Page 95 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

1
Z = ------- – R 2
------
0 LC L 2
or
4
10
1
2S u 43 000 = -------------------------- – --------------------

–
3
2 u 10 C 2 4 u 10 – 6
4
10 3 10 4 4 2 10 + 2S u 4.3 u 10 4 2 u 4 u 10 – 6
--------- = -------------------- + 2S u 4.3 u 10 = -----------------------------------------------------------------------------------
2C 2 4 u 10 – 6 4 u 10 – 6
– 6
4 u 10 – 9
C = 500 ----------------------------------------------------------------------------------- = 6.62 u 10 F = 6.62 nF
2
4 2
4
10 + 2S u 4.3 u 10 u 4 u 10 – 6
Next, we must find the value of C 1 that will make the entire circuit series resonant (minimum
impedance, maximum current) at f = 10 KHz . In the circuit below we let z = – jX C1 ,
1
z = – jX C2 , z = R + jX L , and z LOAD = . 1
2
3
1
C 1
C 2 – jX C2

R 1 jX
– jX ` L + R
+ C1 100 : L 2 mH L
Z v
IN LOAD
1 :
V = 170 0q V
‘
S
Then,
Z IN = z + z __ z + z LOAD
3
2
1
and
Z f = 10 KHz = z + z __ z + z = z + z __ z + 1 (1)

1
2
IN
3
f = 10 KHz LOAD 1 2 3 f = 10 KHz
where z __ z is found with the MATLAB code below.
2
3
f = 10 KHz
format short g; f=10000; w=2*pi*f; C2=6.62*10^( 9); XC2=1/(w*C2); L=2*10^( 3);...
XL=w*L; R1=100; z2=-j*XC2; z3=R1+j*XL; Zp=z2*z3/(z2+z3)
Zp =
111.12 + 127.72i
and by substitution into (1)

Circuit Analysis II with MATLAB Applications 2-29
Orchard Publications