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Quality Factor Q in Parallel Resonance
0p
Y = G = 0.01 : – 1
0
and
t
i = i t = 10cos 5000t mA
G
S
t
t
Next, to compute i and i , we must first find v t . For this example,
L
C
0
t
i 10cos 5000t mA
G
v t = ------------ = --------------------------------------- = 1000cos 5000t mV = cos 5000t V
0
–
G
0.01 : 1
In phasor form,
v t = cos 5000t V V = 10q
0
0
Now,
I L0 = – jB L V = 1 – 90q 0.02 10q = 0.02 – 90q A
0
and in the -domain,
t
t
I L0 = 0.02 – 90q A i = 0.02cos 5000t – 90q A
L0
or
t
i = 20sin 5000t mA
L0
Similarly,
I C0 = jB V = 190q 0.02 10q = 0.02 90q A
0
C
and in the -domain,
t
I C0 = 0.02 90q A i C0 t = 0.02cos 5000t + 90q A
or
i C0 t = – 20sin 5000t mA
We observe that i + i t = 0 as expected.
t
L0 C0
2.4 Quality Factor Q in Parallel Resonance
0p
At parallel resonance,
1
Z C = ---------
0
Z L
and 0
I
S
V = -------
0
G
Then,
I Z C
0
S
I C0 = Z CV = Z C------- = ---------- I S (2.18)
0
0
0
G
G
Also,
Circuit Analysis II with MATLAB Applications 2-9
Orchard Publications
