CHAPTER 4
Exercises
E4.1
The voltage across the circuit is given by Equation 4.8:
v C (t ) = Vi exp( −t / RC )
in which Vi is the initial voltage. At the time t1% for which the voltage
reaches 1% of the initial value, we have
0.01 = exp( −t1% / RC )
Taking the natural logarithm of both sides of the equation, we obtain
ln(0.01) = −4.605 = −t1% / RC
Solving and substituting values, we find t1% = 4.605RC = 23.03 ms.
E4.2
The exponential transient shown in Figure 4.4 is given by
v C (t ) = Vs −Vs exp( −t / τ )
Taking the derivative with respect to time, we have
dv C (t ) Vs
=
exp( −t / τ )
dt
τ
Evaluating at t = 0, we find that the initial slope is VS / τ . Because this
matches the slope of the straight line shown in Figure 4.4, we have shown
that a line tangent to the exponential transient at the origin reaches the
final value in one time constant.
E4.3
(a) In dc steady state, the capacitances act as open circuits and the
inductances act as short circuits. Thus the steady-state (i.e., t
approaching infinity) equivalent circuit is:
From this circuit, we see that ia = 2 A. Then ohm’s law gives the voltage
as v a = Ria = 50 V.
101
(b) The dc steady-state equivalent circuit is:
Here the two 10-Ω resistances are in parallel with an equivalent
resistance of 1/(1/10 + 1/10) = 5 Ω. This equivalent resistance is in
series with the 5-Ω resistance. Thus the equivalent resistance seen by
the source is 10 Ω, and i1 = 20 / 10 = 2 A. Using the current division
principle, this current splits equally between the two 10-Ω resistances,
so we have i2 = i3 = 1 A.
E4.4
(a) τ = L / R2 = 0.1 / 100 = 1 ms
(b) Just before the switch opens, the circuit is in dc steady state with
an inductor current of Vs / R1 = 1.5 A. This current continues to flow in
the inductor immediately after the switch opens so we have i (0 +) = 1.5 A .
This current must flow (upward) through R2 so the initial value of the
voltage is v (0 +) = −R2i (0 +) = −150 V.
(c) We see that the initial magnitude of v(t) is ten times larger than the
source voltage.
(d) The voltage is given by
v (t ) = −
Vs L
exp( −t / τ ) = −150 exp( −1000t )
R1τ
Let us denote the time at which the voltage reaches half of its initial
magnitude as tH. Then we have
0.5 = exp( −1000tH )
Solving and substituting values we obtain
tH = −10 −3 ln(0.5) = 10 −3 ln(2) = 0.6931 ms
102
E4.5
First we write a KCL equation for t ≥ 0.
t
v (t ) 1
+ ∫ v (x )dx + 0 = 2
R
L
0
Taking the derivative of each term of this equation with respect to time
and multiplying each term by R, we obtain:
dv (t ) R
+ v (t ) = 0
dt
L
The solution to this equation is of the form:
v (t ) = K exp( −t / τ )
in which τ = L / R = 0.2 s is the time constant and K is a constant that
must be chosen to fit the initial conditions in the circuit. Since the initial
(t = 0+) inductor current is specified to be zero, the initial current in the
resistor must be 2 A and the initial voltage is 20 V:
v (0+) = 20 = K
Thus, we have
v (t ) = 20 exp( −t / τ )
1
iR = v / R = 2 exp( −t / τ )
t
t
1
iL (t ) = ∫ v (x )dx = [− 20 τ exp( −x / τ )] = 2 − 2 exp( −t / τ )
L0
2
0
E4.6
Prior to t = 0, the circuit is in DC steady state and the equivalent circuit
is
Thus we have i(0-) = 1 A. However the current through the inductor
cannot change instantaneously so we also have i(0+) = 1 A. With the
switch open, we can write the KVL equation:
di (t )
+ 200i (t ) = 100
dt
The solution to this equation is of the form
i (t ) = K 1 + K 2 exp( −t / τ )
in which the time constant is τ = 1 / 200 = 5 ms. In steady state with the
switch open, we have i (∞) = K 1 = 100 / 200 = 0.5 A. Then using the initial
103
current, we have i (0+) = 1 = K 1 + K 2 , from which we determine that
K 2 = 0.5. Thus we have
i (t ) = 1.0 A for t < 0
= 0.5 + 0.5 exp( −t / τ ) for t > 0.
v (t ) = L
di (t )
dt
= 0 V for t < 0
= −100 exp( −t / τ ) for t > 0.
E4.7
As in Example 4.4, the KVL equation is
t
1
Ri (t ) + ∫ i (x )dx + v C (0+) − 2 cos(200t ) = 0
C
0
Taking the derivative and multiplying by C, we obtain
di (t )
RC
+ i (t ) + 400C sin(200t ) = 0
dt
Substituting values and rearranging the equation becomes
di (t )
5 × 10 −3
+ i (t ) = −400 × 10 −6 sin(200t )
dt
The particular solution is of the form
i p (t ) = A cos(200t ) + B sin(200t )
Substituting this into the differential equation and rearranging terms
results in
5 × 10 −3 [− 200A sin(200t ) + 200B cos(200t )] + A cos(200t ) + B sin(200t )
= −400 × 10 −6 sin(200t )
Equating the coefficients of the cos and sin terms gives the following
equations:
− A + B = −400 × 10 −6 and B + A = 0
from which we determine that A = 200 × 10 −6 and B = −200 × 10 −6 .
Furthermore, the complementary solution is iC (t ) = K exp( −t / τ ) , and the
complete solution is of the form
i (t ) = 200 cos(200t ) − 200 sin(200t ) + K exp( −t / τ ) µA
At t = 0+, the equivalent circuit is
104
from which we determine that i (0 +) = 2 / 5000 = 400 µA. Then
evaluating our solution at t = 0+, we have i (0 +) = 400 = 200 + K , from
which we determine that K = 200 µA. Thus the complete solution is
i (t ) = 200 cos(200t ) − 200 sin(200t ) + 200 exp(−t / τ ) µA
E4.8
The KVL equation is
t
1
Ri (t ) + ∫ i (x )dx + v C (0+) − 10 exp(−t ) = 0
C
0
Taking the derivative and multiplying by C, we obtain
di (t )
RC
+ i (t ) + 10C exp( −t ) = 0
dt
Substituting values and rearranging, the equation becomes
di (t )
2
+ i (t ) = −20 × 10 −6 exp( −t )
dt
The particular solution is of the form
i p (t ) = A exp( −t )
Substituting this into the differential equation and rearranging terms
results in
− 2A exp(−t ) + A exp( −t ) = −20 × 10 −6 exp( −t )
Equating the coefficients gives A = 20 × 10 −6. Furthermore, the
complementary solution is iC (t ) = K exp( −t / 2) , and the complete solution
is of the form
i (t ) = 20 exp(−t ) + K exp( −t / 2) µA
At t = 0+, the equivalent circuit is
105
from which we determine that i (0+) = 5 / 10 6 = 5 µA. Then evaluating our
solution at t = 0+, we have i (0 +) = 5 = 20 + K , from which we determine
that K = −15 µA. Thus the complete solution is
i (t ) = 20 exp( −t ) − 15 exp( −t / 2) µA
E4.9
(a) ω 0 =
1
LC
=
1
10
−3
× 10
−7
= 10 5
α=
1
2RC
= 2 × 10 5
ζ =
α
=2
ω0
(b) At t = 0+, the KCL equation for the circuit is
v ( 0 +)
0. 1 =
+ iL (0 +) + Cv ′(0 +)
(1)
R
However, v (0 +) = v (0 −) = 0 , because the voltage across the capacitor
cannot change instantaneously. Furthermore, iL (0+) = iL (0−) = 0 , because
the current through the inductance cannot change value instantaneously.
Solving Equation (1) for v ′(0 +) and substituting values, we find that
v ′(0+) = 10 6 V/s.
(c) To find the particular solution or forced response, we can solve the
circuit in steady-state conditions. For a dc source, we treat the
capacitance as an open and the inductance as a short. Because the
inductance acts as a short v p (t ) = 0.
(d) Because the circuit is overdamped (ζ > 1), the homogeneous solution
is the sum of two exponentials. The roots of the characteristic solution
are given by Equations 4.72 and 4.73:
s 1 = −α − α 2 − ω 02 = −373.2 × 10 3
s 2 = −α + α 2 − ω 02 = −26.79 × 103
Adding the particular solution to the homogeneous solution gives the
general solution:
106
v (t ) = K 1 exp(s1t ) + K 2 exp(s 2t )
Now using the initial conditions, we have
v ( 0 +) = 0 = K 1 + K 2
v ′(0+) = 10 6 = K 1s1 + K 2s 2
Solving we find K 1 = −2.887 and K 2 = 2.887. Thus the solution is:
v (t ) = 2.887[exp(s 2t ) − exp(s1t )]
E4.10
(a) ω 0 =
1
LC
=
1
10 −3 × 10 −7
= 10 5
α=
1
2RC
= 10 5
ζ =
α
=1
ω0
(b) The solution for this part is the same as that for Exercise 4.9b in
which we found thatv ′(0+) = 10 6 V/s.
(c) The solution for this part is the same as that for Exercise 4.9c in
which we found v p (t ) = 0.
(d) The roots of the characteristic solution are given by Equations 4.72
and 4.73:
s1 = −α − α 2 − ω 02 = −10 5
s 2 = −α + α 2 − ω 02 = −10 5
Because the circuit is critically damped (ζ = 1), the roots are equal and
the homogeneous solution is of the form:
v (t ) = K 1 exp(s1t ) + K 2t exp(s1t )
Adding the particular solution to the homogeneous solution gives the
general solution:
v (t ) = K 1 exp(s1t ) + K 2t exp(s1t )
Now using the initial conditions we have
v ( 0 +) = 0 = K 1
v ′(0+) = 10 6 = K 1s1 + K 2
Solving we find K 2 = 10 6 Thus the solution is:
v (t ) = 10 6t exp( −10 5t )
E4.11
(a) ω 0 =
1
LC
=
1
10 −3 × 10 −7
= 10 5
α=
1
2RC
= 2 × 10 4
ζ =
α
= 0.2
ω0
(b) The solution for this part is the same as that for Exercise 4.9b in
which we found that v ′(0+) = 10 6 V/s.
107
(c) The solution for this part is the same as that for Exercise 4.9c in
which we found v p (t ) = 0.
(d) Because we have (ζ < 1), this is the underdamped case and we have
ω n = ω 02 − α 2 = 97.98 × 10 3
Adding the particular solution to the homogeneous solution gives the
general solution:
v (t ) = K 1 exp( −αt ) cos(ω nt ) + K 2 exp( −αt ) sin(ω nt )
Now using the initial conditions we have
v ( 0 +) = 0 = K 1
v ′(0+) = 10 6 = −αK 1 + ω n K 2
Solving we find K 2 = 10.21 Thus the solution is:
v (t ) = 10.21 exp( −2 × 10 4t ) sin(97.98 × 103t ) V
Problems
P4.1
The time constant τ is the interval required for the voltage to fall to
exp(-1) ≅ 0.368 times its initial value. The time constant is given by
τ = RC . Thus to attain a long time constant, we need large values for
both R and C.
P4.2
We have v (t ) = Vi exp( −t / τ ) and w (t ) =
1
2
voltage. The initial stored energy is Wi =
Cv 2 (t ) in which Vi is the initial
1
2
CVi 2 . After two time
constants, we have v (t ) = Vi exp( −2τ / τ ) ≅ 0.1353Vi so that 13.53 percent
of the initial voltage remains.
The energy is w (2τ ) = 1 2 Cv 2 (2τ ) =
1
2
CVi 2 exp( −4) = Wi exp( −4) ≅ 0.0183Wi
so only 1.83% of the initial energy remains.
P4.3
The solution is of the form given in Equation 4.19:
v C (t ) = Vs −Vs exp(− t RC )
RC = 10 5 × 0.01 × 10 −6 = 1 ms
Thus, we have
v C (t ) = 100 − 100 exp(− t 10 −3 )
108
P4.4*
The solution is of the form of Equation 4.17:
v C (t ) =Vs + K 2 exp(− t RC ) = 100 + K 2 exp(− t 10 −3 )
in which K 2 is a constant to be determined. At t = 0 + , we have
v C (0 + ) = −50 = 100 + K 2
Solving, we find that K 2 = −150 and the solution is
v C (t ) = 100 − 150 exp(− t 10 −3 )
P4.5*
The voltage across the capacitor is given by Equation 4.8.
v (t ) = Vi exp( −t / RC )
in which Vi = 100 V is the initial voltage, C = 100 µF is the capacitance,
and R is the leakage resistance.
The energy stored in the capacitance is
1
w = Cv 2 (t ) = 0.5 × 10 −4 × 100 2 exp( −2t / RC )
2
Since we require the energy to be 90% of the initial value after one
minute, we can write
0.9 × 0.5 × 10 −4 × 100 2 ≤ 0.5 × 10 −4 × 100 2 exp( −120 / RC )
109
Solving we determine that RC must be greater than 1139 s. Thus the
leakage resistance must be greater than 11.39 MΩ.
P4.6
The voltage across the capacitor is given by Equation 4.8.
v (t ) = Vi exp( −t / RC )
in which Vi is the initial voltage. Substituting values, we have
v (2 × 10 -3 ) = 10 = Vi exp( −0.002 / 0.0004) = Vi exp( −5)
Vi = 1484 V
P4.7
(a)
RC = 20 ms
vC (t ) = 50 for t < 0
= 50 exp(− t 0.02) = 50 exp(− 50t ) for t > 0
vR (t ) = 0 for t < 0
= 50 exp(− t 0.02) = 50 exp(− 50t ) for t > 0
pR (t ) =
(c)
[v R (t )]
2
R
= 2500 exp(− 100t ) µW
∞
W = ∫ pR (t )dt
0
∞
= ∫ 2500 exp(− 100t )dt
0
= − 25 exp(− 100t ) 0
∞
= 25 µ J
(d) The initial energy stored in the capacitance is
1
W = C [v C (0 )]2
2
1
= × 0.02 × 10 −6 × 502
2
= 25 µJ
110
(b)
P4.8
Equation 4.8 gives the expression for the voltage across a capacitance
discharging through a resistance:
v C (t ) = Vi exp(− t RC )
After one-half-life, we have v C (thalf ) =
Vi
Vi
= Vi exp(− thalf RC ) .
2
2
Dividing by Vi and taking the natural logarithm of both sides, we have
− ln(2) = − thalf RC
and
Solving, we obtain
thalf = RC ln(2) = 0.6931RC = 0.6931τ
P4.9
Prior to t = 0 , we have v (t ) = 0 because the switch is closed. After
t = 0 , we can write the following KCL equation at the top node of the
circuit:
v (t )
dv (t )
+C
= 1 mA
R
dt
Multiplying both sides by R and substituting values, we have
dv (t )
0.01
+ v (t ) = 10
dt
The solution is of the form
v (t ) = K 1 + K 2 exp(− t RC ) = K 1 + K 2 exp(− 100t )
(1)
(2)
Substituting Equation (2) into Equation (1), we eventually obtain
K 1 = 10
The voltage across the capacitance cannot change instantaneously, so we
have
v (0 + ) = v (0 − ) = 0
v (0 + ) = 0 = K 1 + K 2
Thus, K 2 = −K 1 = −10 , and the solution is
v (t ) = 10 − 10 exp(− 100t ) for t > 0
111
P4.10*
The initial energy is
W1 =
1
1
C (Vi )2 = 100 × 10 −6 × 1000 2 = 50 J
2
2
1
C [v (t2 )]2 , which
2
yields v (t2 ) = 707.1 V . The voltage across the capacitance is given by
At t = t2 , half of the energy remains, and we have 25 =
v C (t ) = Vi exp(− t RC ) = 1000 exp(− 10t ) for t > 0
Substituting, we have 707.1 = 1000 exp(− 10t2 ) . Solving, we obtain
ln( 0.7071) = −10t2
t2 = 0.03466 seconds
P4.11*
Because the voltage does not change, the current is zero, and the
voltmeter must be an open circuit. Thus, the resistance of the voltmeter
is infinite.
P4.12
During the charging interval, the time constant is τ 1 = R1C = 40 s, and
the voltage across the capacitor is given by
v C (t ) = 2890[1 − exp( −t / τ 1 )] 0 ≤ t ≤ 28
At the end of the charging interval (t = 28 s), this yields
v C (28) = 2890[1 − exp( −0.7)] = 1455 V
The time constant during the discharge interval is τ 2 = R2C = 50 s.
Working in terms of the time variable t ′ = t − 28, the voltage during the
discharge interval is
v C (t ′) = 1455 exp( −t ′ / τ 2 ) 0 ≤ t ′
At t ′ = 40 − 28 = 12, this yields
1455 exp( −0.24) = 1145 V
P4.13
The initial current is Vi / R = 26185 / 127 = 206.2 A. No wonder one
jumps!!! The time constant is τ = RC = 15.5 ns.
P4.14
The voltage across the resistance and capacitance is
v C (t ) = Vi exp(− t RC )
The initial charge stored on the capacitance is
Qi = CVi
112
The current through the resistance is
iR (t ) =
Vi
exp(− t RC )
R
The total charge passing through the resistance is
∞
Q = ∫ iR (t )dt
0
Vi
∫0 R exp(− t RC )dt
V
∞
= i [− RC exp(− t RC )] 0
R
= CV1
∞
=
P4.15*
v (t ) = V1 exp[−(t − t1 ) / RC ] for t ≥ t1
P4.16
The final voltage for each 1 s interval is the initial voltage for the
succeeding interval.
We have τ = RC = 2 s. For 0 ≤ t ≤ 1, we have
v (t ) = 11 − 11 exp( −t / 2) which yields v (1) = 4.328 V.
For 1 ≤ t ≤ 2, we have v (t ) = 4.328 exp[ −(t − 1) / 2] which yields
v (2) = 2.625 V.
For 2 ≤ t ≤ 3, we have v (t ) = 11 − (11 − 2.625) exp[ −(t − 2) / 2] which yields
v (3) = 5.920 V.
Finally, for 3 ≤ t ≤ 4, we have v (t ) = 5.920 exp[ −(t − 3) / 2] which yields
v (4) = 3.591 V.
P4.17
(a) The voltages across the capacitors cannot change instantaneously.
Thus, v 1 (0 + ) = v 1 (0 − ) = 100 V and v 2 (0 + ) = v 2 (0 − ) = 0 . Then, we can write
v (0 + ) − v 2 (0 + ) 100 − 0
=
i (0 + ) = 1
= 1 mA
R
100 × 10 3
(b) Applying KVL, we have
1
C1
t
− v 1 (t ) + Ri (t ) + v 2 (t ) = 0
1
t
∫ i (t )dt − 100 + Ri (t ) + C ∫ i (t )dt + 0 = 0
0
2 0
Taking a derivative with respect to time and rearranging, we obtain
113
di (t ) 1
+
dt
R
 1
1 
 +
i (t ) = 0
 C1 C2 
(c) The time constant is τ = R
(1)
C 1C 2
= 50 ms .
C1 + C2
(d) The solution to Equation (1) is of the form
i (t ) = K 1 exp(− t τ )
However, i (0 + ) = 1 mA , so we have K 1 = 1 mA and i (t ) = exp(− 20t ) mA .
(e) The final value of v 2 (t ) is
v 2 (∞ ) =
∞
1
C2
∫ i (t )dt + v (0 + )
0
2
t
= 10 6 ∫ 10 −3 exp(− t 0.05)dt + 0
0
= 10 (− 0.05) exp(− t 0.05) ∞0
3
= 50 V
Thus, the initial charge on C 1 is eventually divided equally between
C 1 and C 2 .
P4.18
For a dc steady-state analysis:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the resulting circuit, which consists of dc sources and
resistances.
P4.19
In dc steady state conditions, the voltages across the capacitors are
constant. Therefore, the currents through the capacitances, which are
given by iC = C dv
dt
, are zero. Open circuits also have zero current.
Similarly, the currents through the inductances are constant.
Therefore, the voltages across inductances, which are given by
v = L di
dt
, are zero. Short circuits also have zero voltage.
114
P4.20*
In steady state, the equivalent circuit is:
Thus, we have
i1 = 0
i3 = i2 = 3 A
P4.21*
After the switch opens and the circuit reaches steady state, the 10-mA
current flows through the 1-k Ω resistance, and the voltage is 10 V.
The initial voltage is v C (0 + ) = 0 . The time constant of the circuit is
RC = 10 ms . As a function of time, we have
v C (t ) = 10 − 10 exp(− t RC ) = 10 − 10 exp(− 100t )
Let t99 denote the time at which the voltage reaches 99% is its final
value. Then we have
v C (t99 ) = 9.9 = 10 − 10 exp(− 100t99 )
Solving, we find
t99 = 46.05 ms
P4.22
In steady state with a dc source, the inductance acts as a short circuit
and the capacitance acts as an open circuit. The equivalent circuit is:
115
i4
i3
i2
i1
vC
= (201 V ) (2 kΩ ) = 100.5 mA
=0
= (201 V ) (1 kΩ ) = 201 mA
= i2 + i3 + i 4 = 301.5 mA
= 201 V
P4.23
iL = 6 mA
v x = 18 V
v C = −17 V
P4.24
Prior to t = 0 , the steady-state equivalent circuit is:
and we see that v c = 18 V .
A long time after t = 0 , the steady-state equivalent circuit is:
and we have v c = 18
P4.25
13
= 9.75 V .
13 + 11
With the circuit in steady state prior to t = 0, the capacitor behaves as
an open circuit, the two 2-kΩ resistors are in parallel, and
v C (0−) = (20 mA) × (1 kΩ) = 20 V. Because there cannot be infinite current
in this circuit, we have v C (0+) = 20 V. After the switch opens and the
circuit reaches steady state, we have v C (∞) = (20 mA) × (2 kΩ) = 40 V.
For t ≥ 0, the Thévenin resistance seen by the capacitor is Rt = 2 kΩ and
the time constant is τ = Rt C = 0.2 s. The general form of the solution
for t ≥ 0 is v C (t ) = K 1 + K 2 exp( −t / τ ). However, we know that
116
v C (0+) = 20 = K 1 + K 2 and v C (∞) = 40 = K 1 . Solving we find that K 2 = −20.
Thus, we have
v C (t ) = 20 t ≤ 0
= 40 − 20 exp( −t / τ) t ≥ 0
P4.26
With the circuit in steady state prior to t = 0, the capacitor behaves as
an open circuit, the current is zero, and v C (0−) = 50 V. Because there
cannot be infinite currents in this circuit, we have v C (0+) = 50 V. After
the switch closes and the circuit reaches steady state, the source divides
equally between the two 1-MΩ resistances, and we have v C (∞) = 25 V.
For t ≥ 0, the Thévenin resistance seen by the capacitor is Rt = 500 kΩ
and the time constant is τ = Rt C = 1 s. The general form of the solution
for t ≥ 0 is v C (t ) = K 1 + K 2 exp( −t / τ ). However, we know that
v C (0+) = 50 = K 1 + K 2 and v C (∞) = 25 = K 1 . Solving we find that K 2 = 25.
Thus we have
v C (t ) = 50 t ≤ 0
= 25 + 25 exp( −t / τ ) t ≥ 0
A sketch of the capacitor voltage is:
117
P4.27
iR = 2 mA
v C = 48 V
P4.28*
With the switch in position A and the circuit in steady state prior to
t = 0, the capacitor behaves as an open circuit, v R (0−) = 0, and
10 kΩ
v C (0−) = 30
= 10 V. Because there cannot be infinite
10 kΩ + 20 Ωk
current in this circuit, we have v C (0+) = v R (0+) = 10 V. After switching
and the circuit reaches steady state, we have v R (∞) = 0 V. For t ≥ 0, the
resistance across the capacitor is R = 200 kΩ and the time constant is
τ = RC = 2 s. The general form of the solution for t ≥ 0 is
v R (t ) = K 1 + K 2 exp( −t / τ). However, we know that v R (0+) = 10 = K 1 + K 2 and
v R (∞) = 0 = K 1 . Solving we find that K 2 = 10. Thus, we have
v R (t ) = 0 t < 0
= 10 exp( −t / τ ) t ≥ 0
P4.29
With the switch closed and the circuit in steady state prior to t = 0, the
capacitor behaves as an open circuit, and v C (0−) = 25 V.
Because there
cannot be infinite current in this circuit when the switch opens, we have
v C (0+) = 25 V. After switching and the circuit reaches steady state, we
15 kΩ
have v C (∞) = 25
= 15 V. For t ≥ 0, the Thévenin resistance
10 kΩ + 15 Ωk
118
1
= 6 kΩ, and the time constant
1 / 10 + 1 / 15
is τ = Rt C = 60 ms. The general form of the solution for t ≥ 0 is
seen by the capacitor is Rt =
v C (t ) = K 1 + K 2 exp( −t / τ ). However, we know that v C (0+) = 25 = K 1 + K 2
and v C (∞) = 15 = K 1 . Solving, we find that K 2 = 10. Thus, we have
v C (t ) = 25 t ≤ 0
= 15 + 10 exp( −t / τ ) t ≥ 0
P4.30
The time constant τ is the interval required for the current to fall to
exp(−1) ≅ 0.368 times its initial value. The time constant is given by
τ = L / R . Thus to attain a long time constant, we need a large value for L
and a small value for R.
P4.31
The general form of the solution is x (t ) = A + B exp( −t / τ ) . A is the
steady-state solution for t >> 0. We determine the value of the desired
current or voltage immediately after t = 0, denoted by x (0 +). Then, we
solve x (0+) = A + B for the value of B. Finally, we determine the Thévenin
resistance Rt from the perspective of the energy storage element (i.e.,
the resistance seen looking back into the circuit with the energy storage
element removed) and compute the time constant: τ = Rt C for a
capacitance or τ = L / Rt for an inductance.
P4.32*
In steady state with the switch closed, we have i (t ) = 0 for t < 0
because the closed switch shorts the source.
In steady state with the switch open, the inductance acts as a short
circuit and the current becomes i (∞ ) = 1 A . The current is of the form
i (t ) = K 1 + K 2 exp(− Rt L ) for t ≥ 0
119
in which R = 20 Ω , because that is the Thévenin resistance seen looking
back from the terminals of the inductance with the switch open. Also, we
have
i (0 + ) = i (0 − ) = 0 = K 1 + K 2
i (∞ ) = 1 = K 1
Thus, K 2 = −1 and the current (in amperes) is given by
for t < 0
i (t ) = 0
= 1 − exp(− 20t )
P4.33
for t ≥ 0
The general form of the solution is
iL (t ) = K 1 + K 2 exp(− R t L )
At t = 0 + , we have
iL (0 + ) = iL ((0 − ) = 0 = K 1 + K 2
At t = ∞ , the inductance behaves as a short circuit, and we have
iL (∞ ) = 0.1 = K 1
Thus, the solution for the current is
iL (t ) = 0 for t < 0
= 0.1 - 0.1exp (- 10 6t )
for t > 0
The voltage is
di (t )
dt
= 0 for t < 0
v (t ) = L
= 100 exp(− 10 6t ) for t > 0
P4.34*
The solution is similar to that for Problem P4.33.
iL (t ) = 0.1 − 0.3 exp(− 10 6t ) for t > 0
v (t ) = 300 exp(− 10 6t ) for t > 0
120
P4.35
The solution is of the form
i (t ) = K 1 + K 2 exp(− Rt L )
At t = 0 + , we have
i (0 + ) = 0 = K 1 + K 2
and at t = ∞ , we have
i (∞ ) = (100 V ) (200 Ω )
= 0. 5 = K 1
The time constant is
τ = L / R = 5 ms. Thus, the
solution is
i (t ) = 0.5 − 0.5 exp(− 200t )
The voltage across the inductor
is
v L (t ) = L
P4.36
di (t )
= 100 exp(− 200t )
dt
In steady state, the inductor acts as a short circuit. With the switch
open, the steady-state current is (100 V ) (100 Ω ) = 1 A . With the switch
closed, the current eventually approaches i (∞ ) = (100 V ) (25 Ω ) = 4 A . For
t > 0 , the current has the form
i (t ) = K 1 + K 2 exp(− Rt L )
where R = 25 Ω , because that is the resistance with the switch closed.
Now, we have
i (0 + ) = i (0 − ) = 1 = K 1 + K 2
i (∞ ) = 4 = K 1
Thus, we have K 2 = −3 . The current is
121
i (t ) = 1
t < 0 (switch open)
= 4 − 3 exp(− 12.5t ) t ≥ 0 (switch closed)
P4.37
Before the switch closes, 1 A of current circulates through the source
and the two 10-Ω resistors. Immediately after the switch closes, the
inductor current remains 0 A because infinite voltage is not possible in
this circuit. (Because the inductor current is zero we can consider the
inductor to be an open circuit at t = 0+.) Therefore, the current through
the resistors is unchanged, and i (0 +) = 1 A. In steady state, the inductor
acts as a short circuit, and we have i (∞) = 2 A. The Thévenin resistance
seen by the inductor is 5 Ω because the two 10-Ω resistors are in parallel
when we zero the source and look back into the circuit from the inductor
terminals. Thus, the time constant is τ = L / R = 200 ms. The general
form of the solution is i (t ) = K 1 + K 2 exp( −t / τ ). Using the initial and final
values, we have i (0+) = 1 = K 1 + K 2 and i (∞) = 2 = K 1 which yields K 2 = −1.
Thus, the current is
i (t ) = 1 t ≤ 0
= 2 − exp( −t / τ) t ≥ 0
122
P4.38
Prior to t = 0, the current source is shorted, so we have
i L (t ) = 0 for t < 0
After the switch opens at t = 0, the current i L (t ) increases from zero
headed for 0.5 A. The inductance sees a Thévenin resistance of 8 Ω, and
the time constant is τ = L / 8 = 0.375 s, so we have
iL (t ) = 0.5 − 0.5 exp(−t / 0.375) for 0 < t < 1
At t = 1, the current reaches 0.4653 A. Then, the switch closes, the
source is shorted, and the current decays toward zero. Because the
inductance sees a resistance of 4 Ω, the time constant is τ = L / 4 = 0.75
s.
iL (t ) = 0.4653 exp(−t / 0.75) for 1 < t
P4.39
(a) i (t ) = I i exp(− Rt L ) for t ≥ 0
(b) pR (t ) = Ri 2 (t ) = R (I i )2 exp(− 2Rt L ) for t ≥ 0
∞
(c) W = ∫ pR (t )dt
0
∞
= ∫ R (I i )2 exp (− 2Rt L )dt
0
= R (I i )
2
=
t =∞
 L

 − 2R exp (− 2Rt L )
t =0
1
L (I i )2
2
which is precisely the expression for the energy stored in the inductance
at t = 0 .
P4.40
With the circuit in steady state before the switch opens, the inductor
acts as a short circuit, the current through the inductor is i L (t ) = 0.1 A,
and v R (t ) = 0. Immediately after the switch opens, the inductor current
remains 0.1 A because infinite voltage is not possible in this circuit. Then
the return path for the inductor current is through the 1 kΩ resistor so
v R (0+) = −100 V. After the switch opens, the current and voltage decay
exponentially with a time constant τ = L / R = 0.2 ms. Thus, we have
123
v R (t ) = 0 t < 0
= −100 exp( −t / τ ) t > 0
P4.41
In steady state with the switch closed, the current is i (t ) = 2 A for t < 0.
The resistance of a voltmeter is very high -- ideally infinite. Thus, there
is no path for the current in the inductance when the switch opens, and
di dt is very large in magnitude at t = 0 . Consequently, the voltage
induced in the inductance is very large in magnitude and an arc occurs
across the switch. With an ideal meter and switch, the voltage would be
infinite. The voltmeter could be damaged in this circuit.
P4.42*
The current in a circuit consisting of an inductance L and series
resistance R is given by i (t ) = I i exp( −Rt / L ) in which Ii is the initial
current. The initial energy stored in the inductance is w i = (1 / 2)LI i 2 and
the energy stored as a function of time is
w (t ) = (1 / 2)LI i 2 exp( −2Rt / L)
Thus we require
0.75 × (1 / 2)LI i 2 ≤ (1 / 2)LI i 2 exp[ −2R (3600) / 10]
Solving we determine that we require R ≤ 399.6 µΩ. In practice the only
practical way to attain such a small resistance for a 10-H inductance is to
use superconductors.
P4.43
1. Write the circuit equation and, if it includes an integral, reduce the
equation to a differential equation by differentiating.
2. Form the particular solution. Often this can be accomplished by adding
terms like those found in the forcing function and its derivatives,
including an unknown coefficient in each term. Next, solve for the
unknown coefficients by substituting the trial solution into the
differential equation and requiring the two sides of the equation to be
identical.
124
3. Form the complete solution by adding the complementary solution
x c (t ) = K exp(−t / τ ) to the particular solution.
4. Use initial conditions to determine the value of K.
P4.44*
Applying KVL, we obtain the differential equation:
L
di (t )
+ Ri (t ) = 5 exp(− t ) for t > 0
dt
We try a particular solution of the form:
i p (t ) = A exp(− t )
(1)
(2)
in which A is a constant to be determined. Substituting Equation (2) into
Equation (1), we have
− LA exp(− t ) + RA exp(− t ) = 5 exp(− t )
which yields
5
= −1
R −L
The complementary solution is of the form
ic (t ) = K 1 exp(− Rt L )
A=
The complete solution is
i (t ) = i p (t ) + ic (t ) = − exp(− t ) + K 1 exp(− Rt L )
Before the switch closes, the current must be zero. Furthermore, the
current cannot change instantaneously, so we have i (0 + ) = 0 . Therefore,
we have i (0 + ) = 0 = −1 + K 1 which yields K 1 = 1 . Finally, the current is
given by i (t ) = − exp(− t ) + exp(− Rt L ) for t ≥ 0 .
P4.45
The differential equation is obtained by applying KVL for the node at the
top end of the capacitance:
v C (t ) − v (t )
dv C (t )
+C
=0
R
dt
Rearranging this equation and substituting v (t ) = t , we have
RC
dv C (t )
+ v C (t ) = t for t > 0
dt
We try a particular solution of the form
v Cp (t ) = A + Bt
(1)
(2)
in which A and B are constants to be determined. Substituting Equation
(2) into Equation (1), we have
RCB + A + Bt = t
Solving, we have
125
B =1
A = −RC
Thus, the particular solution is
v Cp (t ) = −RC + t
The complementary solution (to the homogeneous equation) is of the form
v Cc (t ) = K 1 exp(− t RC )
Thus, the complete solution is
v C (t ) = v Cp (t ) + v Cc (t ) = −RC + t + K 1 exp(− t RC )
However, the solution must meet the given initial condition:
v C (0 ) = 0 = −RC + K 1
Thus, K 1 = RC and we have
v C (t ) = v Cp (t ) + v Cc (t ) = −RC + t + RC exp(− t RC )
P4.46*
Write a current equation at the top node:
v (t )
dv C (t )
2 exp( −3t ) = C
+C
R
dt
Substitute the particular solution suggested in the hint:
2 exp(−3t ) ≡
A
exp( −3t ) − 3AC exp(−3t )
R
Solving for A and substituting values of the circuit parameters, we find
A = −10 6. The time constant is τ = RC = 1 s, and the general form of the
solution is
v C (t ) = K 1 exp( −t ) + v p (t ) = K 1 exp( −t ) − 10 6 exp( −3t )
However because of the closed switch, we have v C (0+) = 0. Substituting
this into the general solution we find K 1 = 10 6 . Thus
v C (t ) = 10 6 exp( −t ) − 10 6 exp( −3t ) t > 0
P4.47*
Write a current equation at the top node:
126
5 cos(10t ) =
v (t ) 1 t
+ ∫ v (t )dt + i L (0)
R
L 0
Differentiate each term with respect to time to obtain a differential
equation:
1 dv (t ) v (t )
− 50 sin(10t ) =
+
R dt
L
Substitute the particular solution suggested in the hint:
1
1
− 50 sin(10t ) ≡ [−10A sin(10t ) + 10B cos(10t )] + [A cos(10t ) + B sin(10t )]
R
L
Equating coefficients of sine and cosine terms, we have
− 10A B
− 50 =
+
L
A
R
L
Solving for A and B and substituting values of the circuit parameters, we
find A = 25 and B = −25. The time constant is τ = L / R = 0.1 s, and the
0=
10B
R
+
general form of the solution is
v (t ) = K 1 exp(−t / τ) + v p (t ) = K 1 exp(−t / τ) + 25 cos(10t ) − 25 sin(10t )
However, because the current in the inductor is zero at t = 0+, the 5 A
supplied by the source must flow through the 10-Ω resistor and we have
v (0+) = 50. Substituting this into the general solution we find K 1 = 25 .
Thus
P4.48
v (t ) = 25 exp(−t / τ ) + 25 cos(10t ) − 25 sin(10t ) t ≥ 0
Using KVL, we obtain the differential equation
di (t )
+ Ri (t ) = v (t )
dt
di (t )
+ 300i (t ) = 10 sin(300t )
dt
L
We try a particular solution of the form
i p (t ) = A cos(300t ) + B sin(300t )
in which A and B are constants to be determined. Substituting the
proposed solution into the differential equation yields:
− 300A sin(300t ) + 300B cos(300t ) + 300A cos(300t ) + 300B sin(300t ) =
10 sin(300t )
Equating coefficients of cosines yields
300B + 300A = 0
127
Equating coefficients of sines yields
− 300A + 300B = 10
From these equations, we find that B = 1 60 and A = − 1 60 . The
complementary solution is
ic (t ) = K 1 exp(− Rt L ) = K 1 exp(− 300t )
and the complete solution is
i (t ) = ic (t ) = i p (t )
= K 1 exp(− 300t ) − (1 60 ) cos(300t ) + (1 60 ) sin(300t )
Finally, we use the given initial condition
i (0 ) = 0 = K 1 − 1 60
to determine that K 1 = 1 60 . Thus, the solution for the current in
amperes is
i (t ) = (1 60 ) exp(− 300t ) − (1 60 ) cos(300t ) + (1 60 ) sin(300t )
P4.49
Applying KVL to the circuit, we have
di (t )
+ Ri (t ) = v (t )
dt
di (t )
2
+ 10i (t ) = 10t
dt
L
We try a particular solution of the form i p (t ) = A + Bt in which A and B
are constants to be determined. Substituting the proposed solution into
the differential equation yields 2B + 10A + 10Bt = 10t . From this, we find
that B = 1 and A = −0.2 . The complementary solution is
ic (t ) = K 1 exp(− Rt L ) = K 1 exp(− 5t )
and the complete solution is
i (t ) = ic (t ) + i p (t ) = K 1 exp(− 5t ) − 0.2 + t
Finally, we use the given initial condition:
i (0 ) = 0 = K 1 − 0.2
to determine that K 1 = 0.2 . Thus, the solution is
i (t ) = 0.2 exp(− 5t ) − 0.2 + t
P4.50.
Applying KVL, we obtain the differential equation:
2
di (t )
+ i (t ) = 5 exp(− t ) sin(t ) for t > 0
dt
Because the derivative of the forcing function is
− 5 exp(− t ) sin(t ) + 5 exp( −t ) cos(t )
we try a particular solution of the form:
128
(1)
i p (t ) = A exp(− t ) sin(t ) + B exp( −t ) cos(t )
(2)
in which A and B are constants to be determined. Substituting Equation
(2) into Equation (1), we have
− 2A exp(− t ) sin(t ) − 2B exp(−t ) cos(t ) + 2A exp( −t ) cos(t ) − 2B exp( −t ) sin(t )
+ A exp( −t ) sin(t ) + B exp( −t ) cos(t ) ≡ 5 exp(− t ) sin(t )
which yields
− A − 2B = 5
and
2A − B = 0
Solving we find A = −1 and B = −2
The complementary solution is of the form
ic (t ) = K 1 exp(− Rt L ) = K 1 exp( −t / 2)
The complete solution is
i (t ) = i p (t ) + ic (t ) = − exp( −t ) sin(t ) − 2 exp( −t ) cos(t ) + K 1 exp( −t / 2)
Before the switch closes, the current must be zero. Furthermore, the
current cannot change instantaneously, so we have i (0 + ) = 0 . Therefore,
we have i (0 + ) = 0 = −2 + K 1 which yields K 1 = 2 . Thus, the current is given
by
P4.51
i (t ) = − exp(−t ) sin(t ) − 2 exp(−t ) cos(t ) + 2 exp(−t / 2) for t ≥ 0
Usually, the particular solution includes terms with the same functional
forms as the terms found in the forcing function and its derivatives. In
this case, there are four different types of terms in the forcing function
and its derivatives, namely t sin(t ), t cos(t ), sin(t ), and cos(t ).
Thus, we we are led to try a particular solution of the form
v p (t ) = At sin(t ) + Bt cos(t ) + C sin(t ) + D cos(t )
Substituting into the differential equation, we have
2[A sin(t ) + At cos(t ) + B cos(t ) − Bt sin(t ) + C cos(t ) − D sin(t )] +
At sin(t ) + Bt cos(t ) + C sin(t ) + D cos(t ) ≡ 5t sin(t )
We require the two sides of the equation to be identical. Equating
coefficients of like terms, we have
2A − 2D + C = 0
2B + 2C + D = 0
2A + B = 0
129
A − 2B = 5
Solving these equations, we obtain A = 1 , B = −2 , C = 6 / 5, and D = 8 / 5.
Thus, the particular solution is
v p (t ) = t sin(t ) − 2t cos(t ) +
P4.52
6
8
sin(t ) + cos(t )
5
5
di (t )
+ 2i = 3 exp( −2t )
dt
(b) τ = L / R = 0.5 s
ic (t ) = A exp(−2t )
(a)
(c) A particular solution of the form i p (t ) = K exp(−2t ) does not work
because the left-hand side of the differential equation is identically zero
for this choice.
(d) Subsitituting i p (t ) = Kt exp(−2t ) into the differential equation
produces
− 2Kt exp( −2t ) + K exp(−2t ) + 2Kt exp(−2t ) ≡ 3 exp(−2t )
from which we have K = 3.
(e) Adding the particular solution and the complementary solution we have
i (t ) = A exp( −2t ) + 3t exp( −2t )
However, the current must be zero in the inductor prior to t = 0 because
of the open switch, and the current cannot change instantaneously in this
circuit, so we have i (0 +) = 0. This yields A = 0. Thus, the solution is
i (t ) = 3t exp( −2t )
P4.53
First, we write the differential equation for the system and put it in the
form
d 2 x (t )
dx (t )
+ 2α
+ ω02 (t ) = f (t )
2
dt
dt
Then compute the damping ratio ζ = α / ω0 .
If we have ζ < 1, the system is underdamped, and the complementary
solution is of the form
x c (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
in which ωn = ω02 − α 2 .
If we have ζ = 1, the system is critically damped, and the complementary
solution is of the form
x c (t ) = K 1 exp(s 1t ) + K 2t exp(s 1t )
130
in which s1 is the root of the characteristic equation s 2 + 2αs + ω02 = 0 .
If we have ζ > 1, the system is overdamped, and the complementary
solution is of the form
x c (t ) = K 1 exp(s 1t ) + K 2 exp(s 2t )
in which s1 and s2 are the roots of the characteristic equation
s 2 + 2αs + ω02 = 0 .
P4.54
One way to determine the particular solution is to assume that it is a
constant [ x p (t ) = K ], substitute into the differential equation, and solve
for K.
A second method is to replace the inductors by short circuits the
capacitances by open circuits and solve for the steady-state dc response.
P4.55
We look at a circuit diagram and combine all of the inductors that are in
series or parallel . Then, we combine all of the capacitances that are in
series or parallel. Next, we count the number of energy storage
elements (inductances and capacitances) in the reduced circuit. If there
is only one energy-storage element, we have a first-order circuit. If
there are two, we have a second-order circuit and so forth.
P4.56
The unit step function is defined by
u (t ) = 0 for t < 0
= 1 for t ≥ 0
The unit step function is illustrated in Figure 4.27 in the book.
P4.57
The sketch should resemble the response shown in Figure 4.29 for
ζ = 0.1. Second-order circuits that are severely underdamped ( ζ << 1 )
have step responses that display considerable overshoot and ringing.
P4.58*
Applying KVL to the circuit, we obtain
L
di (t )
+ Ri (t ) + v C (t ) = v s = 50
dt
(1)
For the capacitance, we have
i (t ) = C
dv C (t )
dt
(2)
131
Using Equation (2) to substitute into Equation (1) and rearranging, we
have
d 2v C (t )
dv (t )
+ (R L ) C
+ (1 LC )v C (t ) = 50 LC
2
dt
dt
(3)
d 2v C (t )
dv (t )
+ 4 × 10 4 C
+ 10 8v C (t ) = 50 × 10 8
2
dt
dt
We try a particular solution of the form v Cp (t ) = A , resulting in A = 50 .
Thus, v Cp (t ) = 50 . (An alternative method to find the particular solution
is to solve the circuit in dc steady state. Since the capacitance acts as
an open circuit, the steady-state voltage across it is 50 V.) Comparing
Equation (3) with Equation 4.67 in the text, we find
α=
R
= 2 × 10 4
2L
ω0 =
1
LC
= 10 4
Since we have α > ω0 , this is the overdamped case. The roots of the
characteristic equation are found from Equations 4.72 and 4.73 in the
text.
s1 = − α + α 2 − ω02 = −0.2679 × 10 4
s 2 = − α − α 2 − ω02 = −3.732 × 10 4
The complementary solution is
v Cc (t ) = K 1 exp(s1t ) + K 2 exp(s 2t )
and the complete solution is
v C (t ) = 50 + K 1 exp(s 1t ) + K 2 exp(s 2t )
The initial conditions are
v C (0 ) = 0
and
i (0 ) = 0 = C
Thus, we have
v C (0 ) = 0 = 50 + K 1 + K 2
dv C (t )
dt
t =0
dv C (t )
dt
t =0
= 0 = s1K 1 + s 2K 2
Solving, we find K 1 = −53.87 and K 2 = 3.867 . Finally, the solution is
132
v C (t ) = 50 − 53.87 exp(s1t ) + 3.867 exp(s 2t )
P4.59*
As in the solution to P4.58, we have
v Cp (t ) = 50
α=
R
= 10 4
2L
1
ω0 =
LC
= 10 4
Since we have α = ω0 ,this is the critically damped case. The roots of the
characteristic equation are equal:
s1 = − α + α 2 − ω02 = −10 4
s 2 = − α − α 2 − ω02 = −10 4
The complementary solution is given in Equation 4.75 in the text:
v Cc (t ) = K 1 exp(s 1t ) + K 2t exp(s 1t )
and the complete solution is
v C (t ) = 50 + K 1 exp(s 1t ) + K 2t exp(s 1t )
The initial conditions are
v C (0 ) = 0
and
i(0 ) = 0 = C
Thus, we have
dv C (t )
dt
t =0
v C (0 ) = 0 = 50 + K 1
dv C (t )
= 0 = s1K 1 + K 2
dt t =0
Solving, we find K 1 = −50 and K 2 = −50 × 10 4 . Finally, the solution is
v C (t ) = 50 − 50 exp(s1t ) − (50 × 10 4 )t exp(s1t )
P4.60*
As in the solution to P4.58, we have
v Cp (t ) = 50
α=
ω0 =
R
= 0.5 × 10 4
2L
1
LC
= 10 4
Since we have α < ω0 , this is the under-damped case. The natural
frequency is given by Equation 4.76 in the text:
ωn = ω02 − α 2 = 8.660 × 10 3
The complementary solution is given in Equation 4.77 in the text:
133
v Cc (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
and the complete solution is
v C (t ) = 50 + K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
The initial conditions are
v C (0 ) = 0
and
Thus, we have
i (0 ) = 0 = C
dv C (t )
dt
t =0
v C (0 ) = 0 = 50 + K 1
dv C (t )
= 0 = − αK 1 + ωn K 2
dt t =0
Solving, we find K 1 = −50 and K 2 = −28.86 . Finally, the solution is
v C (t ) = 50 − 50 exp(− αt ) cos(ωnt ) − (28.86) exp(− αt ) sin(ωnt )
P4.61
(a) Using Equation 4.103 from the text, the damping coefficient is
1
α=
= 20 × 10 6
2RC
Equation 4.104 gives the undamped resonant frequency
1
ω0 =
= 10 × 10 6
LC
Equation 4.71 gives the damping ratio
ζ = α ω0 = 2
Thus, we have an overdamped circuit.
(b) Writing a current equation at t = 0 + , we have
v (0 + )
+ iL (0 + ) + Cv ′(0 + ) = 1
R
Substituting v (0 + ) = 0 and iL (0 + ) = 0 , yields
v ′(0 + ) = 1 C = 10 9
(c) Under steady-state conditions, the inductance acts as a short circuit.
Therefore, the particular solution for v (t ) is:
v p (t ) = 0
(d) The roots of the characteristic equation are found from Equations
4.72 and 4.73 in the text.
134
s1 = −α + α 2 − ω02 = −2.679 × 10 6
s 2 = −α − α 2 − ω02 = −37.32 × 10 6
The complementary solution is
v c (t ) = K 1 exp(s1t ) + K 2 exp(s 2t )
and (since the particular solution is zero) the complete solution is
v (t ) = K 1 exp(s 1t ) + K 2 (s 2t )
The initial conditions are
v (0 + ) = 0 and v ′(0 + ) = 10 9
Thus, we have
v (0 + ) = 0 = K 1 + K 2
v ′(0 + ) = 10 9 = s 1K 1 + s 2K 2
Solving, we find K 1 = 28.87 and K 2 = −28.87 . Finally, the solution is
v (t ) = 28.87 exp(s1t ) − 28.87 exp(s 2t )
P4.62
(a) Using Equation 4.103 from the text, the damping coefficient is
1
α=
= 10 × 10 6
2RC
Equation 4.104 gives the undamped resonant frequency:
1
ω0 =
= 10 × 10 6
LC
Equation 4.71 gives the damping ratio:
ζ = α / ω0 = 1
Thus, we have a critically damped circuit.
(b) Writing a current equation at t = 0 + , we have
v (0 + )
+ iL (0 + ) + Cv ′(0 + ) = 1
R
Substituting v (0 + ) = 0 and iL (0 + ) = 0 , yields
v ′(0 + ) = 1 C = 10 9
(c) Under steady-state conditions, the inductance acts as a short circuit.
Therefore, the particular solution for v (t ) is:
v p (t ) = 0
(d) The roots of the characteristic equation are found from Equations
4.72 and 4.73 in the text.
135
s1 = −α + α 2 − ω02 = −10 × 10 6
s 2 = − α − α 2 − ω02 = −10 × 10 6
The complementary solution is given in Equation 4.75 in the text:
v C (t ) = K 1 exp(s1t ) + K 2t exp(s 1t )
and the complete solution is
v (t ) = v C (t ) + v p (t ) = K 1 exp(s1t ) + K 2t exp(s1t )
The initial conditions are
v (0 + ) = 0 and v (0 + ) = 10 9
Thus, we have
v (0 + ) = 0 = K 1
v ′(0 + ) = 10 9 = s 1K 1 + K 2
Solving, we find K 1 = 0 and K 2 = 10 9 . Finally, the solution is
v C (t ) = 10 9t exp(− 10 7t )
P4.63
(a) Using Equation 4.103 from the text, the damping coefficient is
1
α=
= 10 6
2RC
Equation 4.104 gives the undamped resonant frequency:
1
ω0 =
= 10 × 106
LC
Equation 4.71 gives the damping ratio:
ζ = α ω0 = 0.1
Thus, we have an underdamped circuit.
(b) Writing a current equation at t = 0 + , we have
v (0 + )
+ iL (0 + ) + Cv ′(0 + ) = 1
R
Substituting v (0 + ) = 0 and iL (0 + ) = 0 , yields
v ′(0 + ) = 1 C = 10 9
(c) Under steady-state conditions, the inductance acts as a short circuit.
Therefore, the particular solution for v (t ) is:
v p (t ) = 0
136
(d) The natural frequency is given by Equation 4.76 in the text:
ωn = ω02 − α 2 = 9.950 × 10 6
The complementary solution is given in Equation 4.77 in the text:
v C (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
and the complete solution is
v (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
The initial conditions are
v (0 + ) = 0 and v ′(0 + ) = 10 9
Thus, we have
v (0 + ) = 0 = K 1
v ′(0 + ) = 10 9 = − αK 1 + ωn K 2
Solving, we find K 1 = 0 and K 2 = 100.5 . Finally, the solution is
v (t ) = 100.5 exp(− αt ) sin(ωnt )
P4.64
Write a KVL equation for the circuit:
1
di (t )
10 cos(100t ) = L
+ Ri (t ) +
dt
C
t
∫ i (t )dt + vC (0)
0
Differentiate each term with respect to time to obtain a differential
equation:
d 2i (t )
di (t ) i (t )
− 1000 sin(10t ) = L
+R
+
2
dt
dt
C
Substitute the particular solution suggested in the hint to obtain:
− 10 3 sin(10t ) ≡ L[−10 4 A cos(100t ) − 10 4 B sin(100t )]
+ R [−100A sin(100t ) + 100B cos(100t )] +
1
C
[A cos(100t ) + B sin(100t )]
Equating coefficients of sine and cosine terms, we have
− 10 3 = −10 4 BL − 100AR +
0 = −10 4 AL + 100BR +
A
C
B
C
Solving for A and B and substituting values of the circuit parameters, we
find A = 0.2 and B = 0. Thus, the particular solution is
i p (t ) = 0.2 cos(100t )
Using Equations 4.60 and 4.61 from the text, we have
137
α=
ω0 =
R
= 25
2L
1
LC
= 100
Because we have α < ω0 , this is the under-damped case. The natural
frequency is given by Equation 4.76 in the text:
ωn = ω20 − α 2 = 96.82
The complementary solution is given in Equation 4.77 in the text:
ic (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
and the complete solution is
i (t ) = 0.2 cos(100t ) + K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
However, because the current is zero at t = 0+, the voltage across the
inductor must be 10 V which implies that di (0+) / dt = 10. Thus, we can
write
i (0+) = 0 = 0.2 + K 1
di (0+)
= 10 = −αK 1 + ωn K 2
dt
Solving we find K 1 = −0.2 and K 2 = 0.05164. Finally, the solution is:
i (t ) = 0.2 cos(100t ) − 0.2 exp(− αt ) cos(ωnt ) + 0.05164 exp(− αt ) sin(ωnt )
P4.65
As in the solution to P4.64, we have
− 10 3 = −10 4 BL − 100AR +
0 = −10 4 AL + 100BR +
A
C
B
C
Solving for A and B and substituting values of the circuit parameters, we
find A = 0.05 and B = 0. Thus the particular solution is
i p (t ) = 0.05 cos(100t )
Using Equations 4.60 and 4.61 from the text, we have
α=
ω0 =
R
= 100
2L
1
LC
= 100
Since we have α = ω0 ,this is the critically damped case. The roots of the
characteristic equation are equal:
s1 = −α + α 2 − ω20 = −100
s2 = −α − α 2 − ω20 = −100
138
The complementary solution is given in Equation 4.75 in the text:
ic (t ) = K 1 exp(s1t ) + K 2t exp(s1t )
and the complete solution is
i (t ) = 0.05 cos(100t ) + K 1 exp(s1t ) + K 2t exp(s1t )
As in the solution to P4.51, The initial conditions are
i (0+) = 0 = 0.05 + K 1
di (0+)
= 10 = s1K 1 + K 2
dt
Solving we find K 1 = −0.05 and K 2 = 5. Finally, the solution is
i (t ) = 0.05 cos(100t ) − 0.05 exp(− 100t ) + 5t exp(− 100t )
P4.66
As in the solution to P4.64, we have
− 10 3 = −10 4 BL − 100AR +
0 = −10 4 AL + 100BR +
A
C
B
C
Solving for A and B and substituting values of the circuit parameters, we
find A = 0.025 and B = 0. Thus, the particular solution is
i p (t ) = 0.025 cos(100t )
Using Equations 4.60 and 4.61 from the text, we have
α=
ω0 =
R
= 200
2L
1
LC
= 100
Since we have α > ω0 , this is the overdamped case. The roots of the
characteristic equation are found from Equations 4.72 and 4.73 in the
text.
s1 = −α + α 2 − ω20 = −26.79
s 2 = −α − α 2 − ω20 = −373.2
The complementary solution is
ic (t ) = K1 exp(s1t ) + K 2 exp(s2t )
and the complete solution is
i (t ) = 0.025 cos(100t ) + K 1 exp(s1t ) + K 2 exp(s 2t )
As in the solution to P4.51, The initial conditions are
139
i (0+) = 0 = 0.025 + K 1 + K 2
di (0+)
= 10 = s1K 1 + s2K 2
dt
Solving, we find K 1 = 0.00193 and K 2 = −0.02693. Finally, the solution is
i (t ) = 0.025 cos(100t ) + 0.00193 exp(s1t ) − 0.02693 exp(s2t )
P4.67
(a) Applying KCL, we have:
t
1
dv
4
∫v (t )dt + C = 2 sin(10 t )
L
dt
0
Taking the derivative, multiplying by L, and rearranging, we have
d 2v
+ v (t ) = 2L10 4 cos(10 4t )
LC
2
dt
(b) This is a parallel RLC circuit having R = ∞. Using Equation 4.103 from
the text, the damping coefficient is
1
α=
=0
2RC
Equation 4.104 gives the undamped resonant frequency:
1
ω0 =
= 10 4
LC
The natural frequency is given by Equation 4.76 in the text:
ωn = ω 02 − α 2 = 10 4
The complementary solution given in Equation 4.77 becomes:
v C (t ) = K 1 exp(− αt ) cos(ωnt ) + K 2 exp(− αt ) sin(ωnt )
(
)
(
= K 1 cos 10 4t + K 2 sin 10 4t
)
(c)The usual form for the particular solution doesn't work because it has
the same form as the complementary solution. In other words, if we
substitute a trial particular solution of the form
v p (t ) = A cos(10 4t ) + B sin(10 4t ) , the left-hand side of the differential
equation becomes zero and cannot match the forcing function.
(d) When we substitute v p (t ) = At cos(10 4t ) + Bt sin(10 4t ) into the
differential equation, we eventually obtain
− 2A10 −4 sin(10 4t ) + 2B 10 −4 cos(10 4t ) ≡ 200 cos(10 4t )
from which we obtain B = 10 6 and A = 0. Thus, the particular solution is
v p (t ) = 10 6t sin(10 4t )
(e) The complete solution is
v (t ) = v p (t ) + v C (t ) = 10 6t sin(10 4t ) + K 1 cos(10 4t ) + K 2 sin(10 4t )
140
However, we have v (0 ) = 0, which yields K 1 = 0 . Also, by KCL, the current
through the capacitance is zero at t = 0+, so we have
dv (0+)
= 0 = 10 4 K 2
dt
which yields K 2 = 0, so the complete solution is
v (t ) = v p (t ) + v C (t ) = 10 6t sin(10 4t )
141