University of California, Berkeley
EE 42/100
Spring 2012
Prof. A. Niknejad
Problem Set 5
Solutions
Please note that these are merely suggested solutions. Many of these problems can be
approached in different ways.
1. (a) No. v(t) is continuous, as v(0) = (60 − 40e0 ) V = 20 V = v(0− ).
(b) First we observe that i(0− ) = 0, since voltage is constant before the switch is
thrown. After t = 0, we have
i(t) = C
dv(t)
= (0.2 mF)(200e−5t V) = 40e−5t mA
dt
Clearly, i(0+ ) = 40 mA, so it is not continuous, leading to an instantaneous change.
(c) Energy is given by w(0) = 12 Cv 2 (t = 0) = 21 (0.2 mF)(20 V)2 = 40 mJ.
(d) After a long time, v(t) converges to 60 V. Using the same formula as before, we
have w(∞) = 21 Cv 2 (t = ∞) = 12 (0.2 mF)(60 V)2 = 0.36 J.
2. Under DC conditions, all capacitors become open circuits. We redraw the circuit,
along with some nodal voltages:
+ v3 −
20 kΩ
vx
30 kΩ
+
v1
−
+
v2
−
vy
15 kΩ
5 kΩ
+
15 V
+
v4
−
10 kΩ
Since no current flows across the two leftmost resistors, we have v1 = v2 = vx . Also,
v3 = vx − v7 and v4 = vy . To find vx and vy , we can use a simple voltage divider:
vx =
15 + 10
(15 V) = 12.5 V
5 + 15 + 10
10
(15 V) = 5 V
5 + 15 + 10
So v1 = v2 = 12.5 V, v3 = 7.5 V, v4 = 5 V.
vy =
3. The required energy is E = P t = (1.5 kW)(40 × 60s) = 3.6 MJ. So we need a
MJ)
capacitance given by C = V2E2 = 2(3.6
= 180 F. The amount of charge is
(200 V)2
Q = CV = (180 F)(200 V) = 36 kC. As a parallel plate capacitor, it would have an
180×10−6
6
2
=
area of A = Cd
−12 ) = 1.36 × 10 m .
(15)(8.85×10
r 0
4. At the time that the first switch closes, the capacitor sees the first 15 kΩ resistor.
Hence the differential equation is
dvC
vC − 20
+ (200 µ)
=0
15 k
dt
Rearranging the equation into a standard form gives us
vC + 3
dvC
= 20
dt
Hence, we have a solution of the form vC (t) = Ae−t/3 + B, where the first term is the
homogeneous solution and the second is the particular solution. The time constant is
simply the coefficient in front of dvdtC , so τ = RC = 3 s. The initial and final
conditions are v(0) = 0 and v(∞) = 20, so the full solution is
vC (t) = 20 − 20e−t/3 , 0 ≤ t ≤ 10
When switch 2 closes, the equivalent resistance seen by the capacitor is now
(15 kΩ)||(15 kΩ) = 7.5 kΩ. So the time constant is τ = Req C = 1.5 s. The form of the
solution remains the same, but we have different conditions. The initial condition is
equal to the voltage at t = 10 from before: vC (10) = 20 − 20e−10/3 = 19.3 V. The
final condition is vC2 (∞) = 10, as the capacitor is open in steady-state and we have a
voltage divider. Hence our solution is
vC (t) = 10 + 7308e−t/1.5 , t ≥ 10
Graphically, we have a growing exponential to 10 V, followed by a decaying
exponential to 10 V.
20
18
16
14
vC(t)
12
10
8
6
4
2
0
0
2
4
6
8
2
10
t
12
14
16
18
20
5. At the time of the switch action, both capacitors have 20 V across them, as they act
as open circuits. They do not change instantaneously, so they remain at 20 V each
after the switch opens. Thus, the voltage drop across the resistor remains 0, and no
current flows. We would also expect the current to be 0 at steady state at t = ∞, so
i(t) = 0 for all time. You would get this solution even if you were to solve the
governing ODE due to the 0 boundary conditions.
6. The differential equation is
vC (t) + RC
dvC (t)
= v(t) = t
dt
If we use the standard homogeneous solution along with the suggested particular
solution, we get
vC (t) = A + Bt + De−t/RC
The initial condition is that vC (0) = 0, so A + D = 0. Now if we plug in the solution
into our original ODE:
D −t/RC
−t/RC
A + Bt + De
+ RC B −
e
=t
RC
We now match terms to determine the coefficients’ values. The only linear term on
the LHS is Bt, so we must have B = 1. The constant terms are A + RCB = A + RC,
and this must be equal to 0, since there are no constants on the RHS. Thus,
A = −RC and D = −A = RC. The full solution is
vC (t) = −RC + t + RCe−t/RC = t − RC(1 − e−t/RC )
While we need a specific value for RC for an accurate plot, we can sketch a general
characteristic by assigning a value of, say, 1 to RC.
10
9
8
7
vC(t)
6
5
4
3
2
1
0
0
1
2
3
4
5
t
6
7
8
9
10
Notice that the voltage originally exhibits a delay due to the exponential term. As
time passes, the capacitor’s voltage becomes linear and follows that of the source
almost identically after it gets past the initial “inertia” presented by the capacitor.
3
7. (a) Writing a KCL equation at v(t) gives us
vC (t)
dvC (t)
dvC (t)
+ (2 µ)
= i(t) = (10 µ)e−t ⇒ vC (t) +
= 5e−t
500 k
dt
dt
(b) The time constant is τ = RC = 1 s, the coefficient in front of the derivative term.
The complementary (homogeneous) solution thus takes the form v0 (t) = Ae−t .
(c) If we were to use a particular solution of the form vp (t) = Ke−t , we get the same
form as that of the complementary solution. But the complementary solution only
works for a homogenous ODE (i.e., the forcing function is 0):
vC (t) +
dvC (t)
= Ke−t − Ke−t = 0
dt
(d) If we were to try vp (t) = Kte−t instead and plug it into the ODE, we now obtain
Kte−t + Ke−t − Kte−t = 5e−t
So the constant is K = 5. Combining it with the homogeneous solution, we have
vC (t) = (A + 5t)e−t . If we assume that the capacitor is initially uncharged, then
vC (0) = 0 and A = 0. So our full solution is vC (t) = 5te−t .
8. In using superposition, we consider one source at a time. Let’s deal with v(t) first.
Then the current source becomes an open circuit, and the equivalent resistance seen
by the capacitor is 1 kΩ (the two 1 k are in series, and they are in parallel with the
2 k). Hence the time constant is τ = Req C = (1 kΩ)(0.5 µF) = 0.5 ms. The forcing
function (voltage source) is constant, so we seek a solution of the form
vO1 (t) = A + Be−2000t . Using the conditions vO1 (0) = 0 and vO1 (∞) = 0.5 V (the
capacitor becomes an open circuit, and we have a simple voltage divider at vO1 (t)),
the full solution is vO1 (t) = 0.5(1 − e−2000t ).
Next we have the current source, and v(t) becomes a short. To make this easier to
deal with, we can use a source transformation to turn the current source and resistor
into a Thévenin equivalent with a voltage source (alternatively, you can write some
KCL equations).
1 kΩ
1 kΩ
1V +
vO (t)
2 kΩ
0.5 µF
Notice that the circuit is nearly identical to the previous one. So the solution is
(without re-solving it) vO2 (t) = 0.5(1 − e−2000t ).
The full solution is the sum of the two individual components: vO (t) = 1 − e−2000t .
4
9. The golden rules hold, as the op amp is in negative feedback. Defining the node
between the capacitor and resistor in the feedback loop to be vx , we can write KCL
equations at vx and at the inverting input.
C
d(vx − 0) vx − vo
+
=0
dt
R2
0 − vi
d(0 − vx )
+C
=0
R1
dt
We can integrate the second equation to solve for vx and plug it back into the first:
Z
Z
1
1
vi (t)
vo (t)
−
=0
vx (t) = −
vi (t)dt ⇒ −
vi (t)dt −
R1 C
R1
R1 R2 C
R2
Z
R2
1
⇒ vo (t) = − vi (t) −
vi (t)dt
R1
R1 C
10. Since the op amp merely serves as a buffer, we have vi (t) on the left side of the
capacitor. The voltage on the right side is io (t)R by Ohm’s law. Then KCL gives us
io (t) + C
d(io (t)R − vi (t))
=0
dt
dio (t)
dvi (t)
=C
dt
dt
We cannot simplify further since we do not know the form of vi (t). We can infer that
io (t) will have a homogeneous solution Ae−t/RC , but the form of the particular
solution depends on vi (t).
⇒ io (t) + RC
5