ECE 330: Power Circuits and Electromechanics
Summer 2016
Homework 3 Solution
1. Problem 3.10
a. Assuming an ideal (lossless) transformer, the primary side current I1 may be
estimated to be
I1 =
N2 P2
120 4.8 kV A
N2
I2 =
=
= 20 A
N1
N1 V2
240 120 V
(1)
b.
Homework 6
4.1 (c) and (d)
The force of electrical origin is just the partial of coenergy with respect to x.
There is only one current, so we need to only integrate w.r.t. i.
Taking the partial with respect to x:
Since this is a magnetically linear system
ECE 330 homework assignment #4, due in Midterm examination
Problem 1
For the magnetic circuit in Fig. 1 below, the upper part with winding is fixed, and the lower part is
movable (up and down only). Take into account the effect of fringing in the air gap
ECE 330 homework assignment #2 (In class quiz Fri, Sep 13)
Text problem 2.16 (partial answer: I line = 1,255Amps, P+jQ tot = 600 +j450 MVA)
Text problem 2.17
Text problem 2.19
Text problem 2.20
Text problem 2.25
Special problem #1
A balanced, 3phase, 3w
ECE330
Spring 2013
Homework 3
Problem 3.1
Amperes law may be applied to yield
J f nda
H dl =
s
C
2Hg lg + Hc lc = Ni
2lg
Ni
lc
+
,
=
0 Ag iron Ac
where the reluctances are as follows:
lg
0.3cm
At
=
= 3.145 106
0 Ag (4 107 )(3cm + 0.3cm)(2cm + 0.3cm)
Wb
lc
ECE 330 homework assignment #3 (In class quiz Fri, Feb 14)
Text problem 3.1 (and also find the magnetic flux density in the iron)
(partial answer: L = 6.1 mH)
Text problem 3.2 (part b only)
(partial answer: L11=3.125mH, L22=0.78125mH)
Text problem 3.3 (al
ECE 330 homework assignment #2 (In class quiz Fri, Feb 7)
Text problem 2.16 (partial answer: I line = 1,255Amps, P+jQ tot = 600 +j450 MVA)
Text problem 2.17
Text problem 2.19
Text problem 2.20
Text problem 2.25
Special problem #1
A balanced, 3phase, 3wi
)
D
/\*
J("
)
/\:'

f_
l(v t 
,ro A
AJ. =
3_
>i>.1
n.4
d
Il
/vr

c/ =
rD
ltlp
=dz = d
I
t
7n[ zo)
A J,'
(s * ZFn
d = x );t' A
X+2x
\ (.',x) = a:!
!_t ,
(X+2:c)
L
vU,*' ( ,r * , =
,
,f
. \ [ i 5x ) J ,
fW'"(",x) = ztaAnrtr'z
I
z f.oz.)
\
 r\^(
ECE 330: Power Circuits and Electromechanics
Summer 2016
Homework 4 Solution
1. Problem 4.1
c. From homework 5, we note that (i, x) = 0 W 2 N 2 ( g1 +
2gx(i,x)
.
0 W 2 N 2 (g+2x)
1
)i,
2x
thus i(, x) =
We can compute the energy by:
Z
Wm (, x) =
0
x)d
=
i
ECE 330 homework assignment #6 (In class quiz Mon, Mar 17)
Text problem 4.1 (c) and (d) only
Text problem 4.2 (b)
e
Text problem 4.4 but do not compute Wm (partial answer: f = 0*w*w*N*N*i*i/(2x*x) )
Text problem 4.6 but do not compute Wm
e
(partial answe
ECE430
Spring 2011
Homework 9
Problem 5.5
(a)
Choosing the states to be
x=
x1
=
x2
,
the following second order nonlinear system results:
x2
24 sin x1 0.5x2
0.1
x=
.
Setting x = 0, and applying the restriction that = cfw_ R  0 , provides the equilibrium
ECE330 Spring 2013
2.1
(
)
v(t ) = 100 cos 377t + 10 0 V => V =
(
)
i (t ) = cos 377t + 55 0 A => I =
1
2
100
2
10 0 V
55 0 A
Average power: P = VI cos( v i ) =
100 1
(
)
(
22
Power factor: PF = cos( v i ) = cos( 45 0 ) = 0.7071 leading
2.2
(
)
a) v(t ) =
ECE330 Spring 2013
Problem 1
+ I1
V1
Req
Rc
I2/a
jXeq
V2
jXm
Load
_
10:1
Turns ratio a = 2200/220 = 10.
Rc' = a 2 Rc = 19500
'
X m = a 2 X m = 17000
Assume V2 = 2150 0 V,
Load current:
10000
I2 =
cos 1 (0.96 ) = 46.51 16.26 0 A
215
Primary voltage:
I
V
ECE 330: Power Circuits and Electromechanics
Summer 2016
Homework 6 Solution
1. Problem 1
a. The coenergy function for this system is:
Wm0 (ias , ibs , ir , )
Z
Z ias
as (ias , 0, 0, )dias +
=
bs (ias , ibs , 0, )dibs +
Z
ir
bs (ias , ibs , ir , 0, )dibs
ECE 330: Power Circuits and Electromechanics
Spring 2016
Homework 6 Solution
1. Problem 4.1
a. With the permeability of the core assumed to be infinite, the equivalent magnetic
circuit is as shown in Fig. 1, where the reluctances are defined to be
g
0 W 2
ECE330
Power Circuits and Electromechanics
Dr. Nam NguyenQuang
Spring 2013
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 4
1
Electromechanical system Introduction
Magnetic circuits with one moving member will be studied.
Systematic derivation of mat
ECE 330 homework assignment #2, due on Monday, February 18, 2013
Problem 1
A coil of 180 turns is wound uniformly over a wooden ring having a mean circumference of 700
mm, and a uniform cross sectional area of 400 mm2. If the current through the coil is 6
ECE330
Fall 2013
Homework 3
Problem 3.1
Amperes law may be applied to yield
l
H d =
C
n
J f da
s
2Hg lg + Hc lc = Ni
2lg
Ni
lc
=
+
,
0 Ag iron Ac
where the reluctances are as follows:
lg
0.3cm
At
=
= 3.145 106
0 Ag (4 107 )(3cm + 0.3cm)(2cm + 0.3cm)
Wb
ECE330
Fall 2013
Homework 2
Problem 2.16
(a)
Assuming a balanced 3 conguration, the complex power per phase is
S 1 =
S 3 750
=
= 250 MV A.
3
3
Taking the line voltage to be the reference angle, the complex impedance per phase is
2
S 1 =
V
Z
=
(345 kV)2
Z
ECE 330 homework assignment #7 (In class quiz Thursday, March 30)
Text problem 4.14 (partial answer: i=5, x=0 (or +1, 1)
Text problem 4.16
Text problem 4.18
Text problem 4.19
Text problem 4.27
ECE 330 homework assignment #1 (In class quiz Fri, Jan 31)
Text problem 2.1 (Partial answer: 0.707 leading)
Text problem 2.2 (Partial answers: 250, 500, 500, 0, 150, 1900, 150)
Text problem 2.8
Text problem 2.9
Text problem 2.10 (Partial answer: 100 Am
ECE 330 homework assignment #4 (In class quiz Fri, Feb 21)
Text problem 3.2
Text problem 3.10 a) and b) only
(answers: 20A, 12 Ohms)
Text problem 3.12
Text problem 3.14
Text problem 3.17
Special problem
A 480/240 V, 4.8 kVA, 60 Hz, singlephase transforme
ECE330
Spring 2013
Homework 5
Problem 4.1
(a)
With the permeability of the core assumed to be infinite, the equivalent magnetic circuit is as shown in Fig. 1, where
the reluctances are defined to be
g
Rg =
0 W 2
x(t)
,
R x (t) =
0 W 2
and the total reluct
ECE 330 Quiz #2a (HW2)
Name:_
Class for returning: T 9:30 am R 2:00 pm
Special Problem #2
NetID:_
The following threephase, balanced loads are connected across a threephase, wyeconnected
source (60 Hz and 480 V line to line). The nature of the three lo
ECE 330 homework assignment #3
(In class quiz Thu, Feb 9)
Text problem 3.1 (and also find the magnetic flux density in the iron)
(partial answer: L = 6.1 mH)
Text problem 3.2b (part b only)
(partial answer: L11=3.125mH, L22=0.78125mH)
Text problem 3.3 (al
ECE 330 homework assignment #5 (In class quiz Thu Mar 9)
Text problem 4.1 (a) and (b) only
Text problem 4.2 (a) only (partial solution: L1= (1/4)*(0N*N*(Ag0/g0)+(Ag1/g1)
Special problem #1 (see Exam 2 Spring 2004 for solution)
For the structure drawn in F
ECE 330 homework #4 (Due in class quiz Thursday Feb 16, 2017)
Text problem 3.2
Text problem 3.10 a) and b) only
(answers: 20A, 12 Ohms)
Text problem 3.12
Text problem 3.14
Text problem 3.17
Special problem
A 480/240 V, 4.8 kVA, 60 Hz, singlephase transfo
ECE 330 homework assignment #11 (In class quiz Monday Dec 5)
Text problem 7.3
Text problem 7.4
Text problem 7.11
Text problem 7.14
Special problem #1 (see solution Spring 2005 final)
A 460 Volt (line to line), 60 Hz, 3phase, Yconnected, 4 pole Induction