Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.002 { Electronic Circuits
Spring 1999
Homework #2 Solutions
Handout S99-019
Due 2/17/99
Exercise 2.1:
Determine the resistance of each network shown below as viewed from its port.
R1
R2
R1
R2
R1
R3
R2
R3
R1
R2
R3
R3
Network (A)
Network (B)
Network (C)
Network (D)
ANSWER: All four circuits are simple and the equivalent resistances can be read o directly.
(A) Req = R + R + R
(B) Req = R + R2 k R = R + RR22RR33
(C) Req = R k (R + R ) = RR11 RR22 RR33
(D) Req = R k R k R = R2 R3 RR1 R1 R2R3 3 R1 R2
1
2
1
1
1
3
3
2
2
3
3
1
+
(
+
+
+
+
)
+
Exercise 2.2:
For both networks shown below, nd the current through and the voltage across
resistor R1 . Hint: make use of the results of Exercise 2.1.
R2
R1
R2
R1
R3
V +Network (A)
I
R3
Network (B)
ANSWER: Redraw Network (A) as shown below, we see that
VR1 = R R+ Rk Rk R V
1
2
3
1
3
R1R3
V
= R R +R
1 3
1 R2 + R2 R3
IR1 = VR1 =R1
= R R + RRR3 + R R V
1 3
1 2
2 3
Similarly, redraw Network (B) as shown below, we see that
IR1 = R + R
R +R I
VR1 = IR1 R
= R +RRR + R I
3
1
2
3
1
3
1
V +
-
2
R2
R1
1
3
R2
I
R3
R1
R3
Network (A) Redrawn
Network (B) Redrawn
We can further simplify the circuits by applying Norton's Theorem to Network (A) and Thevenin's
Theorem to Network B as shown. Now we see that Network (A) is just a simple current divider
and Network (B) is just a simple voltage divider. We get the same results as before.
Thevenin Equivalent
Norton Equivalent
I=V/R2
R2
R1
V=IR3 +
-
R3
R3
R2
R1
Network (b)
Network (a)
Exercise 2.3:
Following the node method, develop a set of simultaneous equations for each
network shown below that can be used to solve for the three unknown node voltages in the corresponding network. Express these equations in the form
2
e
6
G4 e
e
1
3
7
2 5
=S
3
where G is a 3 3 matrix of conductance terms and S is a 3 1 vector of terms involving the
sources.
e2
I
R1
R3 e
2
e1
V +-
R5 e
3
e1
R6
R4
R2
R3
R1
R4
e3
I
R2
V +-
R5
Network (B)
Network (A)
ANSWER: You could arrive at the conductance matrix by writing down the node equations and
combining similar terms. This method has been worked out in detail in both the class notes and
Problem 2 of Problem Set 1. However, as you get more familiar with this method, you should be
able to \read" o the conductance matrix G and the source matrix S by simply looking at the
circuit. The following outlines the procedure. Keep in mind that the following procedure has to be
modied if there are supernodes in the circuit.
1. G(i, i) = summation of all conductances connected to node i.
2. G(i, j) = negative of the summation of all conductances connected between node i and j for
i 6= j. Note that G is a symmetric matrix.
3. S(i) = summation of: (1) current sources connected to node i, with current going in as positive
and current going out as negative, and (2) terms in the form V G, where G is the conductance
connecting the voltage source V and node i. Again, the sign is determined by following the
sign of the current going from V to node i.
4. See your TA if you need further explanation.
For network (A):
2
6
4
G +G +G
;G
0
;G
G + G + G ;G
0
;G
G +G
1
2
3
3
3
3
4
5
5
5
5
32
76
54
6
3
e
e
e
1
2
2
VG
7
6
5 = 4 ;I
3
1
0
3
For network (B):
2
6
4
G + G ;G
0
;G G + G
;G
0
;G G + G + G
1
2
1
32
1
1
3
3
76
54
3
5
4
3
e
e
e
3
1
7
2 5
3
2
3
0
= 64 I
VG
7
5
4
7
5
Problem 2.1:
Find the Thevenin and Norton equivalents of the following networks, and graph
their i-v relations as viewed from their ports.
R
R
R
i
i
+
v
-
+
v
-
I
R
2R
Network (A)
R
+
V -
Network (B)
4R
i + v
4R
R
R
i + 4R
v4R
R
R
+
V -
Network (C)
I
Network (D)
ANSWER:
Req
i +
VTh
+
-
v
i +
IN
Req
v
-
Norton equivalent
Thevenin equivalent
For each circuit, we want VTh, IN , and Req , as represented in the circuits above. Note that the
current at each port is dened to go into the positive terminal. This is the opposite direction
from our denition of isc , which is the current we nd from the positive terminal to the negative
when we load the terminals with a short circuit. So, VTh = voc (the voltage at the port when it is
open-circuit), IN = isc (the current through the port when it is short-circuit). This tells us further
that the v-intercept of the i-v relation is VTh , while the i-intercept is ;IN . Note that shutting o
the sources and nding the equivalent resistance seen at the port determines the slope of the i-v
relation. Note also that the v-i relations will be of the form
v = Req i + VTh
while another way to relate them is with the inverse function
i = R1 v ; IN
eq
(A) It's easy to nd IN since when we short the terminals, 2R is shorted out of the circuit.
IN = I2
To nd Req , we open the current source, and we get
Req = 2R k (R + R) = R
We nd VTh through the following relation:
VTh = IN Req = IR
2
(B) Here nding VTh is easy since the R at the left end does not contribute any voltage drop.
Therefore,
VTh = V2
We can nd Req by shorting out the voltage source:
Req = R + R k R = 23 R
isc can be found by
IN = VTh=Req = 3VR
(C) VTh is easy to nd because the left and right vertical branches are independent of each other.
Dene node voltages e1 and e2 as in the gure (left), we see that we are working with two
simple voltage dividers.
VTh = e1 ; e2 = 54 V ; 15 V = 53 V
Req
4R
+v+
+
i
e1
e2 R
4R
R
+ V
-
R
4R
Req
4R
R
4R
R
4R
R
Looking at the gure above, center, we see the circuit with the voltage source shut o. It
is dicult to see how to analyze this circuit until it is redrawn as above, right. (If you are
unsure the right drawing is the same as the middle one, label the nodes and trace each path.)
We nd that
Req = R k 4R + R k 4R = 58 R
The current IN therefore equals to
IN = VTh=Req = 38VR
Req
4R
R
+v+
+
i
e1
e2 R
4R
-
Req
R
4R
I
4R
4R
R
R
4R
R
(D) Since the two vertical branches have the same resistance, it must be true that I=2 ows
through each branch. Consequently,
VTh = e ; e = I2 (4R ; R) = 23 IR
1
2
As shown from the gure above, the equivalent resistance is easier to nd in this problem.
Rth = (R + 4R) k (4R + R) = 25 R
The short-circuit current is then
IN = VTH =RTH = 35 I
The i-v relations are plotted below. Keep in mind that in i-v plots the slope is 1=Req .
i
i
i
i
3IR/2
RI/2 v
-I/2
(A)
Problem 2.2:
V/2
-V/3R
(B)
v
3V/5
-3V/8R
(C)
v
v
-3I/5
(D)
This problem analyzes the network shown below by two methods: superposition
and the direct application of the node method. You should compare the work required to analyze
the network by these two methods.
(A) First, use superposition to analyze the network. That is, superpose the two partial node
voltages obtained with only single sources active to nd the total node voltages. Remember
that a zero-valued voltage source is a short circuit, and a zero-valued current source is an open
circuit. Hint: rather than employing the node method twice, once for each partial analysis,
consider employing alternative simpler analyses such as those used in Exercises 2.1 and 2.2.
R1
R3 e
2
e1
V +-
R2
R4
I
ANSWER: Opening the current source, we get
e = (R (R+ R+ )Rk)Rk R+ R V = R (R + RR) (+RR+(RR )+ R + R ) V
3
0
1
4
3
4
2
2
2
1
2
3
3
4
4
1
2
3
4
e = R R+ R e = R (R + R ) +RRR(R + R + R ) V
4
0
2
3
2
0
1
4
2
3
4
4
1
2
3
4
Shorting the voltage source, we get
R (R + R )
e = I Req = I (R + R ) k (R k R ) = R R +R(R
+ R )(R + R ) I
e = e R R+ R = R R + (RR +R RR )(R + R ) I
0 0
3
1
0 0
1
1
2
4
0 0
2
The nal answers are
4
1
3
4
1
2
2
1
2
3
3
3
4
4
1
4
4
1
2
e = e + e = R (RR R+ +R ()RV ++ RR R)(R(R++RR) )I
e = e + e = R RR+R(RV ++ RR R)(RR +I R )
0
1
2
0 0
1
2
1
0
2
3
1
4
2
3
2
0 0
2
1
1
2
e1
1
3
4
3
1
2
4
2
4
1
2
R3 e
2
e1
R3 e
2
I
R2
2
4
4
R1
V +-
1
2
R4
Superposition
R2
R1
R4
2
(B) Second, use the node method to directly analyze the network and determine the two unknown
node voltages in total.
ANSWER: Using the procedure described in Exercise 2.3, we can directly write down the
node equations in matrix form:
"
#"
G + G + G ;G
;G
G +G
1
2
3
3
3
3
#
e
e
1
4
"
= V G0 + I
#
1
2
We can crank out the answer using Cramer's Rule.
det V G0 + I G;+GG e =
;
G
G +G +G
det ;G
G + G )(G + G )
= (G +(VGG++GI)(
G +G );G
1
3
3
4
1
1
2
3
3
3
1
1
2
3
3
4
3
4
3
4
2
3
det G +;GG + G V G0 + I e =
G
+
G
+
G
;
G
det
;G
G + G + I)
= (G + G G+ (GV G
)(G + G ) ; G
2
1
2
3
3
1
2
3
3
3
1
1
2
3
3
4
1
3
3
4
2
3
(C) Compare the solutions to Parts (A) and (B). The two solutions should be the same. In addition,
you should compare the work required to complete the analysis by both methods.
ANSWER: Although the answers from part A and part B look completely dierent, you
should verify that they are the same. The superposition method gives more intuition about
the structure of the circuit. The node method is generic but does not provide much insight
into the circuit. Both methods are relatively easy to apply for simple circuits.
Problem 2.3:
This problem studies the two networks shown below. Both networks contain a
nonlinear resistor having the terminal relation iN = vN2 , where is a constant with units A/V2 .
Assume that 0.
(A) Analyze Network (A) graphically to determine iN and vN in terms of vS and the network
parameters. To do so, note that the voltage source and linear resistor together constrain the
relation between iN and vN , and that the nonlinear resistor also constrains this relation. State
the two constraints, and on a single graph plot both constraints and identify the operating
point for iN and vN . Within what voltage range will vN lie?
R
R
vS
+
-
iN
+
vN
Nonlinear
Resistor
vS1
+
-
R
+
vN
-
iN
Nonlinear
Resistor
vS2
+
-
Network (B)
Network (A)
ANSWER: The nonlinear resistor has to satisfy both its terminal relation and the KVL
around the loop. Therefore the two constraints are:
(
iN = vN2
vS = iN R + vN
The second equation can be written as iN = vS =R ; vN =R. Combining this with the rst
equation in the pair above, we can nd the operating point graphically as shown below. Note
that since the i-v relation of the nonlinear resistor is quadratic, there will be a mathematical
solution in the second quadrant, where the signs of iN and vN dier. What this means is
that at this operating point the non-linear resistor is generating power rather than dissipating
it. Although the concept of a negative resistor exists, we will not consider it in this problem.
Therefore we will restrict ourselves to the solution in the rst quadrant. Here, the constraint
on vN is 0 vN vS .
iN
vS/R
Operating
Point
vS
vN
(B) Analytically solve for vN in Network (A) in terms of vS and the network parameters. Check
that this solution is consistent with the graphical solution from Part (A).
ANSWER: Solving the two equations in part (A) simultaneous we get
p4Rv + 1 ; 1
S
vN =
2R
p
p
Since 4RvS + 1 > 1, we see that vN 0. By expanding the above as power series ( 1 + x =
1 + x=2 ; x2 =4 + ) we can also show that vN vS . This is consistent with the graphical
solution from part (A).
(C) Again in Network (A), let vS = VS + vs and let vN = VN + vn , where VS and VN are constant
large-signal voltages, and vs and vn are varying small-signal voltages. Linearize the solution
from Part (B) to determine VN in terms of VS and vn in terms of vs .
ANSWER: The bias point is:
p4RV + 1 ; 1
S
VN =
2R
Dene Am = vvns as the small signal gain, then Am is just the coecient for the linear term in
the Taylor's series expansion.
@v
1
N
Am = @v = p4RV
S VS
S +1
(D) Consider Network (B) with vS2 = 0. Determine vN in terms of vS1 and the network parameters.
Hint: with an appropriate Thevenin equivalence, make use of the results from Part (B).
ANSWER: By nding the Thevenin equivalence as shown in the gure below, we can directly
employ the equation found in part (b).
pRv + 1 ; 1
S
vN =
R
1
R
+
vN
-
+
-
vS1
R
R/2
Thevenin
+
vN
-
vS1/2 +-
(E) Consider Network (B) with vS1 = 0. Determine vN in terms of vS2 and the network parameters.
Hint: with an appropriate Thevenin equivalence, make use of the results from Part (B). Or,
look for symmetry and make use of the results from Part (D).
ANSWER: since vS2 is symmetric to vS1, we only need to exchange vS2 for vS1 in the
equation in part (D).
pRv + 1 ; 1
S2
vN =
R
(F) Consider Network (B) in general. Determine vN in terms of vS1 and vS2 and the network
parameters. Hint: with an appropriate Thevenin equivalence, make use of the results from
Part (B).
ANSWER: By nding the Thevenin equivalence as shown in the picture below, we can
directly employ the equation found in part (b).
vN = R(vS +RvS ) + 1 ; 1
p
1
R
vS1
+
-
2
R/2
R
+
vN
-
vS2 +
-
Thevenin
(vS1+vS2)/2
+
-
+
vN
-
(G) Explain why the result from Part (F) is not the superposition of the results from Parts (D)
and (E).
ANSWER: Superposition only applies to linear elements. We can use superposition to calculate
everything (including the Thevenin voltage) outside of the nonlinear resistor, but we can not apply
superposition to relate the nonlinear element's input to its output. This is because for nonlinear
systems, the sum of the outputs from individual inputs is usually not equal to the output of the sum
of combined inputs. This is clearly true for this problem: answers in parts D and E do not sum up to
2
the answer in part F. To give another simple example, consider the system function vout = vin
. For
2
2
2
each input vin1 and vin2 , vout1 = vin1 and vout2 = vin2 , but clearly vout = (vin1 +vin2 ) 6= vout1 +vout2 .