Chapter 5
Solution 1.
Apply KCL.
vL
v
+ iL + L = 0
R1 + R2
R3
Next, use the definition of inductor voltage to eliminate the variable v L from the nodal equation:
L
di L
L di L
di L
(R1 + R2 )R3 i = 0
+ iL +
=0
+
L
dt L(R1 + R2 + R3 )
R1 + R2 dt
R3 dt
Substituting numerical values, then obtain the following differential equation:
di L
+ 2.67 ⋅10 6 i L = 0
dt
Solution 2.
Apply KCL vC .
v
v − V1
iC + C + C
=0
R2
R1
Next, use the definition of capacitor current to eliminate the variable
R + R2
V
dvC R1 + R2
V
dv
C C + 1
vC = 1 ⇒
+
vC = 1
dt
C (R1R2 )
CR1
dt
R1R2
R1
iC from the nodal equation:
Substituting numerical values, then obtain the following differential equation:
dvC
+ (4052)vC − 35292 = 0
dt
Solution 3.
Apply KCL at the two node
v − v2
For node #1:
iC + C
=0
R2
v 2 − vC v 2 v 2 − V1
For node #2:
+
+
=0
R2
R3
R1
Solving the system:
R3
R R + R1R3 + R2 R3
vC =
V1 − 1 2
iC
R1 + R3
R1 + R3
R3
RR
v2 =
V1 − 1 3 iC
R1 + R3
R1 + R3
Next, use the definition of capacitor current to eliminate the variable
C (R1R2 + R1R3 + R2 R3) dvC
iC from the nodal equation:
R3
+ vC =
V1
R1 + R3
dt
R1 + R3
dvC
R1 + R3
R3
+
vC =
V1
dt
C (R1R2 + R1R3 + R2 R3 )
C (R1R2 + R1R3 + R2 R3 )
Substituting numerical values, then obtain the following differential equation:
dvC
+ (790)vC − 6876 = 0
dt
Solution 4.
Applying KVL
(R2 + R3)i L + v L + VS2 = 0
Next, using the definition of inductor voltage to eliminate the variable v L from the nodal equation:
di
(R2 + R3)i L + L L + VS 2 = 0
dt
di L (R2 + R3 )
V
+
i L + S2 = 0
dt
L
L
Substituting numerical values, then obtain the following differential equation:
di L
+1.96 ⋅10 5 i L + 76.5 = 0
dt
Solution 11.
Before opening, the switch has been closed for a long time. we treat the inductor as a short circuit. The
voltages across the resistances R3 is equal to zero, since it is in parallel to the short circuit, so all the current
flow through the resistor R1 and R2 :
⎛ 1
1 ⎞
1 ⎞
⎛ 1
⎟⎟ = 12⎜
iL (0) = VS ⎜⎜ +
+
⎟ = 4 mA
R
R
6K
6K
⎝
⎠
2⎠
⎝ 1
After the switch has been opened for a long time, we have again a steady-state condition, and we treat the
inductor as a short circuit. When the switch is open, the voltage source is not connected to the circuit. Thus,
i L (∞) = 0 A.
Solution 12.
Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we
treat the capacitor as an open circuit. When the switch is open, the voltage source is not connected to the
circuit. Thus,
vC (0) = 0 V.
After the switch has been closed for a long time, we have again a steady-state condition, and we treat the
capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance
R2 :
vC (∞) =
R2
1800
V1 =
12 = 8.71 V.
R1 + R2
1800 + 680
Solution 13.
Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we
treat the capacitor as an open circuit. When the switch is open, the voltage source is not connected to the
circuit. Thus,
vC (0) = 0 V.
After the switch has been closed for a long time, we have again a steady-state condition, and we treat the
capacitor as an open circuit. Since the current flowing through the resistance R2 is equal to zero, the
voltage across the capacitor is equal to the voltage across the resistance R3 :
vC (∞) =
R3
1800
V1 =
12 = 8.71 V.
R1 + R3
1800 + 680
Solution 14.
In a steady-state condition we can treat the inductor as a short circuit. Before the switch changes, applying
the KVL
V − VS 2
⇒ i L (0) = S 1
= 0 mA.
R1 + R2
After the switch has changed for a long time, we have again a steady-state condition, and we treat the
inductor as a short circuit. Thus, applying the KVL
VS 2
i L (∞) = −
= −0.39 mA.
R2 + R3
VS1 − VS 2 = (R1 + R2 )i L (0)
Solution 21.
Determine the steady state current through the inductor at t < 0 . If this current is equal to the current
specified, steady-state conditions did exit; otherwise, opening the switch interrupted a transient in progress.
To determine the steady-state current before the switch was opened, replace the inductor with an equivalent
DC short-circuit and compute the steady-state current through the short circuit.
At steady state, the inductor is modeled as a short circuit:
( )
( )
VR 3 0 − = 0
iR3 0 − = 0
Thus, the short-circuit current through the inductor is simply the current through
by applying Kirchoff’s Laws and Ohm’s Law.
V
12
iR 2 0 − = s =
− Vs + iR3 0− R2 = 0
= 2mH
Apply KVL:
R2 6 × 103
R2 which can be found
( )
( )
(0 )+ i (0 )+ i (0 ) = 0, i (0 ) = i (0 ) = 2mH
−
−
−
−
− iR 2 −
Apply KCL:
L
R3
L
R2
The actual steady state current through the inductor is larger than the current specified. Therefore, the
circuit is not in a steady state condition just before the switch is opened.
Solution 22.
To solve this problem, find the steady state voltage across the capacitor before the switch is thrown. Since
the voltage across a capacitor cannot change instantaneously, this voltage will also be the capacitor voltage
immediately after the switch is thrown. At that instant, the capacitor may be viewed as a DC voltage source.
At t = 0− :
Determine the voltage across the capacitor. At steady state, the capacitor is modeled as an open circuit:
( )
( )
iR1 0 − = iR 2 0 − = 0
Apply KVL:
( )
VS1 + 0 − VC 0 − + 0 − VS 2 = 0
( )= V
VC 0
+
−
S1 − VS 2
= −95V
At t = 0 :
( ) ( )
(0 )= i (0 )
VC 0 + = VC 0 − = −95V
+
iR 2 +
R3
Apply KVL:
( )
( )
(0 ) =
( )
VS 2 − iR3 0 + R2 + VC 0 + − iR3 0 + R3 = 0
( )
V +V +
iR3 0 + = S 2 C
R2 + R3
130 − 95
7 × 103 + 23 × 103
= 1.167mA
Solution 23.
At t = 0− , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there
will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause
a current flow through R2 which will dissipate energy until no energy is stored in the capacitor at which
time the current ceases and the voltage across the capacitor is zero. These are the steady state conditions,
i.e.:
( )
( )
iC 0 − = 0
VC 0 − = 0
+
At t = 0 , the switch is closed and the transient starts. Continuity requires:
( ) ( )
VC 0+ = VC 0 − = 0
At this instant, treat the capacitor as a DC voltage source of strength zero i.e. a short-circuit. Therefore, all
of the voltage V1 is across the resistor R1 and the resulting current through R1 is the current into the
capacitor.
Apply KCL: (Sum of the currents out of the top node)
( ) V (0R ) − 0 + V (0R ) − V = 0
(0 ) = VR = 0.6812× 10 = 17.65mA
iC 0 + +
C
+
C
2
iC
+
+
1
1
1
3
1
The CURRENT through the capacitor is NOT continuous but changes from 0 to 17.65mA when the switch
is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE
INSTANTANEOUSLY.
Solution 26.
In steady-state the inductor acts like a short-circuit so it has no voltage across it for t < 0. However, its
current is non-zero and is equal to the current out of the source VS and through RS. At the instant the switch
is changed the current through the inductor is unchanged since the current through an inductor cannot
change instantaneously. Also notice that after the switch is changed the current through R1 is always equal
to the inductor current and the voltage across R1 is always equal to the inductor voltage. Thus, at t = 0+ the
voltage across the inductor must be non-zero. That’s fine since the voltage across an inductor can change
instantaneously (or relatively so.)
Assume a polarity for the voltage across the inductor.
t = 0− : Steady state conditions exist. The inductor can be modeled as a short circuit with: VL 0− = 0
Apply KVL;
− V S + i L (0 − ) R S + V L ( 0 − ) = 0
()
i L (0 − ) =
VS
12
=
= 17.14 A
R S 0.7
( ) ( )
At t = 0 , the transient commences. Continuity requires: iL 0+ = iL 0 −
Apply KVL:
+
( )
( )
( )
( )
iL 0+ R1 + V L 0 + = 0 ⇒ V L 0 + = − iL 0+ R1 = −17.14 × 22 × 10 3 = −337.1kV
Solution 32.
The current source and the capacitor will be in series so the current to the capacitor will be constant at I0.
Therefore, the rate at which charge accumulates on the capacitor will also be constant and, consequently,
the voltage across the capacitor will rise at a constant rate. The integral form of the capacitor i-V
relationship best expresses this accumulation process. The continuity of the voltage across the capacitor
requires:
( ) ( )
VC 0+ = VC 0 − = −7V
iC (t ) = I 0 = 17mA
1 t
1 0
t
∫ iC (t )dt = ( ∫− ∞ iC (t )dt + ∫0 I 0 dt )
C −∞
C
I
I
=VC 0+ + 0 ∫0t dt =VC 0+ + 0 t |t0
C
C
VC (t ) =
( )
= −7 +
( )
17 × 10 − 3
0.55 × 10
−6
t = −7 + 30.91 × 103 t
Solution 35.
At t = 0− :
Assume steady state conditions exist. At steady state the inductor is modeled as a short circuit:
VL (0− ) = 0
The current through the inductor at this point is given directly by Ohm’s Law:
i L (0 − ) =
VG
RG + R1
At t = 0+ :
Continuity of the current through the inductor requires that:
VG
i L (0 + ) = i L (0 − ) =
RG + R1
Note that the voltage across the gap VR was
V R
V R ( 0 + ) = −i L ( 0 + ) R = − G
written as –23 kV since the current from the
RG + R1
inductor flows opposite to the polarity shown for
VG R
12 × 1.7 × 10 3
− RG = −
R1 = −
− 0.37 = 0.5170Ω VR; that is, the actual polarity of the voltage
across R is opposite that shown.
V R (0 + )
− 23 × 10 3
For t > 0:
Determine the Thevenin equivalent resistance as "seen" by the inductor, ie, with respect to the port or
terminals of the inductor:
Req = R1 + R
τ=
L
L
=
Req R1 + R
L = τ ( R1 + R) = 13 × 10 − 3 × (0.5170 + 1.7 × 103 ) = 22.11H
Solution 37.
At t = 0− , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there
will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause
a current flow through R2 which will dissipate energy until no energy is stored in the capacitor at which
time the current ceases and the voltage across the capacitor is zero. Thus, in this circuit, and others like it
that contain no connected sources prior to the switch being thrown, the steady state condition is zero
currents and zero voltages.
( )
( )
iC 0 − = 0
+
VC 0 − = 0
At t = 0 , the switch is closed; the transient starts. Continuity requires:
( ) ( )
VC 0+ = VC 0 − = 0
Since the voltage across the capacitor is zero at this instant, the voltage across the resistor R2 is also zero at
this instant which implies that no current passes through the resistor at t = 0+. Therefore, at t = 0+, the
current out of the source must be the current into the capacitor.
12
V
= 30 A
iC 0 + = 1 =
R1 0.4
The CURRENT through the capacitor is NOT continuous but changes from 0 to 30 A when the switch is
closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE
INSTANTANEOUSLY.
( )
Solution 61.
The conditions that exist at t < 0 have no effect on the long-term DC steady state conditions at t → ∞ . In
the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open
circuit. In this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across
these branches and the current through them are zero. In other words all of the current produced by the 9V
source travels through the inductor in this case.
V
9
i L (∞) = S 2 =
= 31.03 mA
RS 2 290
Of course, this current is also the current traveling through the 290 Ω resistor.
i RS 2 (∞) = i L (∞) = 31.03 mA
And since the voltage across the inductor in the long-term DC steady state is zero (short circuit)
0 + VC (∞) + i R2 (∞)R2 = 0
VC (∞) = 0
Solution 63.
As t → ∞, the circuit will return to DC steady state conditions. In the long-term DC steady state the
inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the
current through R2 is zero and thus the voltage across the capacitor must be equal to the voltage across R1.
Furthermore, the current through the inductor and R1 is simply VS/(RS + R1).
VS
170
iC (∞) = 0
i L (∞) = i R1 (∞) =
=
≅ 18.3 mA
RS + R1 7000 + 2300
VR1 (∞) = i L (∞) R1 = 18.28 ×10−3 × 2.3×10 3 = 42.04 V
and
VC (∞) = VR1 (∞) = 42.04 V