Page 93 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

5.
R 1 L 1 mH
`
jZL
1 :
C
Z IN 10 : R 2
1
---------- 1 PF
jZC

a. It is important to remember that the relation Z = 1 e LC applies only to series RLC and
0
parallel GLC circuits. For any other circuit we must find the input impedance Z IN , set the
imaginary part of Z IN equal to zero, and solve for Z 0 . Thus, for the given circuit

1 1jZCe 10 + jZL ˜
Z IN = R + ---------- jZC R + jZL = 1 + ------------------------------------------------
__
2
1
1 ZC
10 +
j ZL –
e
10 + j ZL – 1 ZC + 10 jZC + LC 10 j ZL – 1 ZC
–

e

e
e
e
= -------------------------------------------------------------------------------------------- -----------------------------------------------
˜
10 + j ZL – 1 ZC 10 j ZL – – 1 ZC
e
e
100 + j10 ZL – 1 ZC + 100 jZC + 10L C – j10 ZL – 1 ZC
e
e

e
e

= -------------------------------------------------------------------------------------------------------------------------------------------------------------------
100 + ZL – 1 ZC 2
e
2
1 ZC
–
1 ZC
jL C ZL –
ZL –
1 ZC –
10 ZC ZL –
e

e
e

e
e
+ -------------------------------------------------------------------------------------------------------------------------------------------------------
100 + ZL – 1 ZC 2
e
–
–

e
e
e
e
= 100 10L C + + ZL1 ZC 2 10 ZC ZL – 1 ZC
------------------------------------------------------------------------------------------------------------------------------------------
100 + ZL – 1 ZC 2
e
1 ZC
–
jL C ZL –

100
e
e
jZC e
+ ---------------------------------------------------------------------------------
100 + ZL – 1 ZC 2
e
For resonance, the imaginary part of Z IN must be zero, that is,
100 jL 1 ·
§
------------ – ----- Z L – ---------- = 0
jZ C C © 0 Z C ¹
0
0
j 100
1 ·
§
– ---- --------- + L Z L – ---------- = 0
C Z 0 © 0 Z C ¹
0
L
100 Z L – ---------- = 0
2
--------- +
Z 0 Z C
0 0
2
2
–
L CZ + 100C L = 0
0
Circuit Analysis II with MATLAB Applications 2-27
Orchard Publications