Homework 4
Solutions:
P 4.61 [a]
20k
120V
10k
i0
[b]
Check:
P 4.62 [a] First remove the 16 and 260 resistors:
Next use a source transformation to convert the 1 A current source and 40 resistor:
which simplifies to
[b] Return to the original circuit with v
'ILEL
{3' HitMA: Inlo;
m IL= "1
TE.
LL'UL
W; H; 1.] vi. D '7 ® 10:. Va. "' V
I III1W Itch. I. M: IJtuLiralih:iu\)
= unrumj/VT u ' (my a: 4: p r
I I a J I 1 I I
13¢
95%.
L4 31) _
13km Lg: smr
In: =1; V anwu
hi ~L Tm camn ma. 1" 3 rFM
iv gtlull
o*.
fl wor[( 7
J ol . l ; , , 5
14,7^\
V,l
V;,l
\',r.2
n
=
Vo,",3
l .
=
 ,z)Ax. l5
s;"611
,
tr J\I
l/rr
< r s s u w l a V s > > . 7V
3
o
a
*
r )

^
l q
TI
, l
t,h
v , F Lrt, v
)
l " l L 9 . q .c t 0 5 1
L." Iui ,5
q
sc " o n l o T ; ^ ] , 'u3= z Vs;
Homework 3
Solutions:
P 3.50 Redraw the circuit, replacing the detector branch with a short circuit.
P 3.56 [a] Calculate the values of the Yconnected resistors that are equivalent to the 10, 40 , and 50 connected resistors:
;
;
Replacing the R2R3R4 delt
Homework #2
Solutions:
P 2.8
First, 10va = 5 V, so va = 0.5 V. Then recognize that each of the three branches is connected
between the same two nodes, so each of these branches must have the same voltage drop. The
voltage drop across the middle branch is
Homework #1
Solutions:
P 1.5
P 1.8
P 1.9
Therefore,
Thus, average velocity
P 1.14
Assume we are standing at box A looking toward box B. Then, using the
passive sign convention
, since the current i is flowing into the
terminal of the voltage v. Now we ju
Ohms Law
+ V I
V
IR
R
Parallel Resistors
P=VI
P = V2/R
P = I2 R
two resistors: R1 R2 = R1 R2 / (R1 + R2)
more than two resistors: 1/Req = 1/R1 + 1/R2 + 1/R3 +
Voltage Divider
+
VA
_
Current Divider
R2
R1
IA
+
V2 = VA R2/(R 1 + R2)
+
V1 = VA R1/(R 1+ R2)
Final Test Practice,
EE2001, Spring 2012, U of Minnesota
EE2001,
1
May 4th, 2012
Electrical and Computer Engineering Department
University of Minnesota
Final Review Problems
Instructions
1. Closed books and lecture notes exam. Two letter size papers with
import random
suits = 'CSDH'
values = '23456789TJQKA'
deck = [0 for x in range(52)]
hand = [0 for x in range(5)]
z = 0
flushes = 0
percent = 0
for x in range(len(suits):
for y in range(len(values):
deck[z] = (values[y]+suits[x])
z += 1
#print(deck)
a = de
MidTermI Practice,
EE2001, Spring 2012, U of Minnesota
EE2001,
Spring 2012
Electrical and Computer Engineering Department
University of Minnesota
MidtermI Problems
Instructions
1. Closed books and lecture notes exam. A letter size crap sheet
with equat
MidTermII Practice,
EE2001, Spring 2012, U of Minnesota
EE2001,
1
Spring 2012
Electrical and Computer Engineering Department
University of Minnesota
MidtermII Practice
Instructions
Closed books and lecture notes exam for MidtermII. A letter
size crap
Quiz two
EE2001, U of Minnesota, March 2, 2012
EE2001,
1
March 2, 2012
Electrical and Computer Engineering Department
University of Minnesota
Quiz #2 Problems
Instructions
There are 2 problems (page: 2) in this test; Answer all of them.
Student Name:
Stud
Quiz one
EE2001, U of Minnesota, Feb 3, 2012
EE2001,
1
Feb 3, 2012
Electrical and Computer Engineering Department
University of Minnesota
Quiz #1 Problems
Instructions
There are 2 problems (page: 2,3) in this test; Answer all of them.
Student Name:
Studen
Quiz three
EE2001, U of Minnesota, March 23, 2012
EE2001,
1
March 23, 2012
Electrical and Computer Engineering Department
University of Minnesota
Quiz #3 Problems
Instructions
There are 2 problems (page: 2) in this test; Answer all of them.
Student Name:
Quiz four
EE2001, U of Minnesota, April 13, 2012
EE2001,
1
April 13, 2012
Electrical and Computer Engineering Department
University of Minnesota
Quiz #4 Problems
Instructions
There are 2 problems (page: 2) in this test; Answer all of them.
Student Name:
S
F E a A I lS t l n 8 : n t : L r [ r l u t n ^ ' ] t
Feb15,2Al?
EE2001,
Deporim'zni
Eleciricolond aomPuierEngineering
of
Universlty /v\inhesolo
MidiermlProblems
Instructions
A
is
ond
books lecturenotesexdn Colculoior permiiled'
1. Closed
is
diogrdn/iable
MidTermII Practice,
EE2001, Spring 2012, U of Minnesota
EE2001,
1
Spring 2012
Electrical and Computer Engineering Department
University of Minnesota
MidtermII Practice
Instructions
1. Closed books and lecture notes exam for MidtermII. A letter
size cr
MidtermII,
EE2001, Spring 2012, U of Minnesota
EE2001,
1
March 28, 2012
Electrical and Computer Engineering Department
University of Minnesota
MidtermII Problems
Instructions
1. Closed books and lecture notes exam. Calculator is permitted. A
lettersize
Problem 3.20  Corrected Solution
For solution 3.20, there is one error. The values for Nd and Na are exchanged
in the solutions, as a result it is giving the incorrect answer. The correct answer
for the Current should be 85.82A with the given set of valu
Final Test Practice,
EE2001, Spring 2012, U of Minnesota
EE2001,
1
May 4th, 2012
Electrical and Computer Engineering Department
University of Minnesota
Final Review Problems
Instructions
1. Closed books and lecture notes exam. Two letter size papers with