Summary All Solutions for EPS

Electrical Power Systems Questions

, Week 1
Chapter 2 (par 2.2 Uli 2.8 :
par 2.9)

2.1 Current Source
2.2
Voltage Source
2.3 Resistance
2.4
Capritor
2.5 Inductor


2.6 Current direction is the direction
defined as
of the Movement
of positive charges

2.7 Ilt) = 25in ( 101T t)


[ =
dat dg = idt
g.
{dq ! = id



G H)
=
ft( 25in kort) dt =
2 { Sinnoh dt


= 2 [ -

Cos ( 10Mt) .
# A) to =
[ FIT COS ( toot))to

Tot COS (10Mt) 90
q Ct)
= -
t




2.8 Ilt) = 12 t i =
dat dq = idt



{dq { = idt {dq { =
12T dt



G) I. =L H q# =
A
'


q
= ne

'


ne = 6T
'
n = %

2.9 Ilt) = 5 i =
dat dq = idt



§dq
t

=
! 5dt



[ 9) ! [5T )! =
= 5T



qtt) =
5T
IJ

, 4 5


2.10
{ dq dt



q = [ 51-750--5.5-5.0=25

Answer : 25 Coulomb


2.14 P= UI
WE
=
5-2=10

=p 10-1=109 t =
-




2.21
T 9

iz
^
3A Liao 7--31-1.2
is = 4A
7A




222
r -3A 2AM Is % '



<
Is = -1A 9

5A


Is =
-

31-2 = -1A
"




(5 =
-2A

Ig : 5A -1A -2A tig =D




ig-8A.in
-8 tig

,
is in
2.23 in ixnistiz = 10mA




is -_ 8mA

is = 2mA

VII. R = 0,002.3000 = GV

, 2.24

[i -_ 0




} in
>
=
÷ .no
is
Irtis
-_
in


2.29 [ no
Va -
8-61-6=0
Va = -8W



•




2.30 Er -_ 0
e-
-61-14
zon

-12 TVN
Vr -_ kV


I =L -

-
0,2A




2.31 Info
-


Vat 51-5=0
VD -- 10W
1 2

[ Vn -0
-




-

Vctzto -10=0