21
E E 2315 Lecture 12 - Natural Response of Parallel RLC Circuits
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E E 2315
Lecture 12 - Natural Response of Parallel RLC
Circuits
Natural Response of Parallel RLC Circuit (1/5)
L R C
t = 0
+v-
iL
KCL for t 0: 0 L
dv vC idt R
but
then 00
10
t
L
dv vC vdt idt R L
Natural Response of Parallel RLC Circuit (2/5)
L R C
t = 0
+v-
iL
2
2
10
d v dv vCdt R dt L
Differentiate KCL:
From experience with 1st order problems:
Natural Response of Parallel RLC Circuit (3/5)
L R C
t = 0
+v-
iL
into differentiated KCLatVeSubstitute
equation: 20 at at ata Va CVe Ve e
R L
Divide out atVe
Natural Response of Parallel RLC Circuit (4/5)
L R C
t = 0
+v-
iL
Solve for a: 2 2
1 1 1
2 4a
RC R C LC
2 1o LC
Let and
Natural Response of Parallel RLC Circuit (5/5)
2 21
2 22
o
o
a
a
Then
• Three types of response:– real and unequal (both negative)– real and equal (negative)– complex conjugate pair
Overdamped Parallel RLC (1/5)
62.5 mH 1 +v-t = 0
iL 10 mF
0 6Li A
150
2 1 0.01
1
2
50 2500 1600 20
50 2500 1600 80
a
a
Overdamped Parallel RLC (2/5)
62.5 mH 1 +v-t = 0
iL 10 mF
20 801 2( ) t t
Li t k e k e
1 2(0) 0 20 80v k k then 1 24k k
Overdamped Parallel RLC (3/5)
62.5 mH 1 +v-t = 0
iL 10 mF
1 2 20 6 3Li k k k
20 80
20 80
8 2
10 10
t tL
t t
i t e e
v t e e
Overdamped Parallel RLC (4/5)
0 0.05 0.1 0.15 0.20
1
2
3
4
5
6Inductor Current
iL t( )
t
Overdamped Parallel RLC (5/5)
0 0.05 0.1 0.15 0.25
4
3
2
1
0Capacitor Voltage
v t( )
t
Critically Damped Parallel RLC (1/4)
40 mH 1 +v-t = 0
iL 10 mF
150
2 1 0.01
50 501 2
t tLi t g te g e
50 50 501 1 2
150 50
25t t tLdiv t L g te g e g e
dt
0 10Li A
Critically Damped Parallel RLC (2/4)
40 mH 1 +v-t = 0
iL 10 mF
Use initial conditions to find g1 and g2.1
2(0) 0 225
gv g 1 250g g
50 50500 10t tLi t te e
1 500g
Critically Damped Parallel RLC (3/4)
0 0.05 0.1 0.15 0.20
2
4
6
8
10Inductor Current
iL t( )
t
Critically Damped Parallel RLC (4/4)
0 0.05 0.1 0.15 0.28
6
4
2
0Capacitor Voltage
v t( )
t
Underdamped Parallel RLC (1/6)
8 mH 1 +v-t = 0
iL 10 mF 0 20Li A
150
2 1 0.01
2 1
12,5000.08 0.01o
1
2
50 100
50 100
a j
a j
Underdamped Parallel RLC (2/6)
8 mH 1 +v-t = 0
iL 10 mF
50 cos 100 sin 100tL c si t e I t I t
Ldiv t Ldt
50
50
50 cos 100 sin 1001
125 100 cos 100 sin 100
tc s
ts c
e I t I tv t
e I t I t
0 20Li A
Underdamped Parallel RLC (3/6)
8 mH 1 +v-t = 0
iL 10 mF
Using initial conditions: 0 20L ci A I
and 10 0 50 100
125 c sv V I I
then
0 20Li A
Underdamped Parallel RLC (4/6)
8 mH 1 +v-t = 0
iL 10 mF
50( ) 20cos(100 ) 10sin(100 )tLi t e t t
0 20Li A
Underdamped Parallel RLC (5/6)
0 0.05 0.1 0.15 0.25
0
5
10
15
20Inductor Current
iL t( )
t
Underdamped Parallel RLC (6/6)
0 0.05 0.1 0.15 0.212
10
8
6
4
2
0
2
4Capacitor Voltage
v t( )
t
