University of California, Berkeley
EE 42/100
Spring 2012
Prof. A. Niknejad
Problem Set 2
Solutions
Please note that these are merely suggested solutions. Many of these problems can be
approached in different ways.
1. In problems like this, you may find it helpful to redraw the circuit to ignore all the
irrelevant terminals and such that everything is rectangular. Note that since we are
finding Req from terminals a and b in each of these, we must simplify the circuits
toward these terminals.
(a) Terminals c and d are connected together, while e and f are irrelevant. If you
redraw the circuit as below, you can see that the two resistors in the first
column are in parallel as, they are connected together at both pairs of terminals
(due to the short). This is also true for the second column of resistors. So each
of the pairs can be redrawn as one resistor with 21 R.
d
R
Ra b
e
d
R
a
f
R
R
R
R
R
R
R
R
R
b
R/2
a
R
R
b
R/2
c
c
It is easy to see that all the resistors are in series, so Req = 3R.
(b) Terminals e and f are shorted. Intuitively, Req should only include the two inner
resistors, since the other four are shorted out. That is, any current flowing from
a to b would only go through 2R of resistance and proceed between e and f via
the short.
d
R
R
R a b
e
R
f
R
R
2R
R
a
R
R
e
R
f
R
R
b
R
a
b
R
a
R
R
2R
c
(c) The short between c and e gets rid of the one resistor between the two nodes.
After that, everything else simplifies as before.
b
d
R
Ra b
e
d
R
a
f
R
R
R
R
2R
R
R
e
f
c
R
b
R
a
R
b
a
b
8R/3
R
R
c
2. The key to this problem is that the ladder is infinite. Using this fact, we can conclude
that the equivalent resistance Rin should not change if we remove one “rung” of the
ladder. So we can replace the ladder, save for the first rung, with one resistor Rin .
R
R in
R
R in
R
Now we can write an equation in terms of R and Rin and solve.
Rin = R + (R||Rin ) + R = 2R +
RRin
R + Rin
2
Rin
− 2RRin − 2R2 = 0
Since this is a quadratic equation, there are two solutions. We know that resistance
cannot be negative, so we discard the negative solution. The positive solution is
√
Rin = (1 + 3)R
3. We have two unknown nodes, so we’ll have two nodal equations, one at node 1 and
the other at node 2.
V1 − V2 V1
+
−8=0
200
30
V2 − V1 V2
+
−3=0
200
20
Solving this system gives us V1 = 218.4 V and V2 = 74.4 V.
We can also do this using superposition. Here we split the problem into two
subcircuits, each one with a different current source zeroed out. Recall that the
zeroed out current source becomes an open circuit.
200Ω
1
30Ω
8A
2
20Ω
2
V1 − V2 V1
+
−8=0
200
30
V2 − V1 V2
+
=0
200
20
This first circuit gives us V1 = 211.2 V and V2 = 19.2 V.
200Ω
1
2
-3A
20Ω
30Ω
V1 − V2 V1
+
=0
200
30
V2 − V1 V2
+
−3=0
200
20
The solutions are V1 = 7.2 V and V2 = 55.2 V. Clearly, the sum of these with the
solutions for the first subcircuit yields the same answers as the original circuit.
4. There are five nodes in this circuit, labeled below. Placing the reference node at vE
reduces the number of unknowns to three, since now vE = 0 V and vD = 5 V.
50Ω
A
50Ω
18Ω
5V
75Ω
1.2A
30Ω
E
+
D
C
B
The three nodal equations are the following:
VA − VC VA − VB VA − 5
+
+
= 0 → 43VA − 25VB − 9VC = 45
50
18
50
VB − VA VB − VC
+
− 1.2 = 0 → −25VA + 31VB − 6VC = 540
18
75
VC − VA VC − VB VC
+
+
= 0 → −3VA − 2VB + 10VC = 0
50
75
30
The solutions are VA = 32.8 V, VB = 47.6 V, and VC = 19.3 V. The branch currents
can be found using Ohm’s law and KCL: IAC = 0.27 A, IAB = −0.82 A, IBC = 0.38 A,
IAD = 0.56 A, ICE = 0.64 A, IDE = 0.56 A.
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5. The circuit has four unknown nodes, labeled below.
+
5V +
−
500Ω
VA
VB
1V
+
100Ω
10 Vx
VC
+
Vx
−
80Ω
505Ω
Vx
90Ω
In addition, we must also write a supernode equation at the 1 V source. The
dependent current source does not present a problem, since we treat Vx as one of the
unknowns. The nodal equations at VA and Vx are
VA − 5 VA VA − VB
+
+
− 10Vx = 0
100
80
500
Vx − VC Vx
+
=0
505
90
The two equations at the supernode are
VB − VA VC − Vx
+
=0
500
505
VB − VC = 1
The solution to these equations is VA = 0.966 V, VB = 0.981 V, VC = −0.019 V,
Vx = −0.003 V.
6. Note that the left part of the circuit is independent of the right part. Since there is no
complete loop connecting the two parts, no current can flow between them; the wire
connecting them only serves as a common ground. The equivalent resistance on the
left is 3 + 12||12 = 9 Ω. So the total current from Vs is just V9s . Also, we have that Ix
is just half of that, since the two parallel branches have equal resistance. So Ix = V18s .
On the right half, the two resistors are in parallel, so the equivalent resistance is just
s
2 Ω. Since the current going through them is given by AIx = AV
, the voltage across
18
AVs
Vout
them is Vout = IReq = 9 . Solving for A and plugging in Vin = 9, we have that
A = 81.
7. The steps to finding the equivalent circuits are to find Voc , Isc , and/or Req . Since
there is a dependent source in this circuit, it is easier to find the first two. Voc is
simply the voltage difference between a and b. By inspection, this is just
Voc = −(100 kΩ)βIx = −(100 kΩ)β
Vs
= −10βVs
10 kΩ
Isc is found by shorting the terminals and solving for Iab . Note that shorting out the
s
100 kΩ resistor will leave us with Iab = −βIx = −β 10VkΩ
(note the sign!).
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s
Thus, our desired values are VT h = −10βVs and IN = − 10βVkΩ
. Also, the Thévenin and
VT h
Norton resistance is just given by RT h = RN = IN = 100 kΩ.
8. (a) The maximum power transfer theorem tells us that the load resistance should be
equal to the Thévenin resistance of the circuit. In this case, it is trivial to see
that that RL = RT h = Rs . If you want to review the derivation of this result,
please consult the lecture notes or textbook.
(b) We want the power across the load, PL , to be equal to 80% of the power
delivered by the voltage source, Ps . The former can be found using PL = I 2 RL ,
whereas the latter is Ps = IVs . In this circuit,
I=
Vs
Rs + RL
So putting everything together, we have
Vs2 RL
Vs2
=
0.8
(Rs + RL )2
Rs + RL
Solving for RL gives us RL = 4Rs .
(c) Power is dependent on both voltage and current. While the answer to part (b)
gives us a higher voltage across RL and a greater fraction of the total power, the
total power is actually much smaller in this case. This is because the current is 5
times smaller due to the greater resistance, while the voltage across RL is only
1.6 times larger.
9. Intuitively we know that closing the switch puts R2 and R3 in parallel, lowering the
effective resistance. So it must be in the open case that vo takes the greater voltage
value, 26 V. In this scenario, the circuit looks like the following:
50 Ω
+
39V +
−
R2
+
vo = 26V
−
Solving for this circuit, using voltage divider, nodal analysis, or otherwise, gives us
R2 = 100 Ω.
Now if we close the switch, vo = 24 V. The value of R2 remains unchanged, so all
that’s remaining is to solve for R3 .
50 Ω
+
39V +
−
100 Ω
5
+
vo
−
R3
We can write a nodal equation at vo = 24 V:
vo − 39
vo
vo
+
+
=0
50
100 R3
The solution is R3 = 400 Ω.
10. (a) When the bridge is balanced, no current flows across the galvanometer. By KCL,
the current across R1 and R2 are the same, as well as the current across R3 and
R4 . In addition, the voltages va and vb must be equal due to the 0 current.
+
R1
I1
a
Vs
R3
b
G
R2
I2
Rx
By KVL, the voltage drops across R1 and R3 must then be the same. The same
reasoning goes for R2 and Rx .
I1 R1 = I2 R3
I1 R2 = I2 Rx
Substituting for the currents and solving for Rx , we have Rx =
R2 R3
.
R1
(b) Given our values for R1 and R3 , we can measure Rx = 2R2 . Since the highest
value that R2 can take is 1000 Ω, the maximum resistance that can be measured
for Rx is just 2000 Ω. As R2 comes in increments of 10 Ω, the measurement can
be made with an accuracy to within 20 Ω.
(c) The Thévenin equivalent is found by finding the voltage and equivalent
resistance across the open terminals a and b. Note that here we cannot assume
the bridge is balanced, as we are looking at the general case. voc is simply the
difference in potential between nodes a and b, both of which can be found via a
voltage divider:
R2
va =
Vs
R1 + R2
Rx
Vs
vb =
R3 + Rx
R2
Rx
VT h = va − vb =
−
Vs
R1 + R2 R3 + Rx
The equivalent resistance can be found by zeroing out the voltage source,
making it a short, and finding Req .
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R1
R3
a
b
R2
Rx
We have R1 and R2 in parallel with each other, and R3 and Rx in parallel with
each other. These two pairs are in series, so the equivalent resistance is
RT h = (R1 ||R2 ) + (R3 ||Rx ).
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