The Series RLC Circuit — Question 7

Section 31-8 · Physics pages 5124–5125

Question 7 Source resistance reduces both resonant current and loaded quality factor A source with rms open-circuit voltage Vs and internal resistance Rs drives an external series RLC circuit with resistor R. All losses are resistive and frequency-independent. ∼ vs(t) R + Rs C L q i = ˙q Schematic; labels, not drawing lengths, specify the data. (a) Find the resonant current and the quality factor of the complete loop. (b) Find the fraction of total average dissipation delivered to the external resistor and the terminal rms voltage at resonance.
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Question 7 - Solution Source resistance reduces both resonant current and loaded quality factor ∼ vs(t) R + Rs C L q i = ˙q (a) Method and calculation Rtotal = R + Rs, ω0 = 1/ √ LC. I0 = Vs/(R + Rs), Qloaded = ω0L/(R + Rs) . (b) Interpretation and further calculation PR = I2 0R, Ptotal = I2 0(R + Rs). PR/Ptotal = R/(R + Rs), Vterminal = VsR/(R + Rs) . Physical check Using only the external resistor would overestimate current and underestimate bandwidth.

Question 7

Source resistance reduces both resonant current and loaded quality factor

A source with rms open-circuit voltage Vs and internal resistance Rs drives an external series RLC circuit with resistor R. All losses are resistive and frequency-independent.

Diagram shown in the original SVG worksheet.

  1. Find the resonant current and the quality factor of the complete loop.

  2. Find the fraction of total average dissipation delivered to the external resistor and the terminal rms voltage at resonance.

Question 7 - Solution

Source resistance reduces both resonant current and loaded quality factor

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Rtotal=R+Rs,ω0=1/LC.R_{\mathrm{total}}=R+R_s,\quad\omega_0=1/\sqrt{LC}.I0=Vs/(R+Rs),𝒬loaded=ω0L/(R+Rs).\boxed{I_0=V_s/(R+R_s),\quad\mathcal Q_{\mathrm{loaded}}=\omega_0L/(R+R_s)}.

(b) Interpretation and further calculation

PR=I02R,Ptotal=I02(R+Rs).P_R=I_0^2R,\quad P_{\mathrm{total}}=I_0^2(R+R_s).PR/Ptotal=R/(R+Rs),Vterminal=VsR/(R+Rs).\boxed{P_R/P_{\mathrm{total}}=R/(R+R_s),\quad V_{\mathrm{terminal}}=V_sR/(R+R_s)}.

Physical check

Using only the external resistor would overestimate current and underestimate bandwidth.