Electric Circuits
Fall 2015
Solution #5
RULES:
 Please try to work on your own. Discussion is permissible, but identical
submissions are unacceptable!
 Please show all intermediate steps: a correct solution without an explanation will
get zero credit.
 Please submit on time. NO late submission will be accepted.
 Please prepare your submission in English only. No Chinese submission will be
accepted.
5.1 [10%] Suppose the two capacitors in Fig. 1 are initially uncharged. Then we connect a
15V source as shown. Find the voltages 𝑣1 and 𝑣2 at steady state by considering the
conservation of charge on each plate.
Fig. 1
Solution:
Since both capacitors are initially uncharged, conservation demands that the net charge at
each node remain 0 after the voltage source is connected. We apply this principle to the
node between the two capacitors. Since we are looking at the negative plate of cap 1 and
the positive plate of cap 2, we must have that
−𝑞1 + 𝑞2 = 0 [3pts]
Remember that the charge on parallel-plate capacitors are equal and opposite. Another
relation we can construct using KVL is the following:
𝑞
𝑞
𝑣1 + 𝑣2 = 𝐶1 + 𝐶2 = 15 [3pts]
1
2
The above two equations give us a system that we can solve. The solutions are 𝑞1 = 𝑞2 =
0.126 mC [2pts]. With this we find that the voltages are
𝒗𝟏 = 𝟏𝟎. 𝟓 𝑽 [1pt]
𝒗𝟐 = 𝟒. 𝟓 𝑽 [1pt]
5.2 [8%] Determine 𝐿𝑒𝑞 that may be used to represent the inductive network of Fig. 2 at the
terminals.
Fig. 2
1
Electric Circuits
Fall 2015
Solution #5
Solution:
_
+
+
+
_
_
Let the voltage across terminal a-b be 𝑣𝑎𝑏 , and the input current be i. We know that
𝑑𝑖
𝑣𝑎𝑏 = 𝐿𝑒𝑞 𝑑𝑡 [2]
(1)
[2] By applying KVL, we can find that
𝑣𝑎𝑏 = 𝑣4𝐻 + 𝑣1
(2)
𝑑𝑖
𝑣1 = 2 𝑑𝑡 + 𝑣2
(3)
[2] For the 3H and 5H inductor branches,
𝑣1 = 3
𝑑𝑖1
𝑑𝑡
𝑣2 = 5
𝑑𝑖2
𝑑𝑡
3
𝑑(𝑖1 +𝑖2 )
𝑑𝑡
⇒ 𝑣1 + 5 𝑣2 = 3
𝑑𝑖
= 3 𝑑𝑡
(4)
Solving equation (3) and (4),
𝑣1 =
21 𝑑𝑖
8 𝑑𝑡
(5)
5 𝑑𝑖
𝑣2 = 8 𝑑𝑡
Since
𝑑𝑖
𝑣4𝐻 = 4 𝑑𝑡
(6)
[2] Substituting (5) (6) into (2),
𝑣𝑎𝑏 = 4
𝑑𝑖
𝑑𝑡
⇒ 𝑳𝒆𝒒 =
+
𝟓𝟑
𝟖
21 𝑑𝑖
8 𝑑𝑡
=
53 𝑑𝑖
8 𝑑𝑡
= 𝟔. 𝟔𝟐𝟓 H
5.3 [10%] The current in a 4 H inductor is
𝑖 = 10 A,
𝑡 ≤ 0;
𝑡
𝑖 = (𝐵1 cos 4𝑡 + 𝐵2 sin 4𝑡)𝑒 −2 A, 𝑡 ≥ 0.
The voltage across the inductor (passive sign convention) is 60 V at t = 0. Calculate the
power at the terminals of the inductor at t = 1s. State whether the inductor is absorbing or
2
Electric Circuits
Fall 2015
Solution #5
delivering power.
Solution:
We know that 𝑝(𝑡) = 𝑣(𝑡)𝑖(𝑡), we can first get 𝑣(𝑡):
𝑑𝑖
𝑑
𝑣(𝑡) = 4 𝑑𝑡 = 4 𝑑𝑡 (𝐵1 cos 4𝑡 + 𝐵2 sin 4𝑡)𝑒 −𝑡/2 [2]
𝑡
⟹ 𝑣(𝑡) = 𝑒 −2 [(−16𝐵1 − 2𝐵2 ) sin 4𝑡 + (−2𝐵1 + 16𝐵2 ) cos 4𝑡] V [2]
Since 𝑖(0) = 10A, 𝑣(0) = 60V,
𝑖(0) = 𝐵1 = 10𝐴, 𝑣(0) = −2𝐵1 + 16𝐵2 = 60𝑉 ⟹ 𝐵2 = 5 [1]
𝑡
𝑖(𝑡) = (10 cos 4𝑡 + 5 sin 4𝑡)𝑒 −2 A, 𝑡 ≥ 0
⟹
𝑡
𝑣(𝑡) = (−170 sin 4𝑡 + 60 cos 4𝑡)𝑒 −2 V, 𝑡 ≥ 0
⟹ 𝑝(𝑡) = (−850𝑠𝑖𝑛2 4𝑡 + 600𝑐𝑜𝑠 2 4𝑡 − 1400 sin 4𝑡 cos 4𝑡)𝑒 −𝑡 W [3]
At t = 1s,
𝒑(𝟏) = (−𝟖𝟓𝟎𝒔𝒊𝒏𝟐 𝟒 + 𝟔𝟎𝟎𝒄𝒐𝒔𝟐 𝟒 − 𝟏𝟒𝟎𝟎 𝐬𝐢𝐧 𝟒 𝐜𝐨𝐬 𝟒)𝒆−𝟏 = −𝟑𝟑𝟗. 𝟓𝟕 W [1]
Since p(1) < 0, the inductor is delivering power at t = 1s. [1]
5.4 [8%] a) Show that the two coupled coils in Fig. 3 can be replaced by a single coil having
an inductance of 𝐿𝑎𝑏 = 𝐿1 + 𝐿2 + 2𝑀. (Hint: Express 𝑣𝑎𝑏 as a function of 𝑖𝑎𝑏 .)
b) Show that if the connections to the terminals of the coil labeled 𝐿2 are reversed,
𝐿𝑎𝑏 = 𝐿1 + 𝐿2 − 2𝑀.
Fig. 3
Solution:
a)
b)
+
+
_
_
_
_
+
+
+
_
_
+
a) 𝑣𝑎𝑏 = 𝐿𝑎𝑏
𝑑𝑖
𝑑𝑡
= 𝐿1
𝑑𝑖1
𝑑𝑡
+ 𝐿2
_
+
_
+
_
+
+
𝑑𝑖2
𝑑𝑡
+𝑀
𝑑𝑖2
𝑑𝑡
+𝑀
𝑑𝑖1
𝑑𝑡
= (𝐿1 + 𝐿2 + 2𝑀)
_
𝑑𝑖
𝑑𝑡
⟹ 𝐿𝑎𝑏 = 𝐿1 + 𝐿2 + 2𝑀 [4]
𝑑𝑖
b) 𝑣𝑎𝑏 = 𝐿𝑎𝑏 𝑑𝑡 = 𝐿1
𝑑𝑖1
𝑑𝑡
+ 𝐿2
𝑑𝑖2
𝑑𝑡
−𝑀
𝑑𝑖2
𝑑𝑡
−𝑀
𝑑𝑖1
𝑑𝑡
𝑑𝑖
= (𝐿1 + 𝐿2 − 2𝑀) 𝑑𝑡
⟹ 𝐿𝑎𝑏 = 𝐿1 + 𝐿2 − 2𝑀 [4]
3
Electric Circuits
Fall 2015
Solution #5
5.5 [10%] Consider the circuit in Fig. 4. Find 𝑖(𝑡) for 𝑡 < 0 and 𝑡 > 0.
Fig. 4
Solution:
+
_
1. 𝑡 < 0
Before t = 0, the circuit has reached steady state so that the capacitor acts like an open
circuit. Apply nodal analysis for node A (let its node voltage be 𝑣𝐴 ), we can easily obtain
80 − 𝑣𝐴
𝑣𝐴
+ 0.5𝑖 = 𝑖 =
40
80
Then we get 𝑣𝐴 = 64V and i = 0.8A [2]
2. 𝑡 > 0
Let the current through the capacitor after the switch opened to be 𝑖1 .
KCL:
𝑖1 + 0.5𝑖 = 𝑖
[2] For the capacitor,
𝑖1 = −C
𝑑𝑣1
,
𝑑𝑡
𝑣1 = (30 + 50)𝑖
𝑑𝑖
⟹ −240 = 0.5𝑖
𝑑𝑡
[2] Therefore,
1
𝑑𝑖
𝑖
1
= − 480 𝑑𝑡.
[4] Solving this equation,
𝑡
ln(𝑖) = − 480 + 𝐶0 , where 𝐶0 is a constant number.
t = 0, i = 0.8A, substituting this and we get 𝐶0 = ln(0.8),
𝑖
𝑡
ln (0.8) = − 480
𝒕
⟹ 𝒊(𝐭) = 𝟎. 𝟖𝒆−𝒕/𝟒𝟖𝟎 A, t >0, or 𝒊(𝐭) = 𝟎. 𝟖𝒆−𝟒𝟖𝟎 𝒖(𝒕) A,
where 𝑢(𝑡) is unit step signal. (Other correct methods are also accepted.)
4
Electric Circuits
Fall 2015
Solution #5
5.6 [8%] For the op amp circuit of Fig. 5, let 𝑅1 = 10 kΩ, 𝑅𝑓 = 20 kΩ, 𝐶 = 20 𝜇F, and
v(0) = 1 V. Find 𝑣𝑜 .
Fig. 5
Solution:
For ideal op amp, we know that 𝑣3 = 0. Then 𝑣2 = 𝑣, and 𝑣1 = 4𝑢(𝑡)
At node 2,
𝑣1 −𝑣2
𝑅1
⟹
4𝑢(𝑡)−𝑣
𝑅1
𝑑𝑣
= C 𝑑𝑡
𝑑𝑣
= C 𝑑𝑡 , v(0)=1V
⟹ − ln(4u(t) − v) =
𝑡
𝑅1 𝐶
− ln(3)
⟹ 𝑣(𝑡) = 4𝑢(𝑡) − 3𝑒 −𝑡/𝑅1 𝐶 V [3]
At node 3,
C
𝑑𝑣 𝑣3 − 𝑣𝑜 −𝑣𝑜
=
=
𝑑𝑡
𝑅𝑓
𝑅𝑓
𝑑𝑣
⟹ 𝑣𝑜 = −𝑅𝑓 C 𝑑𝑡 = −
3𝑅𝑓
𝑅1
𝑒 −𝑡/𝑅1 𝐶 [3]
Substitute the values,
𝒗𝒐 = −𝟔𝒆−𝟓𝒕 𝑽, 𝒕 > 𝟎
[2]
or
𝒗𝒐 = −𝟔𝒆−𝟓𝒕 𝒖(𝒕) 𝑽
5.7 [10%] In the circuit shown in Fig. 6, the switch makes contact with position b just before
breaking contact with position a. As already mentioned, this is known as a
make-before-break switch and is designed so that the switch does not interrupt the
current in an inductive circuit. The interval of time between “making” and “breaking” is
assumed to be negligible. The switch has been in the position for a long time. At t = 0 the
switch is thrown from position a to position b.
a) Determine the initial current in the inductor.
b) Determine the time constant of the circuit for t > 0.
5
Electric Circuits
Fall 2015
Solution #5
c) Find i, 𝑣1 , and 𝑣2 for t ≥ 0.
d) What percentage of the initial energy stored in the inductor is dissipated in the 72 Ω
resistor 15 ms after the switch is thrown from position a to position b?
Fig. 6
Solution:
a) Before the switch is thrown from a, the circuit is in steady state and the inductor acts
like a short circuit. Therefore, the initial current through it is 𝑰𝟎 =24/12 = 2 A. [2]
b) The time constant is 𝛕 =
𝑳
𝑹𝒆𝒒
=
𝟏.𝟔
𝟖𝟎
= 𝟎. 𝟎𝟐 s. [2]
c) 𝒊(𝒕) = 𝑰𝟎 𝒆−𝒕/𝝉 = 𝟐𝒆−𝟓𝟎𝒕 𝒖(𝒕) A
𝒗𝟏 (𝒕) = 𝐋
𝒅𝒊
𝒅𝒕
= −𝟏𝟔𝟎𝒆−𝟓𝟎𝒕 𝒖(𝒕) V
𝒗𝟐 (𝒕) = −𝟕𝟐𝒊(𝐭) = −𝟏𝟒𝟒𝒆−𝟓𝟎𝒕 𝒖(𝒕) V [3]
1
d) At 0s, the initial energy stored in the inductor is w(0) = 2 𝐿𝑖(0)2 = 3.2 J
1
At 15 s, the energy stored in the inductor is w(15) = 2 𝐿𝑖(15)2 = 0.714 J
Therefore, the percentage of the initial energy dissipated in the 72 Ω resistor 15 ms is
(3.2−0.714)
3.2
∗
72
72+8
= 69.92% [3]
5.8 [8%] The switch in the circuit in Fig. 7 has been in position 1 for a long time. At t = 0, the
switch moves instantaneously to position 2. Find 𝑣𝑜 (𝑡) for t ≥ 0+ .
Fig. 7
Solution:
Initially the current through the inductor is
240
4
∗
12+4+40//10 5
= 8 A. [2]
After the switch moved to position 2, the circuit can be equivalent to an inductor in
series with 1 resistor, 𝑅𝑒𝑞 = (4+6)//40 +10 = 18 Ω. Let 𝑖(𝑡) be the current through the
𝐿
inductor when t > 0. Then 𝜏 = 𝑅 = 0.004 𝑠 [2]
𝑒𝑞
6
Electric Circuits
Fall 2015
Solution #5
Since 𝑖(𝑡) = 𝐼0 𝑒 −𝑡/𝜏 = 8𝑒 −250𝑡 𝑢(𝑡) A, we can find the current through the 40 Ω resistor
to be
10
𝑖(𝑡)
10+40
= 1.6𝑒 −250𝑡 𝑢(𝑡). [2]
Therefore,
𝒗𝒐 (𝒕) = 𝟏. 𝟔𝒆−𝟐𝟓𝟎𝒕 𝒖(𝒕) ∗ 𝟒𝟎 = 𝟔𝟒𝒆−𝟐𝟓𝟎𝒕 𝒖(𝒕) 𝐕, 𝐭 ≥ 𝟎+ [2]
5.9 [8%] The voltage across a 0.2 mF capacitor was 20V until a switch was opened at t = 0,
causing the voltage to vary with time as 𝑣(𝑡) = (60 − 40𝑒 −5𝑡 ) V for t > 0.
a) Did the switch action result in an instantaneous change in v(t)? Why or why not?
b) Did the switch action result in an instantaneous change in the current i(t)? Why
or why not?
c) How much energy was initially stored in the capacitor at t = 0?
d) How much energy will be stored in the capacitor at t = ∞?
Solution:
a) No, 𝑣0 = 60 − 40 = 𝑣(0− ). The voltage v(t) across a capacitor is continuous.
b) First we observe that i(0− ) = 0, since voltage is constant before the switch is
thrown. After t = 0, we have
𝑑𝑖
𝑖(𝑡) = C = (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.
1
1
c) Energy is given by 𝑤(0) = 2 𝐶𝑣 2 (𝑡 = 0) = 2 (0.2 mF)(20𝑉)2 = 40 mJ
d) After a long time, v(t) converges to 60 V. Using the same formula as before, we have
𝑤(∞) = 0.36 J
[2*4]
5.10 [10%] In Fig. 8, suppose that both switches have been open for a long time prior to
t = 0. Then switch 1 closes at t = 0, followed by switch 2 at t = 10 s. Use MATLAB to
plot 𝑣𝐶 (𝑡) for 𝑡 ≥ 0, assuming that 𝑣𝐶 (0) = 0. (Your MATLAB script should be
attached!)
Fig. 8
Solution:
At the time that the first switch closes, the capacitor sees the first 15 k resistor.
Hence the differential equation is
𝑣𝐶 −20
+
15𝑘
(200𝜇)
𝑑𝑣𝐶
𝑑𝑡
=0
[2]
Rearranging the equation into a standard form gives us
7
Electric Circuits
Fall 2015
Solution #5
𝑑𝑣𝐶
= 20
𝑑𝑡
Hence, we have a solution of the form 𝑣𝐶 = 𝐴𝑒 −𝑡/3 + 𝐵, where the first term is the
homogeneous solution and the second is the particular solution. The time constant is
𝑣𝐶 + 3
simply the coefficient in front of
𝑑𝑣𝐶
,
𝑑𝑡
so τ = RC = 3 s. The initial and final conditions
are v(0) = 0 and v(∞) = 20, so the full solution is
𝑡
𝑣𝐶 (t) = 20 − 20𝑒 −3 , 0 ≤ 𝑡 ≤ 10
[3]
When switch 2 closes, the equivalent resistance seen by the capacitor is now
(15 k)||(15 k) = 7.5 k. So the time constant is τ = RC = 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) = 19.3 𝑉. The final condition is v(∞) = 10, as the
capacitor is open in steady-state and we have a voltage divider. Hence our solution is
𝑡
𝑣𝐶 (t) = 10 + 7308𝑒 −1.5 , 𝑡 ≥ 10 [3]
Graphically, we have a growing exponential to 10V, followed by a decaying
exponential to 10V. [2]
5.11 [10%] Suppose the voltage source in the circuit of Fig. 9 is defined by a ramp function,
such that v(t) = 0 for t < 0 and v(t) = t for 𝑡 ≥ 0. If 𝑣𝐶 (0) = 0, derive an expression for
𝑣𝐶 (𝑡) for 𝑡 ≥ 0 and use MATLAB to sketch it to scale versus time. Consider trying a
particular solution of the form 𝑣𝐶 (𝑡)= A + Bt. (Your MATLAB script should be
attached!)
Fig. 9
Solution:
The differential equation is 𝑣𝐶 (t) + RC
𝑑𝑣𝐶
𝑑𝑡
= 𝑣(t) = 𝑡
8
Electric Circuits
Fall 2015
Solution #5
If we use the standard homogeneous solution along with the suggested particular solution,
we get
𝑣𝐶 (t) = 𝐴 + 𝐵𝑡 + 𝐷𝑒 −𝑡/𝑅𝐶 [2]
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
𝐴 + 𝐵𝑡 + 𝐷𝑒 −𝑡/𝑅𝐶 + 𝑅𝐶 (𝐵 − 𝑅𝐶 𝑒 −𝑅𝐶 ) = 𝑡 [3]
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
𝑡
𝑡
𝑣𝐶 (t) = −𝑅𝐶 + 𝑡 + 𝑅𝐶𝑒 −𝑅𝐶 = 𝑡 − 𝑅𝐶 (1 − 𝑒 −𝑅𝐶 ) [3]
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. [2]
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.
9