Chapter 8 – The Complete Response of RL and RC Circuit
Exercises
Ex. 8.3-1
Before the switch closes:
After the switch closes:
2
= 8 Ω so τ = 8 ( 0.05 ) = 0.4 s .
0.25
Finally, v (t ) = voc + (v (0) − voc ) e− t τ = 2 + e−2.5 t V for t > 0
Therefore R t =
Ex. 8.3-2
Before the switch closes:
After the switch closes:
8-1
2
6
= 8 Ω so τ = = 0.75 s .
0.25
8
1
1
Finally, i (t ) = isc + (i (0) − isc ) e−t τ = + e−1.33 t A for t > 0
4 12
Therefore R t =
Ex. 8.3-3
At steady-state, immediately before t = 0:

12
 10  
i ( 0) = 
 
 = 0.1 A
 10 + 40   16+ 40||10 
After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be
so I sc = 0.3 A, Rt = 40 Ω, τ =
Finally:
L
20
1
s
=
=
40
2
Rt
i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A
8-2
Ex. 8.3-4
At steady-state, immediately before t = 0
After t = 0, we have:
so Voc = 12 V, Rt = 200 Ω, τ = Rt C = (200)(20 ⋅10 −6 ) = 4 ms
Finally:
v(t ) = (12 − 12) e −t 4 + 12 = 12 V
Ex. 8.3-5
Immediately before t = 0, i (0) = 0.
After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the
inductor current:
I sc = 0.2 A, Rt = 45 Ω, τ =
L 25 5
=
=
Rth 45 9
So i (t ) = 0.2 (1 − e−1.8t ) A
Now that we have the inductor current, we can calculate v(t):
d
i (t )
dt
= 8(1 − e −1.8t ) + 5(1.8)e −1.8t
v(t ) = 40 i (t ) + 25
= 8 + e −1.8t V for t > 0
8-3
Ex. 8.3-6
At steady-state, immediately before t = 0
so i(0) = 0.5 A.
After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get
I sc = 93.75 mA, Rt = 640 Ω,
τ =
L
.1
1
s
=
=
Rt 640 6400
i (t ) = 406.25 e −6400t + 93.75 mA
Finally:
v (t ) = 400 i (t ) + 0.1
Ex. 8.4-1
d
i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t )
dt
= 37.5 − 97.5e −6400t V
τ = ( 2×103 )(1×10−6 ) = 2 ×10−3 s
vc (t ) = 5 + (1.5−5 ) e −500 t
V
vc (0.001) = 5 − 3.5e − 0.5 = 2.88 V
So vc (t ) will be equal to vT at t = 1 ms if v = 2.88 V.
T
8-4
Ex. 8.4-2
iL (0) = 1 mA, I sc = 10 mA 

 ⇒ iL ( t ) = 10 − 9 e
L
Rt =500 Ω, τ =

500
vR (t ) = 300 iL ( t ) = 3 − 2.7 e
−
500
t
L
−500
t
L
mA
V
We require that vR ( t ) = 1.5 V at t = 10 ms = 0.01 s . That is
1.5 = 3 − 2.7e
Ex. 8.6-1
0 < t < t1
−
500
(0.01)
L
⇒ L=
5
= 8.5 H
0.588
v(t ) = v(∞)+Ae− t / RC where v(∞) = (1 A)(1 Ω ) = 1 V
v(t ) = 1 + A e− t /(1)(.1) = 1 + A e−10 t
Now v(0− ) = v(0+ ) = 0 = 1 + A ⇒ A = −1
∴ v(t ) =1−e−10t V
t > t1
t1 = 0.5 s,
v(t ) = v(t1 ) e
−
t −.5
(1)(.1)
= v(0.5) e−10(t −.5)
Now v(0.5) 1 − e −10(0.5) = 0.993 V
∴ v(t ) = 0.993 e−10(t −0.5) V
8-5
Ex. 8 6-2
t < 0 no sources ∴ v(0− ) = v(0+ ) = 0
0 < t <t1
− t
v (t ) = v (∞ ) + A e
− t / RC
= v (∞ ) + A e
2 × 1 0 5 (1 0 − 7 )
where for t = ∞ (steady-state)
∴capacitor becomes an open ⇒ v(∞) = 10 V
v(t ) = 10 + Ae −50t
Now v(0) = 0 = 10 + A ⇒ A = −10
∴ v(t ) = 10(1− e −50t ) V
t > t1 , t1 = .1 s
v(t ) = v(.1) e−50(t −.1)
where v(.1) = 10(1− e −50(.1) ) = 9.93 V
∴ v(t ) = 9.93e−50( t −.1) V
Ex. 8.6-3
For t < 0 i = 0.
For 0 < t < 0.2 s
KCL: −5 + v / 2 + i = 0 
 di
 + 10i = 50
di
also: v = 0.2
 dt
dt
∴ i (t ) = 5+ Ae−10t
i (0) = 0 = 5+ A ⇒ A = −5
so we have i (t ) = 5 (1−e−10t ) A
For t > 0.2 s
i (0.2) = 4.32 A ∴ i(t ) = 4.32e−10(t −.2) A
8-6
Ex. 8.7-1
vs ( t ) = 10sin 20t V
 d v( t ) 
KVL: −10sin 20t +10 .01
 + v( t ) = 0
dt 

d v( t )
⇒
+10 v( t ) = 100sin 20t
dt
Natural response: vn (t ) = Ae−t τ where τ = Rt C ∴ vn (t ) = Ae−10t
Forced response: try v f (t ) = B1 cos 20t + B2 sin 20 t
Plugging v f (t ) into the differential equation and
equating like terms yields: B1 = − 4 & B2 = 2
Complete response: v(t ) = vn (t ) + v f (t )
v(t ) = Ae −10t − 4 cos 20t + 2 sin 20t
Now v(0− ) = v(0+ ) = 0 = A − 4 ∴ A = 4
∴ v(t ) = 4 e −10t − 4 cos 20t + 2sin 20 t V
Ex. 8.7-2
is ( t ) = 10 e −5t
for t > 0
KCL at top node: −10e −5t + i ( t ) + v( t ) /10 = 0
Now v ( t ) = 0.1
Natural response: in (t ) = Ae −t τ
di ( t )
di ( t )
⇒
+100 i ( t ) = 1000 e −5t
dt
dt
where τ = L R t ∴ in (t ) = Ae −100t
Forced response: try i f (t ) = Be −5t & plug into the differential equation
−5 Be −5t + 100 Be −5t = 1000e −5t
Complete response: i (t ) = Ae
−
−100 t
+ 10.53e
⇒ B = 10.53
−5t
+
Now i (0 ) = i (0 ) = 0 = A + 10.53 ⇒ A = −10.53
∴ i (t ) = 10.53 (e −5t − e −100t ) A
Ex. 8.7-3
When the switch is closed, the inductor current is iL = vs / R = vs . When the switch opens, the
inductor current is forced to change instantaneously. The energy stored in the inductor
instantaneously dissipates in the spark. To prevent the spark, add a resistor (say 1 kΩ) across the
switch terminals.
8-7
Problems
Section 8.3: The Response of a First Order Circuit to a Constant Input
P8.3-1
Here is the circuit before t = 0, when the switch is
open and the circuit is at steady state. The open
switch is modeled as an open circuit.
A capacitor in a steady-state dc circuit acts like an
open circuit, so an open circuit replaces the
capacitor. The voltage across that open circuit is the
initial capacitor voltage, v (0).
By voltage division
v (0) =
6
(12 ) = 4 V
6+6+6
Next, consider the circuit after the switch closes. The
closed switch is modeled as a short circuit.
We need to find the Thevenin equivalent of the part
of the circuit connected to the capacitor. Here’s the
circuit used to calculate the open circuit voltage, Voc.
Voc =
6
(12 ) = 6 V
6+6
Here is the circuit that is used to determine Rt.
A short circuit has replaced the closed switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.
( 6 )( 6 ) = 3 Ω
Rt =
6+6
Then
τ = R t C = 3 ( 0.25 ) = 0.75 s
Finally,
v ( t ) = Voc + ( v ( 0 ) − Voc ) e−t / τ = 6 − 2 e−1.33 t V for t > 0
8-8
P8.3-2
Here is the circuit before t = 0, when the switch is
closed and the circuit is at steady state. The closed
switch is modeled as a short circuit.
An inductor in a steady-state dc circuit acts like an
short circuit, so a short circuit replaces the
inductor. The current in that short circuit is the
initial inductor current, i(0).
i ( 0) =
12
=2 A
6
Next, consider the circuit after the switch opens.
The open switch is modeled as an open circuit.
We need to find the Norton equivalent of the part
of the circuit connected to the inductor. Here’s the
circuit used to calculate the short circuit current,
Isc.
I sc =
12
=1 A
6+6
Here is the circuit that is used to determine Rt.
An open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by an short circuit.
( 6 + 6 )( 6 ) = 4 Ω
Rt =
( 6 + 6) + 6
Then
L 8
τ=
= =2 s
Rt 4
Finally,
i ( t ) = I sc + ( i ( 0 ) − I sc ) e −t / τ = 1 + e −0.5 t A for t > 0
8-9
P8.3-3
Before the switch closes:
After the switch closes:
−6
= 3 Ω so τ = 3 ( 0.05 ) = 0.15 s .
−2
Finally, v (t ) = voc + (v (0) − voc ) e− t τ = −6 + 18 e−6.67 t V for t > 0
Therefore R t =
P8.3-4
Before the switch closes:
After the switch closes:
8-10
−6
6
= 3 Ω so τ = = 2 s .
−2
3
t
−
10
Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 + e−0.5 t A for t > 0
3
Therefore R t =
P8.3-5
Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Thevenin equivalent circuit to get:
Therefore τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s .
Next, v C (t ) = voc + (v (0) − voc ) e
−
t
τ
= 10 − 5 e−12.5 t V for t > 0
Finally, v 0 (t ) = vC (t ) = 10 − 5 e −12.5 t V for t > 0
8-11
P8.3-6
Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Norton equivalent circuit to get:
Therefore τ =
5
= 0.25 ms .
20 × 103
Next, i L (t ) = isc + (i L (0) − isc ) e
Finally, vo ( t ) = 5
−
t
τ
= 0.5 − 0.25 e−4000 t mA for t > 0
d
i L ( t ) = 5 e −4000 t V
dt
for t > 0
P8.3-7
At t = 0− (steady-state)
Since the input to this circuit is constant, the
inductor will act like a short circuit when the
circuit is at steady-state:
for t > 0
iL ( t ) = iL ( 0 ) e − ( R L ) t = 6 e−20t A
8-12
P8.3-8
Before the switch opens, the circuit will be at steady state. Because the only input to this circuit
is the constant voltage of the voltage source, all of the element currents and voltages, including
the capacitor voltage, will have constant values. Opening the switch disturbs the circuit.
Eventually the disturbance dies out and the circuit is again at steady state. All the element
currents and voltages will again have constant values, but probably different constant values than
they had before the switch opened.
Here is the circuit before t = 0, when
the switch is closed and the circuit is at
steady state. The closed switch is modeled as
a short circuit. The combination of resistor
and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit
replaces the switch and the resistor R. A
capacitor in a steady-state dc circuit acts like
an open circuit, so an open circuit replaces
the capacitor. The voltage across that open
circuit is the capacitor voltage, vo(t).
Because the circuit is at steady state, the value of the capacitor voltage will be constant.
This constant is the value of the capacitor voltage just before the switch opens. In the absence of
unbounded currents, the voltage of a capacitor must be continuous. The value of the capacitor
voltage immediately after the switch opens is equal to the value immediately before the switch
opens. This value is called the initial condition of the capacitor and has been labeled as vo(0).
There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equal
to the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore
vo ( 0 ) = Vs
The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives
vo ( 0 ) = 2 + 8 e0 = 10 V
Consequently,
Vs = 10 V
8-13
Next, consider the circuit after the switch
opens. Eventually (certainly as t →∞) the
circuit will again be at steady state. Here is
the circuit at t = ∞, when the switch is open
and the circuit is at steady state. The open
switch is modeled as an open circuit. A
capacitor in a steady-state dc circuit acts like
an open circuit, so an open circuit replaces
the capacitor. The voltage across that open
circuit is the steady-state capacitor voltage,
vo(∞). There is no current in the horizontal
resistor and vo(∞) is equal to the voltage
across the vertical resistor. Using voltage
division,
10
vo ( ∞ ) =
(10 )
R + 10
The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives
vo ( ∞ ) = 2 + 8 e −∞ = 2 V
Consequently,
10
(10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω
R + 10
−t τ
Finally, the exponential part of vo(t) is known to be of the form e
where τ = R t C and
2=
Rt is the Thevenin resistance of the part of the circuit connected to the capacitor.
Here is the circuit that is used to determine Rt.
An open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.
R t = 10 +
so
( 40 )(10 ) = 18
40 + 10
Ω
τ = R t C = 18 C
From the equation for vo(t)
t
−0.5 t = −
⇒ τ =2s
τ
Consequently,
2 = 18 C ⇒ C = 0.111 = 111 mF
8-14
P8.3-9:
Before the switch closes, the circuit will be at steady state. Because the only input to this circuit
is the constant voltage of the voltage source, all of the element currents and voltages, including
the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting
out the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All
the element currents and voltages will again have constant values, but probably different constant
values than they had before the switch closed.
The inductor current is equal to the current in the 3 Ω resistor. Consequently,
− 0.35 t
vo (t ) 6 − 3 e
− 0.35 t
=
= 2− e
A when t > 0
i (t ) =
3
3
In the absence of unbounded voltages, the current in any inductor is continuous. Consequently,
the value of the inductor current immediately before t = 0 is equal to the value immediately after
t = 0.
Here is the circuit before t = 0, when the
switch is open and the circuit is at steady
state. The open switch is modeled as an open
circuit. An inductor in a steady-state dc
circuit acts like a short circuit, so a short
circuit replaces the inductor. The current in
that short circuit is the steady state inductor
current, i(0). Apply KVL to the loop to get
R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0
⇒ i (0) =
24
R1 + R 2 + 3
The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives
i ( 0 ) = 2 − e0 = 1 A
Consequently,
1=
24
⇒ R1 + R 2 = 21
R1 + R 2 + 3
Next, consider the circuit after the switch
closes. Here is the circuit at t = ∞, when the
switch is closed and the circuit is at steady
state. The closed switch is modeled as a
short circuit. The combination of resistor
and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit
replaces the switch and the resistor R1.
8-15
An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the
inductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to
the loop to get
24
R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) =
R2 + 3
The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives
i ( ∞ ) = 2 − e −∞ = 2 A
Consequently
2=
24
⇒ R2 = 9 Ω
R2 + 3
Then
R1 = 12 Ω
Finally, the exponential part of i(t) is known to be of the form e
−t τ
where τ =
L
and
Rt
Rt is the Thevenin resistance of the part of the circuit that is connected to the inductor.
Here is shows the circuit that is used to determine Rt.
A short circuit has replaced combination of resistor
R1 and the closed switch. Independent sources are set
to zero when calculating Rt, so the voltage source
has been replaced by an short circuit.
R t = R 2 + 3 = 9 + 3 = 12 Ω
so
τ=
L
L
=
R t 12
From the equation for i(t)
−0.35 t = −
t
τ
⇒ τ = 2.857 s
Consequently,
2.857 =
L
⇒ L = 34.28 H
12
8-16
P8.3-10
First, use source transformations to obtain the equivalent circuit
for t > 0:
for t < 0:
So iL ( 0 ) = 2 A, I sc
1
L
1
s
= 0, Rt = 3 + 9 = 12 Ω, τ =
= 2 =
24
Rt 12
and iL ( t ) = 2e−24t
t >0
Finally v ( t ) = 9 iL ( t ) = 18 e−24t
t >0
8-17
Section 8-4: Sequential Switching
P8.4-1
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to
get:
Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc =
5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore
v (t ) = voc + (v (0) − voc ) e−t τ = 5 + 5e−5 t V for 0 < t < 1.5 s
At t =1.5 s, v (1.5) = 5 + 5e (
τ = 8 × 0.05 = 0.4 s . Therefore
−0.05 1.5)
= 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω so
v (t ) = voc + (v (1.5) − voc ) e
−( t −1.5) τ
= 10 − 5 e
−2.5 ( t −1.5 )
V for 1.5 s < t
Finally
−5 t
for 0 < t < 1.5 s
 5 + 5 e V
v (t ) = 
−2.5 ( t −1.5)
for 1.5 s < t
V
10 − 5 e
P8.4-2
Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:
Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2
12
A and Rt = 6 Ω so τ = = 2 s . Therefore
6
i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s
8-18
At t =1.5 s, i (1.5) = 2 + e
−0.5 (1.5)
= 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ =
Therefore
i (t ) = isc + (i (1.5) − isc ) e
−( t −1.5 ) τ
= 3 − 0.53 e
−0.667 ( t −1.5 )
12
= 1.5 s .
8
V for 1.5 s < t
Finally

2 + e −0.5 t A
for 0 < t < 1.5 s
i (t ) = 
−0.667 ( t −1.5 )
for 1.5 s < t
A
3 − 0.53 e
P8.4-3
At t = 0−:
KVL : − 52 + 18 i + (12 8) i = 0
⇒
i (0− ) =104 39 A
 6 
+
∴ iL = i 
 = 2 A = iL (0 )
 6+ 2 
For 0 < t < 0.051 s
iL (t ) = iL (0) e −t τ
where τ = L R t
R t = 6 12 + 2 = 6 Ω
iL (t ) = 2 e −6t A
 6 +12 
−6 t
∴ i (t ) = iL (t ) 
 =6e A
 6 
For t > 0.051 s
i L ( t ) = i L (0.051) e − ( R
L ) ( t − 0.051)
i L (0.051) = 2 e − 6 (.051) = 1.473 A
i L ( t ) = 1.473 e − 14 ( t − 0.051) A
i ( t ) = i L ( t ) = 1.473 e − 14 ( t − 0.051) A
8-19
P8.4-4
At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 µF
capacitor is 3 V. The charge stored by the capacitor is
q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C
0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at
t = 0+). The voltage across the parallel capacitors is determined by considering charge
conservation:
q ( 0+ ) = (100 µ F) v ( 0+ ) + (400 µ F) v ( 0+ )
v (0
+
) = 100 ×10
( )
q ( 0+ )
−6
+ 400 ×10−6
=
q ( 0− )
500 × 10−6
=
300 × 10−6
500 × 10−6
v 0+ = 0.6 V
10 ms < t < l s: Combine 100 µF & 400 µF in parallel to obtain
v(t ) = v ( 0+ ) e − (t −.01) RC
3
= 0.6e − (t −.01) (10 ) (5 x10
v(t ) = 0.6 e−2(t −.01) V
−4 )
P8.4-5
For t < 0:
Find the Thevenin equivalent of the
circuit connected to the inductor (after
t >0). First, the open circuit voltage:
40
=1 A
20 + 20
voc = 20 i1 − 5 i1 = 15 V
i1 =
8-20
Next, the short circuit current:
20 i 1 = 5 i 1 ⇒ i 1 = 0
i sc + 0 =
40
⇒ i sc = 2 A
20
Then
voc 15
= = 7.5 Ω
i sc
2
Replace the circuit connected to the
inductor by its Norton equivalent
circuit. First
L 15 × 10−3
1
τ=
=
=
Rt
7.5
500
Next
i ( t ) = 2 − 2 e − 500 t A
t >0
Rt =
After t = 0, the steady state inductor
current is 2 A. 99% of 2 is 1.98.
1.98 = 2 − 2 e − 500 t
P8.4-6
⇒ t = 9.2 ms
v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s
∴ v ( t ) = 5 e −10 t V for t > 0
2.5 = 5 e −10 t1
i (t1 ) =
v (t1 )
100 ×10
3
t 1 = 0.0693 s
=
2.5
= 25 µ A
100 × 103
8-21
Section 8-5: Stability of First Order Circuits
P8.5-1
This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the
inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor
is calculated as
100

v T (400− R) 100
iT 
=
100+ 400
 ⇒ Rt =
iT
100+ 400

vT = 400 i (t ) − R i (t ) 
i (t ) =
The circuit is stable when R < 400 Ω.
P8.5-2
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
Ohm’s law: v ( t ) = 1000 iT ( t )
KVL: A v ( t ) + vT ( t ) − v ( t ) − 4000 iT ( t ) = 0
∴ vT ( t ) = (1 − A ) 1000 iT ( t ) + 4000 iT ( t )
vT ( t )
= ( 5 − A ) ×1000
iT ( t )
The circuit is stable when A < 5 V/V.
Rt =
P8.5-3
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
v (t )
Ohm’s law: i ( t ) = − T
6000
v (t )
KCL: i ( t ) + B i ( t ) + i T ( t ) = T
3000
 v ( t )  vT ( t )
∴ i T ( t ) = − ( B + 1)  − T
+
 6000  3000
( B + 3) vT ( t )
=
6000
vT ( t ) 6000
Rt =
=
iT ( t ) B + 3
The circuit is stable when B > −3 A/A.
8-22
P8.5-4
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
(1000 )( 4000 ) i (t ) = 800 i (t )
v(t ) =
T
T
1000 + 4000
vT (t ) = v(t ) − Av(t ) = (1 − A ) v(t )
vT (t )
= 800 (1 − A)
iT (t )
The circuit is stable when A < 1 V/V.
Rt =
8-23
Section 8-6: The Unit Step Response
P8.6-1
The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the
capacitor is replaced by an open circuit.
The value of the capacitor voltage at time t = 0,
will be equal to the steady state capacitor voltage before
the input changes. At time t = 0 the output voltage is
−a 0
vo ( 0 ) = A + B e ( ) = A + B
The steady-state circuit for t < 0.
Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives
A+ B = 8 V
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the
capacitor is replaced by an open circuit.
The value of the capacitor voltage at time t = ∞, will be
equal to the steady state capacitor voltage after the input
changes. At time t = ∞ the output voltage is
vo ( ∞ ) = A + B e
The steady-state circuit for t > 0.
− a (∞)
=A
Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives
A = -7 V
Therefore
B = 15 V
8-24
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.
1
=τ = Rt C
a
Here is the circuit used to calculate Rt.
Rt = 6 Ω
Therefore
a=
1
1
= 2.5
−3
s
( 6 ) ( 66.7 ×10 )
(The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .)
Putting it all together:
8 V
for t ≤ 0

vo ( t ) = 
− 2.5 t
V for t ≥ 0
−7 + 15 e
P8.6-2
The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Capacitors act like open circuits when the input is
constant and the circuit is at steady state. Consequently,
the capacitor is replaced by an open circuit.
The steady-state circuit for t < 0.
The value of the capacitor voltage at time t = 0,
will be equal to the steady state capacitor voltage
before the input changes. At time t = 0 the output
voltage is
−a 0
vo ( 0 ) = A + B e ( ) = A + B
Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives
8-25
A+ B =
6
( 3) = 2 V
3+ 6
Capacitors act like open circuits when the input is
constant and the circuit is at steady state. Consequently,
the capacitor is replaced by an open circuit.
The value of the capacitor voltage at time t = ∞, will be
equal to the steady state capacitor voltage after the
input changes. At time t = ∞ the output voltage is
vo ( ∞ ) = A + B e
The steady-state circuit for t > 0.
− a (∞)
=A
Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives
A=
6
(6) = 4 V
3+ 6
Therefore
B = −2 V
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.
1
=τ = Rt C
a
Here is the circuit used to calculate Rt.
Rt =
( 3)( 6 ) = 2
3+ 6
Ω
Therefore
a=
1
( 2 )(.5 )
=1
1
s
(The time constant is τ = ( 2 )( 0.5 ) = 1 s .)
Putting it all together:
2 V
for t ≤ 0

vo ( t ) = 
− t
 4 − 2 e V for t ≥ 0
8-26
P8.6-3
The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Inductors act like short circuits when the input is
constant and the circuit is at steady state.
Consequently, the inductor is replaced by a short
circuit.
The value of the inductor current at time t = 0,
will be equal to the steady state inductor current before
the input changes. At time t = 0 the output current is
The steady-state circuit for t < 0.
io ( 0 ) = A + B e
− a ( 0)
= A+ B
Consequently, the inductor current is labeled as A + B.
Analysis of the circuit gives
A+ B =
−7
= −1.4 A
5
Inductors act like short circuits when the input is
constant and the circuit is at steady state.
Consequently, the inductor is replaced by a short
circuit.
The value of the inductor current at time t = ∞, will be
equal to the steady state inductor current after the
input changes. At time t = ∞ the output current is
The steady-state circuit for t > 0.
io ( ∞ ) = A + B e
− a (∞)
=A
Consequently, the inductor current is labeled as A.
Analysis of the circuit gives
A=
6
= 1.2 A
5
Therefore
B = −2.6 V
8-27
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit
connected to the inductor.
1
L
=τ =
a
Rt
Here is the circuit used to calculate Rt.
Rt =
( 5 ) ( 4 ) = 2.22
5+ 4
Ω
Therefore
a=
2.22
1
= 1.85
1.2
s
1.2
= 0.54 s .)
2.22
(The time constant is τ =
Putting it all together:
−1.4 A
for t ≤ 0

io ( t ) = 
− 1.85 t
A for t ≥ 0
1.2 − 2.6 e
P8.6-4
v (t ) = 4u (t ) − u (t − 1) − u (t − 2) + u (t − 4) − u (t − 6)
P8.6-5
Assume that the circuit is at steady
state at t = 1−. Then
0

vs ( t ) = 4
0

t <1
1<t < 2
t >2
v ( t ) = 4 − 4 e − (t −1) V for 1 ≤ t ≤ 2
τ = R C = ( 5 ×105 )( 2 × 10−6 ) = 1 s
so v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V
and
v ( t ) = 2.53 e − (t − 2) V for t ≥ 2
∴
0

v(t ) = 4− 4e − ( t −1)
2.53e − (t − 2)

t ≤1
1≤t ≤ 2
t ≥2
8-28
P8.6-6
The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0.
For 0 < t < 0.5 s the switch is open:
v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 ×
1 1
=
s
6 2
so v ( t ) = 10 e − 2 t V
In particular, v ( 0.5 ) = 10 e
For t > 0.5 s the switch is closed:
− 2 ( 0.5)
= 3.679 V
v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω,
1
6
τ = 2× =
so
1
s
3
v ( t ) = 10 + ( 3.679 − 10 ) e
= 10 − 6.321 e
− 3 ( t − 0.5)
− 3 ( t − 0.5 )
V
V
P8.6-7
Assume that the circuit is at steady state before t = 0. Then the initial inductor current is
i(0−) = 0 A.
For 0 < t < 1 ms:
The steady state inductor current will be
30
i ( ∞ ) = lim i ( t ) =
( 40 ) = 24 A
t →∞
30 + 20
The time constant will be
50 × 10−3
1
τ=
= 10−3 =
s
30 + 20
1000
The inductor current is i ( t ) = 24 (1 − e −1000 t ) A
In particular, i ( 0.001) = 24 (1 − e −1 ) = 15.2 A
For t > 1 ms
Now the initial current is i(0.001) = 15.2 A and
the steady state current is 0 A. As before, the time
constant is 1 ms. The inductor current is
i ( t ) = 15.2 e
−1000 ( t − 0.001)
A
8-29
The output voltage is
480 (1 − e −1000 t ) V t < 1 ms
v ( t ) = 20 i ( t ) = 
−1000 ( t − 0.001)
V t > 1 ms
303 e
P8.6-8
For t < 0, the circuit is:
After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent
circuit to get:
vc ( t ) = −15 + ( −6 − ( −15 ) ) e
− t / ( 4000×0.00005 )
= −15 + 9 e − 5 t V
8-30
Section 8-7 The Response of an RL or RC Circuit to a Nonconstant Source
P8.7-1
Assume that the circuit is at steady state before t = 0:
KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 A
Then vc (0− ) = −12 ix = 22 V = vc (0+ )
After t = 0:
KVL : 12i (t ) − 8e −5t + v ( t ) = 0
x
c
dv ( t )
1 dvc ( t )
KCL : −ix ( t )−2ix ( t ) + (1 36) c
= 0 ⇒ ix ( t ) =
dt
108 dt
 1 dvc ( t ) 
−5t
∴ 12 
+ v (t ) = 0
 − 8e
c
108 dt 
dv (t )
c + 9v (t ) = 72e−5t ⇒ v (t ) = Ae −9t
c
cn
dt
Try v (t ) = Be −5t & substitute into the differential equation ⇒ B = 18
cf
∴ v (t ) = Ae −9t + 18 e−5t
c
v (0) = 22 = A + 18 ⇒ A = 4
c
∴ v (t ) = 4e−9t + 18e−5t V
c
8-31
P8.7-2
Assume that the circuit is at steady state before t = 0:
iL (0+ ) = iL (0− ) =
12
= 3A
4
After t = 0:
v( t ) −12
v( t )
+ iL ( t ) +
= 6 e −2t
4
2
di ( t )
also : v ( t ) = (2 / 5) L
dt
di ( t ) 
3
iL ( t ) + (2 / 5) L  = 3 + 6 e −2t
4
dt 
KCL :
diL ( t ) 10
+ iL ( t ) = 10 + 20 e −2t
3
dt
∴ in (t ) = Ae
− (10 / 3)t
, try i f (t ) = B + Ce−2t , substitute into the differential equation,
and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t
∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15
∴ iL (t ) = −15e − (10 / 3) t + 3 + 15e−2t
Finally, v(t ) =( 2 / 5 )
diL
= 20 e − (10 / 3) t −12 e −2t V
dt
P8.7-3
Assume that the circuit is at steady state before t = 0:
 6 
Current division: iL (0− ) = −5 
 = −1 mA
 6 + 24 
8-32
After t = 0:
KVL: − 25sin 4000 t + 24iL ( t ) + .008
di L ( t )
25
+3000i L ( t ) =
sin4000t
dt
.008
diL ( t )
=0
dt
in (t ) = Ae −3000t , try i f (t ) = B cos 4000t + C sin 4000t , substitute into the differential equation
and equate like terms ⇒ B = −1/ 2, C = 3 / 8 ⇒ i f (t ) = −0.5cos 4000 t + 0.375 sin 4000 t
iL (t ) = in (t ) + i f (t ) = Ae −3000t − 0.5cos 4000 t + 0.375 sin 4000 t
iL (0+ ) = iL (0− ) =−10−3 = A− 0.5 ⇒ A ≅ 0.5
∴ iL (t ) = 0.5 e −3000t − 0.5cos 4000 t + 0.375 sin 4000 t mA
but v(t ) = 24iL (t ) = 12 e −3000t − 12 cos 4000t + 9sin 4000t V
P8.7-4
Assume that the circuit is at steady state before t = 0:
Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get:
dvc ( t ) 

KVL: − 10 cos 2t + 15  1
+v t =0 ⇒
30 dt  c ( )


dvc ( t )
+ 2vc ( t ) = 20 cos 2t
dt
vn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to get
B = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2t
Now vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e−2t + 5cos 2t + 5sin 2t V
8-33
P8.7-5
Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) =
0 A. After t = 0, we have:
di ( t )
KVL : − 10sin100t + i ( t ) + 5
+ v (t ) = 0
dt
v( t )
Ohm's law : i ( t ) =
8
dv( t )
∴
+18 v( t ) = 160sin100t
dt
∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equation
and equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t
∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 t
v(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55
so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t V
P8.7-6
Assume that the circuit is at steady state
before t = 0.
vo ( t ) = −vc ( t )
vC (0+ ) = vC (0− ) = −10 V
After t = 0, we have
v (t )
8 e−5 t
i (t ) = s
=
= 0.533 e−5 t mA
15000
15000
The circuit is represented by the differential
dv ( t ) vC ( t )
equation: i ( t ) = C C
+
. Then
dt
R
( 0.533 ×10 ) e
−3
−5 t
= ( 0.25 ×10−6 )
dvc ( t )
+ (10−3 ) vc ( t ) ⇒
dt
dvc ( t )
+ 4000 vc ( t ) = 4000 e−5t
dt
Then vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get
(
d B e−5t
dt
) + 4000 ( B e ) = 4000 e
−5t
−5t
⇒ B=
4000
= −1.00125 ≅ −1
−3995
8-34
vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae −4000t
vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e−2t − 11 e−4000t V
Finally
vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0
P8.7-7
From the graph iL (t ) = 1 t mA . Use KVL to get
4
(1) iL (t ) + 0.4
diL (t )
= v1 (t ) ⇒
dt
diL (t )
+ 2.5 iL (t ) = 2.5 v1 (t )
dt
Then
di  1 
t + 2.5  1 t  = 2.5 v1 (t ) ⇒
4 
dt  4 
v1 = 0.1+ 0.25t V
P8.7-8
Assume that the circuit is at steady state before
t = 0.
v (0+ ) = v (0− ) =
2
30 = 10 V
4+ 2
After t = 0 we have
KVL :
 1 d v( t ) 
5 d v( t )
+ v (t ) + 4 
−i  = 30
2 dt
 2 dt


1 d v( t ) 
−3t
2 i ( t ) + 4 i ( t ) −
 + 30 = e
2
dt


The circuit is represented by the differential equation
d v( t )
6
6
2
(10 + e −3t )
v (t ) =
+
19
19
3
dt
Take vn ( t ) = Ae
−( 6 /19 ) t
. Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get
8-35
−3Ce −3t +
6
60 4 −3t
( B + Ce −3t ) =
+ e
19
19 19
Equate coefficients to get
B = 10 , C = −
4
4
⇒ v f ( t ) = e −3t + Ae− (6 /19) t
51
51
Then
v ( t ) = vn ( t ) + v f ( t ) = 10 −
4 −3t
e + Ae− (6 /19) t
51
Finally
vc (0+ ) = 10 V, ⇒ 10 = 10 −
∴ vc (t ) = 10 +
4
+A ⇒
51
A=
4
51
4 − (6 /19) t −3t
(e
−e ) V
51
P8.7-9
We are given v(0) = 0. From part b of the
figure:
5t 0 ≤ t ≤ 2 s
vs ( t ) = 
t > 2s
10
Find the Thevenin equivalent of the part of the
circuit that is connected to the capacitor:
The open circuit voltage:
The short circuit current:
(ix=0 because of the short across the right 2 Ω
resistor)
Replace the part of the circuit connected to the
capacitor by its Thevenin equivalent:
KVL:
dv( t )
+ v ( t ) − vs ( t ) = 0
dt
dv( t ) v ( t ) vs ( t )
+
=
dt
2
2
2
vn ( t ) = Ae−0.5 t
8-36
For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation and
equating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0,
we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) .
At t = 2 s, v( 2 ) = 10e−1 = 3.68 .
Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation and
equating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae
determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e
− (t −2) / 2
− (t − 2) / 2
. Using v ( 2 ) = 3.68 , we
.
P8.7-10
 d v (t ) 
KVL: − kt + Rs C C  + vC ( t ) = 0
dt 

d vC ( t )
k
1
vC ( t ) =
t
⇒
+
Rs C
Rs C
dt
vc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t
& plug into D.E. ⇒ B1 +
1
k
t thus B0 = − kRs C , B1 = k .
[ B0 + B1t ] =
Rs C
Rs C
Now we have vc (t ) = Ae − t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C.
∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF get
vc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )]
v(t) and vC(t) track well on a millisecond time scale.
8-37
Spice Problems
SP 8-1
8-38
SP 8-2
8-39
SP 8-3
v(t ) = A + B e −t / τ
for t > 0
⇒ 7.2 = A + B 
 ⇒ B = −0.8 V
8.0 = v(∞) = A + B e −∞ ⇒ A = 8.0 V 
0.05
 8 − 7.7728 
7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⇒ −
= ln 
 = −1.25878
0.8
τ


0.05
⇒ τ=
= 39.72 ms
1.25878
7.2 = v(0) = A + B e 0
Therefore
v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0
8-40
SP 8-4
i (t ) = A + B e −t / τ
0 = i (0) = A + B e
0
⇒ 0 = A+ B
4 × 10−3 = i (∞) = A + B e −∞
⇒
A = 4 × 10−3
2.4514 ×10 = v(5 ×10 ) = ( 4 ×10
−3
⇒ −
Therefore
for t > 0
−6
5 × 10−6
τ
−3

−3
 ⇒ B = −4 × 10 A
A 
) − ( 4 ×10 ) e
−3
(
)
− 5×10−6 / τ
 ( 4 − 2.4514 ) × 10−3 
= ln 
 = −0.94894
4 × 10−3


5 × 10−6
⇒ τ=
= 5.269 µ s
0.94894
i (t ) = 4 − 4 e −t / 5.269×10
−6
mA for t > 0
8-41
Verification Problems
VP 8-1
First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitor
voltage is vc = 4 V.
We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot.
Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
( 2000 )( 4000 ) = 4 kΩ so the time constant is
the circuit connected to the capacitor is R t =
2000 + 4000
3
4
2


τ = R t C =  × 103  ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is
3
3

vc (t ) = 4 e− t
0.67
+4 V
where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then
vc (t1 ) =
 1.33 
−

4 e  .67 
+ 4 = 4.541 ~ 4.5398 V
So the plot is correct.
VP 8-2
The initial and steady-state inductor currents shown on the plot agree with the values obtained
from the circuit.
Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
( 2000 )( 4000 ) = 4 kΩ so the time constant is
the circuit connected to the inductor is R t =
2000 + 4000
3
L
5
15
ms . Thus inductor current is
τ=
=
=
R t 4 ×103
4
3
iL (t ) − 2 e− t 3.75 + 5 mA
where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then
iL (t1 ) =
 3.75 
−

−2 e  3.75 
+ 5 = 4.264 mA ≠ 4.7294 mA
so the plot does not correspond to this circuit.
8-42
VP 8-3
Notice that the steady-state inductor current does not depend on the inductance, L. The initial
and steady-state inductor currents shown on the plot agree with the values obtained from the
circuit.
After t = 0
So I sc = 5 mA and τ =
The inductor current is given by iL (t ) = −2e −1333t
has units of Henries. Let t 1 = 3.75 ms, then
L
L
1333
+ 5 mA , where t has units of seconds and L
4.836 = iL (t1 ) = −2 e −(1333)⋅(0.00375) L + 5 = −2e−5 L + 5
so
4.836−5
= e −5 L
−2
and
L=
−5
=2 H
 4.836−5 
ln 

 −2 
is the required inductance.
VP 8-4
First consider the circuit. When t < 0 and the circuit is at steady-state:
For t > 0
So
Voc =
R2
R1 R2
RRC
( A + B) , Rt =
and τ = 1 2
R1 + R2
R1 + R2
R1 + R2
8-43
Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor
voltage is (Voc =) 4 V, so
vC (t ) = − 6e −t τ + 4
At t 1 = 1.333 ms
3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4
so
τ =
−0.001333
= 0.67 ms
 −4+ 3.1874 
ln 

−6


Combining the information obtained from the circuit with the information obtained from the plot
gives
R2
R2
R1 R2C
A = −2,
( A + B ) = 4,
= 0.67 ms
R1 + R2
R1 + R2
R1 + R2
There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is
one convenient way. Pick R = 3000 and R = 6000. Then
1
1
2
2
2A
= −2 ⇒ A = −3
3
2( A+ B)
= 4 ⇒ B −3 = 6 ⇒ B = 9
3
2
1
µF = C
2000 ⋅ C = ms ⇒
3
3
Design Problems
DP 8-1
R3
6
=
12 ⇒ R1 + R 2 = R 3 .
Steady-state response when the switch is open:
R1 + R 2 + R 3
Steady-state response when the switch is open: 10 =
10 ms = 5 τ = ( R 1 || R 3 ) C =
R3
6
R3
R1 + R 3
12
⇒ R1 =
R3
5
.
C
Let C = 1 µF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ.
8-44
DP 8-2
12
steady state response when the switch is open : 0.001 = R + R
1
2
steady state response when the switch is open: 0.004 =
12
R1
⇒ R + R = 12 kΩ .
1
2
⇒ R1 = 3 kΩ .
Therefore, R 2 = 9 kΩ.
 L 
L
10 ms = 5 τ = 5 
⇒ L = 240 H
=
 R1 + R 2  2400


DP 8-3
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
10−6
(a) ∆t = 5 Rt C ⇒ C =
= 4 pF
5 50×103
(b) ∆ t = 5 ( 50×103 )( 2×10−6
(
)
) = 0.5 s
DP 8-4
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
∆t
−∆t τ
When the switch is open: 5 e
= (1 − k ) 5 ⇒ ln (1− k ) = −
⇒ ∆ t = −τ ln (1− k )
When the switch is open: 5 − 5 e
(a) C =
−∆t τ
τ
= k 5 ⇒ ∆ t = −τ ln (1− k )
10−6
= 6.67 pF
− ln (1−.95 ) ( 50×103 )
(b) ∆ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 s
DP 8-5
i (0) =
R1
20
×
40 R1 R1 + 40
40 +
40 + R1
8-45
For
t > 0:
i (t ) = i (0) e
−t
τ
where τ =
L
10−2
=
R t 40+ R 2
At t < 200 µ s we need i ( t ) > 60 mA and i ( t ) <180 mA
First let's find a value of R 2 to cause i (0) < 180 mA.
1
t
A = 166.7 mA so i (t ) = 0.1667 e− τ .
6
Next, we find a value of R 2 to cause i (0.0002) > 60 mA.
Try R 2 = 40 Ω . Then i (0) =
10−2
1
s.
= 0.2 ms =
50
5000
i (0.0002) = 166.7×10−3 e −5000×0.0002 = 166.7×10−3 e −1 = 61.3 mA
Try R 2 = 10Ω, then τ =
DP 8-6
The current waveform will look like this:
We only need to consider the rise time:
−t
Vs
A
τ
−t
τ
iL (t ) =
(1 − e ) =
(1 − e )
R+2
R+2
where
L 0.2 1
s
τ =
=
=
3 15
Rt
A
∴ iL (t ) = (1 − e −15t )
3
Now find A so that iL2 R fuse ≥ 10 W during 0.25 ≤ t ≤ 0.75 s
∴ we want [iL2 (0.25)]R fuse = 10 W ⇒
A2
(1−e −15(.25) ) 2 (1) =10 ⇒ A = 9.715 V
9
8-46