Preparatory Work The impedance (ZL) of the parallel branches in figure 2 are evaluated as 1 1 = + = 1 Zequiv ZL Zc (1) Notice that they combine in the same way as the case of parallel resistors. Recall that the complex impedances for the capacitor and the inductor are j Żc and ŹL = jwL WC The magnitude of the impedances for each branch are 1 wC Žequiv |2c|= and |ZL| = wL However, you cannot just add the magnitude of the impedances of each branch to obtain the total impedance, the reason is that they are different in phase. If we substitute the values of the complex impedances in equation (??) then (jwL) (c) j(wL-) = ŻLŻC ŹL + 2C ZLZC ZL + Zc The total impedance of the circuit iz = = Lw w²LC -1° -j- Lw Ztotal = R + Zequiv=R-²LC -1° Question 1: Find the analytic expression for the amplitude |2total and the phase shift of total

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Preparatory Work
The impedance (ZL) of the parallel branches in figure 2 are evaluated as
1
1 1
= +
ZLZC
Zequiv ZL Zc ZL + Zc
(1)
Notice that they combine in the same way as the case of parallel resistors. Recall that the complex
impedances for the capacitor and the inductor are
j
wC
The magnitude of the impedances for each branch are
1
WC
Żc
==
|ZC| =
and |2₂| = wL
However, you cannot just add the magnitude of the impedances of each branch to obtain
the total impedance, the reason is that they are different in phase. If we substitute the
values of the complex impedances in equation (??) then
Żequiv
ŻLŻC
ŹL + 2C
The total impedance of the circuit iz
=
and ŹL = jwL
Question 2: What happens when w =
(jwL) (c)
j(wL-c)
Źtotal = R + Żequiv =R-j
=-j
Lw
w²LC-1
Lw
w2LC-11
Question 1: Find the analytic expression for the amplitude |2total and the phase shift of total
Answer to Question 1
1/√CL?
Answer to Question 2
Transcribed Image Text:Preparatory Work The impedance (ZL) of the parallel branches in figure 2 are evaluated as 1 1 1 = + ZLZC Zequiv ZL Zc ZL + Zc (1) Notice that they combine in the same way as the case of parallel resistors. Recall that the complex impedances for the capacitor and the inductor are j wC The magnitude of the impedances for each branch are 1 WC Żc == |ZC| = and |2₂| = wL However, you cannot just add the magnitude of the impedances of each branch to obtain the total impedance, the reason is that they are different in phase. If we substitute the values of the complex impedances in equation (??) then Żequiv ŻLŻC ŹL + 2C The total impedance of the circuit iz = and ŹL = jwL Question 2: What happens when w = (jwL) (c) j(wL-c) Źtotal = R + Żequiv =R-j =-j Lw w²LC-1 Lw w2LC-11 Question 1: Find the analytic expression for the amplitude |2total and the phase shift of total Answer to Question 1 1/√CL? Answer to Question 2
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