Chapter 8, Solution 1.
(a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is
shown in Figure (a).
6Ω 6Ω
+
VS −
6Ω +
+
v 10 µF
vL 10 H
−
(a) −
(b)
i(0-) = 12/6 = 2A, v(0-) = 12V
At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V
(b) For t > 0, we have the equivalent circuit shown in Figure (b).
vL = Ldi/dt or di/dt = vL/L
Applying KVL at t = 0+, we obtain,
vL(0+) – v(0+) + 10i(0+) = 0
vL(0+) – 12 + 20 = 0, or vL(0+) = -8
Hence, di(0+)/dt = -8/2 = -4 A/s
Similarly, iC = Cdv/dt, or dv/dt = iC/C
iC(0+) = -i(0+) = -2
dv(0+)/dt = -2/0.4 = -5 V/s
(c) As t approaches infinity, the circuit reaches steady state.
i(∞) = 0 A, v(∞) = 0 V
Chapter 8, Solution 2.
(a) At t = 0-, the equivalent circuit is shown in Figure (a).
25 kΩ 20 kΩ
iR + iL
+
80V −
60 kΩ v
−
(a)
25 kΩ 20 kΩ
iR iL
+
80V −
(b)
60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA.
By the current division principle,
iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA
vC(0-) = 0
At t = 0+,
vC(0+) = vC(0-) = 0
iL(0+) = iL(0-) = 1.5 mA
80 = iR(0+)(25 + 20) + vC(0-)
iR(0+) = 80/45k = 1.778 mA
But, iR = i C + iL
1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA
(b) vL(0+) = vC(0+) = 0
But, vL = LdiL/dt and diL(0+)/dt = vL(0+)/L = 0
diL(0+)/dt = 0
Again, 80 = 45iR + vC
0 = 45diR/dt + dvC/dt
But, dvC(0+)/dt = iC(0+)/C = 0.278 mohms/1 µF = 278 V/s
Hence, diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45
diR(0+)/dt = -6.1778 A/s
Also, iR = iC + iL
diR(0+)/dt = diC(0+)/dt + diL(0+)/dt
-6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s
(c) As t approaches infinity, we have the equivalent circuit in Figure
(b).
iR(∞) = iL(∞) = 80/45k = 1.778 mA
iC(∞) = Cdv(∞)/dt = 0.
Chapter 8, Solution 3.
At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR(0-) =
0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V.
(a) At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,
the inductor current must still be equal to 0A, the capacitor has a voltage equal to
–10V. Since it is in series with the +10V source, together they represent a direct
short at t = 0+. This means that the entire 2A from the current source flows
through the capacitor and not the resistor. Therefore, vR(0+) = 0 V.
(b) At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s,
iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s. Now for the value of
dvR(0+)/dt. Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.
40 Ω 40 Ω
+ +
iL
+ vC + vC
2A
vR 10 Ω − vR −
− +
− 10 Ω +
−
10V −
10V
(a) (b)
(c) As t approaches infinity, we end up with the equivalent circuit shown in
Figure (b).
iL(∞) = 10(2)/(40 + 10) = 400 mA
vC(∞) = 2[10||40] –10 = 16 – 10 = 6V
vR(∞) = 2[10||40] = 16 V
Chapter 8, Solution 4.
(a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in
Figure (a).
i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V.
Hence, i(0+) = i(0-) = 5A
v(0+) = v(0-) = 25V
3Ω
i
+
+ v 5Ω
40V −
−
(a)
3Ω 0.25 H
i + vL − iC iR
4A
+ 0.1F
40V −
5Ω
(b)
(b) iC = Cdv/dt or dv(0+)/dt = iC(0+)/C
For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b).
Since i and v cannot change abruptly,
iR = v/5 = 25/5 = 5A, i(0+) + 4 =iC(0+) + iR(0+)
5 + 4 = iC(0+) + 5 which leads to iC(0+) = 4
dv(0+)/dt = 4/0.1 = 40 V/s
Chapter 8, Solution 5.
(a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0).
iL(0-) = 0 and vC(0-) = 0.
For t = 0+, 4u(t) = 4. Consider the circuit below.
A iL 1H
i iC + vL −
+ +
4A
4 Ω vC 0.25F 6Ω v
− −
Since the 4-ohm resistor is in parallel with the capacitor,
i(0+) = vC(0+)/4 = 0/4 = 0 A
Also, since the 6-ohm resistor is in series with the inductor,
v(0+) = 6iL(0+) = 0V.
(b) di(0+)/dt = d(vR(0+)/R)/dt = (1/R)dvR(0+)/dt = (1/R)dvC(0+)/dt
= (1/4)4/0.25 A/s = 4 A/s
v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0
Therefore dv(0+)/dt = 0 V/s
(c) As t approaches infinity, the circuit is in steady-state.
i(∞) = 6(4)/10 = 2.4 A
v(∞) = 6(4 – 2.4) = 9.6 V
Chapter 8, Solution 6.
(a) Let i = the inductor current. For t < 0, u(t) = 0 so that
i(0) = 0 and v(0) = 0.
For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0.
vR(0+) = Ri(0+) = 0 V
Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V.
(1)
(b) Since i(0+) = 0, iC(0+) = VS/RS
But, iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS) (2)
From (1), dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt (3)
vR = iR or dvR/dt = Rdi/dt (4)
But, vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0 (5)
From (4) and (5), dvR(0+)/dt = 0 V/s
From (2) and (3), dvL(0+)/dt = dv(0+)/dt = Vs/(CRs)
(c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor
acts like a short circuit.
vR(∞) = [R/(R + Rs)]Vs
vL(∞) = 0 V
Chapter 8, Solution 7.
− 4 ± 4 2 − 4x 4
s2 + 4s + 4 = 0, thus s1,2 = = -2, repeated roots.
2
v(t) = [(A + Bt)e-2t], v(0) = 1 = A
dv/dt = [Be-2t] + [-2(A + Bt)e-2t]
dv(0)/dt = -1 = B – 2A = B – 2 or B = 1.
Therefore, v(t) = [(1 + t)e-2t] V
Chapter 8, Solution 8.
− 6 ± 6 2 − 36
s2 + 6s + 9 = 0, thus s1,2 = = -3, repeated roots.
2
i(t) = [(A + Bt)e-3t], i(0) = 0 = A
di/dt = [Be-3t] + [-3(Bt)e-3t]
di(0)/dt = 4 = B.
Therefore, i(t) = [4te-3t] A
Chapter 8, Solution 9.
− 10 ± 10 − 10
s2 + 10s + 25 = 0, thus s1,2 = = -5, repeated roots.
2
i(t) = [(A + Bt)e-5t], i(0) = 10 = A
di/dt = [Be-5t] + [-5(A + Bt)e-5t]
di(0)/dt = 0 = B – 5A = B – 50 or B = 50.
Therefore, i(t) = [(10 + 50t)e-5t] A
Chapter 8, Solution 10.
− 5 ± 25 − 16
s2 + 5s + 4 = 0, thus s1,2 = = -4, -1.
2
v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A
dv/dt = (-4Ae-4t - Be-t)
dv(0)/dt = 10 = – 4A – B = –3A or A = –10/3 and B = 10/3.
Therefore, v(t) = (–(10/3)e-4t + (10/3)e-t) V
Chapter 8, Solution 11.
−2± 4−4
s2 + 2s + 1 = 0, thus s1,2 = = -1, repeated roots.
2
v(t) = [(A + Bt)e-t], v(0) = 10 = A
dv/dt = [Be-t] + [-(A + Bt)e-t]
dv(0)/dt = 0 = B – A = B – 10 or B = 10.
Therefore, v(t) = [(10 + 10t)e-t] V
Chapter 8, Solution 12.
(a) Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF
(b) Critically damped when C = 6 mF
(c) Underdamped when C < 6mF
Chapter 8, Solution 13.
Let R||60 = Ro. For a series RLC circuit,
1 1
ωo = = = 5
LC 0.01x 4
For critical damping, ωo = α = Ro/(2L) = 5
or Ro = 10L = 40 = 60R/(60 + R)
which leads to R = 120 ohms
Chapter 8, Solution 14.
This is a series, source-free circuit. 60||30 = 20 ohms
1 1
α = R/(2L) = 20/(2x2) = 5 and ωo = = = 5
LC 0.04
ωo = α leads to critical damping
i(t) = [(A + Bt)e-5t], i(0) = 2 = A
v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}
v(0) = 6 = 2B – 10A = 2B – 20 or B = 13.
Therefore, i(t) = [(2 + 13t)e-5t] A
Chapter 8, Solution 15.
This is a series, source-free circuit. 60||30 = 20 ohms
1 1
α = R/(2L) = 20/(2x2) = 5 and ωo = = = 5
LC 0.04
ωo = α leads to critical damping
i(t) = [(A + Bt)e-5t], i(0) = 2 = A
v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}
v(0) = 6 = 2B – 10A = 2B – 20 or B = 13.
Therefore, i(t) = [(2 + 13t)e-5t] A
Chapter 8, Solution 16.
At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V
For t > 0, we have a source-free RLC circuit.
1 1
α = R/(2L) = (40 + 60)/5 = 20 and ωo = = = 20
−3
LC 10 x 2.5
ωo = α leads to critical damping
i(t) = [(A + Bt)e-20t], i(0) = 0 = A
di/dt = {[Be-20t] + [-20(Bt)e-20t]},
but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]
Hence, B = -9.6 or i(t) = [-9.6te-20t] A
Chapter 8, Solution 17.
i(0) = I0 = 0, v(0) = V0 = 4 x15 = 60
di(0) 1
= − (RI0 + V0 ) = −4(0 + 60) = −240
dt L
1 1
ωo = = = 10
LC 1 1
4 25
R 10
α= = = 20, which is > ωo .
2L 2 1
4
s = −α ± α 2 − ωo = −20 ± 300 = −20 ± 10 3 = −2.68, − 37.32
2
i( t ) = A1e − 2.68t + A 2e −37.32 t
di(0)
i(0) = 0 = A1 + A 2 , = −2.68A1 − 37.32A 2 = −240
dt
This leads to A1 = −6.928 = −A 2
(
i( t ) = 6.928 e −37.32 t − e − 268t )
1 t
Since, v( t ) = ∫ i( t )dt + 60, we get
C 0
v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V
Chapter 8, Solution 18.
When the switch is off, we have a source-free parallel RLC circuit.
1 1 1
ωo = = = 2, α= = 0.5
LC 0.25 x1 2 RC
α < ωo
→ underdamped case ω d = ω o − α 2 = 4 − 0.25 = 1.936
2
Io(0) = i(0) = initial inductor current = 20/5 = 4A
Vo(0) = v(0) = initial capacitor voltage = 0 V
v(t ) = e −αt ( A1 cos ω d t + A2 sin ω d t ) = e −0.5αt ( A1 cos1.936t + A2 sin 1.936t )
v(0) =0 = A1
dv
= e −0.5αt (−0.5)( A1 cos1.936t + A2 sin 1.936t ) + e −0.5αt (−1.936 A1 sin 1.936t + 1.936 A2 cos1.936t )
dt
dv(0) (V + RI o ) ( 0 + 4)
=− o =− = −4 = −0.5 A1 + 1.936 A2
→ A2 = −2.066
dt RC 1
Thus,
v(t ) = −2.066e −0.5t sin 1.936t
Chapter 8, Solution 19.
For t < 0, the equivalent circuit is shown in Figure (a).
10 Ω i
+ +
+ i
120V − v L C v
− −
(a) (b)
i(0) = 120/10 = 12, v(0) = 0
For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α.
1 1
ωo = = = 0.5 = ωd
LC 4
i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A
v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],
which leads to -v(0)/L = 0 = B
Hence, i(t) = 12cos0.5t A and v = 0.5
However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V
Chapter 8, Solution 20.
For t < 0, the equivalent circuit is as shown below.
2Ω
i
12
+ −
vC
− +
v(0) = -12V and i(0) = 12/2 = 6A
For t > 0, we have a series RLC circuit.
α = R/(2L) = 2/(2x0.5) = 2
ωo = 1/ LC = 1 / 0.5x 1 4 = 2 2
Since α is less than ωo, we have an under-damped response.
ωd = ωo − α 2 = 8 − 4 = 2
2
i(t) = (Acos2t + Bsin2t)e-2t
i(0) = 6 = A
di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt
di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0
Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A
Chapter 8, Solution 21.
By combining some resistors, the circuit is equivalent to that shown below.
60||(15 + 25) = 24 ohms.
12 Ω t=0 6Ω i
3H
+
24V −
24 Ω +
v (1/27)F
−
At t = 0-, i(0) = 0, v(0) = 24x24/36 = 16V
For t > 0, we have a series RLC circuit. R = 30 ohms, L = 3 H, C = (1/27) F
α = R/(2L) = 30/6 = 5
ωo = 1 / LC = 1 / 3x1 / 27 = 3, clearly α > ωo (overdamped response)
s1,2 = − α ± α 2 − ωo = −5 ± 5 2 − 3 2 = -9, -1
2
v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B (1)
i = Cdv/dt = C[-Ae-t - 9Be-9t]
i(0) = 0 = C[-A – 9B] or A = -9B (2)
From (1) and (2), B = -2 and A = 18.
Hence, v(t) = (18e-t – 2e-9t) V
Chapter 8, Solution 22.
α = 20 = 1/(2RC) or RC = 1/40 (1)
ωd = 50 = ωo − α 2 which leads to 2500 + 400 = ωo2 = 1/(LC)
2
Thus, LC 1/2900 (2)
In a parallel circuit, vC = vL = vR
But, iC = CdvC/dt or iC/C = dvC/dt
= -80e-20tcos50t – 200e-20tsin50t + 200e-20tsin50t – 500e-20tcos50t
= -580e-20tcos50t
iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 µF
R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms
L = 1/(2900x11.21) = 30.76 H
Chapter 8, Solution 23.
Let Co = C + 0.01. For a parallel RLC circuit,
α = 1/(2RCo), ωo = 1/ LC o
α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF
ωo = 1/ 0.5x 0.5 = 6.32 > α (underdamped)
Co = C + 10 mF = 50 mF or 40 mF
Chapter 8, Solution 24.
For t < 0, u(-t) 1, namely, the switch is on.
v(0) = 0, i(0) = 25/5 = 5A
For t > 0, the voltage source is off and we have a source-free parallel RLC circuit.
α = 1/(2RC) = 1/(2x5x10-3) = 100
ωo = 1/ LC = 1 / 0.1x10 −3 = 100
ωo = α (critically damped)
v(t) = [(A1 + A2t)e-100t]
v(0) = 0 = A1
dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000
But, dv/dt = [(A2 + (-100)A2t)e-100t]
Therefore, dv(0)/dt = -5000 = A2 – 0
v(t) = -5000te-100t V
Chapter 8, Solution 25.
In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0.
2Ω 1H
io(t)
t=0, note this is a +
+ make before break 8Ω
30V (1/4)F vo(t)
− switch so the
inductor current is −
not interrupted.
Figure 8.78 For Problem 8.25.
At t = 0-, vo(0) = (8/(2 + 8)(30) = 24
For t > 0, we have a source-free parallel RLC circuit.
α = 1/(2RC) = ¼
ωo = 1/ LC = 1 / 1x 1 4 = 2
Since α is less than ωo, we have an under-damped response.
ωd = ωo − α 2 = 4 − (1 / 16) = 1.9843
2
vo(t) = (A1cosωdt + A2sinωdt)e-αt
vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0.
dvo/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt
at t = 0, we get dvo(0)/dt = 0 = -αA1 + ωdA2
Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843 = 3.024
vo(t) = (24cosωdt + 3.024sinωdt)e-t/4 volts
Chapter 8, Solution 26.
− 2 ± 4 − 20
s2 + 2s + 5 = 0, which leads to s1,2 = = -1±j4
2
i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2
i(0) = 2 = = 2 + A1, or A1 = 0
di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1
i(t) = 2 + sin4te-t A
Chapter 8, Solution 27.
− 4 ± 16 − 32
s2 + 4s + 8 = 0 leads to s = = −2 ± j2
2
v(t) = Vs + (A1cos2t + A2sin2t)e-2t
8Vs = 24 means that Vs = 3
v(0) = 0 = 3 + A1 leads to A1 = -3
dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t
0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3
v(t) = [3 – 3(cos2t + sin2t)e-2t] volts
Chapter 8, Solution 28.
The characteristic equation is s2 + 6s + 8 with roots
− 6 ± 36 − 32
s1, 2 = = −4,−2
2
Hence,
i (t ) = I s + Ae −2t + Be −4t
8I s = 12
→ I s = 1.5
i (0) = 0
→ 0 = 1.5 + A + B (1)
di
= −2 Ae − 2t − 4 Be − 4t
dt
di(0)
= 2 = −2 A − 4 B
→ 0 = 1 + A + 2 B (2)
dt
Solving (1) and (2) leads to A=-2 and B=0.5.
i (t ) = 1.5 − 2e −2t + 0.5e −4t A
Chapter 8, Solution 29.
(a) s2 + 4 = 0 which leads to s1,2 = ±j2 (an undamped circuit)
v(t) = Vs + Acos2t + Bsin2t
4Vs = 12 or Vs = 3
v(0) = 0 = 3 + A or A = -3
dv/dt = -2Asin2t + 2Bcos2t
dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V
(b) s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4
i(t) = (Is + Ae-t + Be-4t)
4Is = 8 or Is = 2
i(0) = -1 = 2 + A + B, or A + B = -3 (1)
di/dt = -Ae-t - 4Be-4t
di(0)/dt = 0 = -A – 4B, or B = -A/4 (2)
From (1) and (2) we get A = -4 and B = 1
i(t) = (2 – 4e-t + e-4t) A
(c) s2 + 2s + 1 = 0, s1,2 = -1, -1
v(t) = [Vs + (A + Bt)e-t], Vs = 3.
v(0) = 5 = 3 + A or A = 2
dv/dt = [-(A + Bt)e-t] + [Be-t]
dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3
v(t) = [3 + (2 + 3t)e-t] V
Chapter 8, Solution 30.
s1 = −500 = −α + α 2 − ω o , s 2 = −800 = −α − α 2 − ω o
2 2
R
s1 + s 2 = −1300 = −2α
→ α = 650 =
2L
Hence,
R 200
L= = = 153.8 mH
2α 2 x650
1
s1 − s 2 = 300 = 2 α 2 − ω o ω o = 623.45 =
2
→
LC
1
C= = 16.25µF
(632.45) 2 L
Chapter 8, Solution 31.
For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent
circuit is shown in Figure (b). By KVL,
v(0+) = v(0-) = 40, i(0+) = i(0-) = 1
By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+)
= 0 which leads to vL(0+) = 40x1 + 40 = 80
vL(0+) = 80 V, vC(0+) = 40 V
40 Ω 10 Ω i1 40 Ω 10 Ω
+ + +
i + +
v 50V vL v 50V
− −
− −
0.5H −
(a) (b)
Chapter 8, Solution 32.
For t = 0-, the equivalent circuit is shown below.
2A
i
+ −
v
6Ω
i(0-) = 0, v(0-) = -2x6 = -12V
For t > 0, we have a series RLC circuit with a step input.
α = R/(2L) = 6/2 = 3, ωo = 1/ LC = 1 / 0.04
s = − 3 ± 9 − 25 = −3 ± j4
Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]
where Vf = final capacitor voltage = 50 V
v(t) = 50 + [(Acos4t + Bsin4t)e-3t]
v(0) = -12 = 50 + A which gives A = -62
i(0) = 0 = Cdv(0)/dt
dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5
v(t) = {50 + [(-62cos4t – 46.5sin4t)e-3t]} V
Chapter 8, Solution 33.
We may transform the current sources to voltage sources. For t = 0-, the equivalent
circuit is shown in Figure (a).
10 Ω i i 10 Ω
+ 1H
+
+ v 5Ω
+
30V − v 4F 30V −
−
−
(a) (b)
i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V
For t > 0, we have a series RLC circuit.
α = R/(2L) = 5/2 = 2.5
ω o = 1 / LC = 1 / 4 = 0.25, clearly α > ωo (overdamped response)
s1,2 = − α ± α 2 − ω 2 = −2.5 ± 6.25 − 0.25 = -4.95, -0.05
o
v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20.
v(0) = 10 = 20 + A1 + A2 (1)
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