H for the inductor and 1nF for the capacitor,solving for ω1 and ω2 yield ω1=498.7562 and ω2=200498.7562. So BW = 20,000? Did I do it right? Jul 11, 2018 at 10:51
• I meant BW=200,000 Jul 11, 2018 at 10:58