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Corresponding to the upper sign, there is a single real positive root: s  2 R R 1 2338:3 rad=s or !h 2L 2L LC and corresponding to the lower sign, the single real positive root s  2 R R 1 2138:3 rad=s or !l 2L 2L LC
fh 372:1 Hz
fl 340:3 Hz
Derive the Q of (a) the series RLC circuit, (b) the parallel RLC circuit.
(a) In the time domain, the instantaneous stored energy in the circuit is given by 1 q2 Ws Li2 2 2C For a maximum,   dWs di q dq di q i L i vL vC 0 Li dt C dt dt C dt Thus, the maximum stored energy is Ws at i 0 or Ws at vL vC 0, whichever is the larger. Now the capacitor voltage, and therefore the charge, lags the current by 908; hence, i 0 implies q Qmax and  2 Q2 1 1 I I2 2 Ws ji 0 max CVC max C max max2 2 2 2C !C 2C! On the other hand, vL vC 0 implies vL vC 0 and i Imax (see the phasor diagram, Fig. 1237), so that
2 Ws jvL vC 0 1 LImax 2
It follows that Ws max 8 2 < Imax =2C!2 :
2 LImax =2
!0
! ! !0 Consequently,
2 The energy dissipated per cycle (in the resistor) is Wd Imax R=!.  W 1=!CR ! !0 Q 2 s max !L=R ! ! !0 Wd
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[CHAP. 12
Fig. 12-37 (b) For the parallel combination with applied voltage v t , 1 2 1 2 q Ws LIL 2 2C C and If v 0, then qC 0 and iL IL max Vmax !L dWs di q LiL L C iC v iL iC 0 dt dt C
giving
Wsjv 0
2 Vmax 2L!2
If iL iC 0, then (see Fig. 12-38) iL iC 0 and qC CVmax , giving
2 Ws jiL iC 0 1 CVmax 2
Fig. 12-38 ( Therefore Ws max
2 Vmax =2L!2 2 CVmax =2
!a
! ! !a Consequently,  ! !a ! ! !a
The energy dissipated per cycle in R is Wd  Ws max R=L! Q 2 !CR Wd
2 Vmax =R!.
A three-element series circuit contains R 10 , L 5 mH, and C 12:5 mF. Plot the magnitude and angle of Z as functions of ! for values of ! from 0.8 !0 through 1:2 !0 .
p !0 1= LC 4000 rad/s. At !0 , XC 1 20  4000 12:5 10 6
XL 4000 5 10 3 20 
Z 10 j XL XC 10 j0 
CHAP. 12]
FREQUENCY RESPONSE, FILTERS, AND RESONANCE
The values of the reactances at other frequencies are readily obtained. A tabulation of reactances and impedances appear in Fig. 12-39(a), and Fig. 12-39(b) shows the required plots. ! 3200 3600 4000 4400 4800 XL 16 18 20 22 24 XC 25 22.2 20 18.2 16.7 10 j9 10 j4:2 10 10 j3:8 10 j7:3 (a) Z 13:4 428 10:8 22:88 10 08 10:7 20:88 12:4 36:28
Fig. 12-39
Show that !0
p !l !h for the series RLC circuit.
By the results of Problem 12.5, 0s 10s 1  2  2 R 1 R A@ R 1 RA 1 !2 !l !h @ 0 2L LC 2L 2L LC 2L LC
Compute the quality factor of an RLC series circuit, with R 20 , L 50 mH, and C 1 mF, using (a) Q !0 L=R, (b) Q 1=!0 CR, and (c) Q !0 = .
1 !0 p 4472 rad=s 0:05 10 6 s s  2  2 R R 1 R R 1 4276:6 rad=s !h 4676:6 rad=s !l 2L 2L LC 2L 2L LC and !h !l 400 rad/s. a b c !0 L 4472 0:050 11:2 R 20 1 1 Q 11:2 !0 CR 4472 10 6 20 Q Q !0 4472 11:2 400
FREQUENCY RESPONSE, FILTERS, AND RESONANCE
[CHAP. 12
12.10 A coil is represented by a series combination of L 50 mH and R 15 . Calculate the quality factor at (a) 10 kHz, (b) 50 kHz.
a c Qcoil !L 2 10 103 50 10 3 209 R 15   50 1047 Qcoil 209 10
12.11 Convert the circuit constants of Problem 12.10 to the parallel form (a) at 10 kHz, (b) at 250 Hz.
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