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Problem set #3, EE 223, 1/30/2003 – 2/06/
Chapter 8, Problem 68.
Find vC for t > 0 in the circuit of Fig. 8.93.
-20 0 20 40 60 80 100 120 140
0
1
2
t = (^) τ
tim e (μs )
voltage (V)
Chapter 8, Problem 72.
The switch in Fig. 8.97 has been at A for a long time. It is moved to B at t = 0, and back to A at t = 1 ms. Find R 1 and R 2 so that vC (1 ms) = 8 V and vC (2 ms) = 1 V.
-1 -0.5 0 0.5 1 1.5 2 2.5 3
0
1
2
3
4
5
6
7
8
9
10 τ 1
τ 2
time (ms)
voltage (V)
/ 106 /33.
30,
OC: 0 1, 3 1 2V
SC: 3V 0.06A
R / 2/ 0.06 33.
2(1 )V, 0
th
x x x x oc
x x x sc
th oc sc t R C t c oc t
v v v v v
v v v i
v i
v v e e
e t
− −
−
(^6 )
1 1
(^6 )
10 /( 100)
1000/( 100) 1000/( 100)
1 1 10 /( 100) 3 1000 2
2 2
0 1ms: 9(1 )
1 8 9(1 ), 9
1000 2.197, R 355. R 100
1ms: 8 , 10 1 8 (R 100)
1000 2.079, R 480.9 100 380. R 100
c t R c
R R
t R c
t v
t v e
e e
t v e t t e
− +
− + − +
− ′ + − −
Chapter 9, Problem 3.
A parallel RLC circuit is found to have a
natural resonant frequency of ω 0 = 70.71 ×
12
rad/s. It is known that the inductance L=2 pH,
( a ) compute C;
( b ) determine the value of resistance R that will
lead to an exponential damping coefficient of 5 Gs
( c ) determine the neper frequency of the
circuit;
( d ) compute s 1 and s 2 ;
( e ) calculate the damping ratio of the circuit.
Parallel RLC with ωo = 70.71 × 10
12 rad/s. L = 2 pH.
(a)
(b)
(c) α is the neper frequency: 5 Gs
(d)
(e)
9 5 12
α × − ζ = = = × ω (^) o ×
Chapter 9, Problem 6.
In the circuit of Fig. 9.29, let L = 5 H, R = 8Ω, C = 12.5 mF, and v (
) = 40 V. Find ( a ) v ( t ) if i (
) = 8 A ; ( b ) i ( t ) if i (^) C (
) = 8 A.
5H, R 8 , C 12.5mF, (0 ) 40V
L = = Ω = v =
(a)
(b)
2 12 2
12 2 12
So 100.0 aF (70.71 10 ) (2 10 )
−
ω = = ×
× ×
o LC
C
9 1
10 18
So 1 M (10 ) (100 10 )
−
−
α = = ×
×
s RC
R
2 2 9 12 1 1
2 2 9 12 1 2
−
−
= −α + α − ω = − × + ×
= −α − α − ω = − × − ×
o
o
S j s
S j s
2
2 8 1,2 1 2
1 2
1 2 1 2 2 2 1
(0 ) 8A: 5, 16,
2RC 2 8 12.5 LC
4 5 25 16 2, 8 ( ) A A
40 A A (0 ) (0 ) 80( 8 5) 1040
/ 2A 8A 520 A 4A 3A 480, A 160,A 120
o
t t o
L
i
s v t e e
v i
v s
v t
− −
= α= = = ω = = × ×
ω = =− ± − =− − ∴ = +
2 8 160 V, 0
t t e e t
− −
2 8 3 4
3 4
3 4 3 4
2 8 4 4 3
(0 ) 8A Let ( ) A A ; (0 ) 5A R 8
(0 ) A A (0 ) (0 ) 8 5 13A;
40 (0 ) 2A 8A 8 A/ 4 A 4A 5
3A 13 4, A 3, A 16 ( ) 16 3 A, 0
− −
t t c R
R c
t t
v i i t e e i
i i i
i s
i t e e t
