Sample Exam

Tuesday, 26 March 2019

3:22 PM

 

 

Machine generated alternative text:
Q
1
Q
2
Q
3
Q
4
fu
n
Q
5
Q
6
Q
7
1
.
 Given th
e
phase an
g
2
.
 What are 
3
.
 What is t
h
4
.
 The curr
e
n
ction if th
e
5
.
  The circ
u
techniq
u
done?  
6
.
 Define c
o
7
.
 The net
w
is the tot
a
E
e
 voltage(
v
t
g
le in degr
e
 the names 
h
e average 
e
nt in a 40 
m
e
 frequenc
y
u
it of the F
i
u
e you will 
o
mplex po
w
w
ork in Fig.
Q
a
l impedan
c
E
LEC
2
)120cos
t
e
e.   
of the ima
g
power abs
o
m
H inducto
y
 of the curr
e
i
g.Q5 is co
n
choose to 
f
w
er and pro
v
Q
7 is part 
o
c
e of the ci
r
2
134
Mid
Total Ma
r
Duration:
(
314/
4
t
g
inary parts 
o
rbed by th
e
r is 
12
I
e
nt is 50 H
z
n
sists of tw
o
f
ind current 
v
e that       
f the sche
m
r
cuit at 2 k
H
S
 Circ
u
d
-session tes
t
r
ks: 25 
 45 minutes
4
) V
, deter
m
of impeda
n
e
 inductor a
n
60
A. Ex
p
z
.
o
 voltage s
o
i
x
 in the ci
r
Fig. Q5
m
atic descri
b
H
z?
Fig. Q
7
*
2
rmsrms
VI
u
its an
d
t
 sample 
m
ine the fr
e
n
ce and ad
m
n
d capacit
o
p
ress the vo
l
    
o
urces of di
f
r
cuit and de
b
ing an ind
u
7
*
1
2
VI
d
 Sign
a
e
quency of 
t
         
m
ittance? 
o
r in an AC 
l
tage across
f
ferent freq
u
scribe brie
f
u
strial elect
r
a
ls
the voltage 
        
circuit? Ex
p
s
 the induct
o
           
u
encies. W
h
f
ly in word
s
r
onic sensi
n
    
in Hertz a
n
      (1 m
a
(1 m
a
p
lain briefl
y
(1 m
a
o
r as a time
(2ma
r
h
at circuit 
a
s
 how it wil
l
(2 m
a
           (2 
m
n
g device. 
W
           (3 
m
n
d
a
r
k
)
a
r
k
)
y
.
a
r
k
)
r
ks)
a
nalysis 
l
 be 
a
r
k
)
m
arks)
W
hat
m
arks)

 

 

 

The imaginary part of impedance is reactance

The imaginary part of admittance is susceptance

 

 

The average power absorbed is

The power absorbed by the capacitor and inductor is the reactive power (imaginary component)

 

 

 


 A

 A

 

 

Machine generated alternative text:
Q
Q
9
Q
1
8.
Sketch 
t
load.
Find the 
Assume 
9
.
 Three 
relation 
o
(b)
i
1
(t) 
a
1
0.
 Find th
e
load im
p
maxim
u
t
he power t
r
value of pa
r
that the loa
d
branch c
u
o
f the three 
a
nd
i
3
(t)
?A
e
 Thevenin 
p
edance co
n
u
m power?  
r
iangle to s
h
r
allel capaci
t
d
 is supplied 
u
rrents i
n
currents in 
A
lso show t
h
equivalent
c
n
nected at 
t
h
ow the po
w
t
ance neede
d
by a 240-V 
a netw
o
.  . 
a phasor d
i
h
ese angles 
i
c
ircuits at t
e
t
erminal 
a-
b
w
er factor 
c
d
 to correct a 
(rms), 50-H
z
o
rk are
Write thre
e
i
agram. W
h
i
n the phas
o
e
rminals 
a
-
b
b
 for maxi
m
Fig. Q1
0
c
orrection
w
load of 140 
z
 line. 
    
e
 currents i
n
h
at is the p
h
o
r diagram. 
b
for the ci
r
m
um power 
0
w
hen a shun
t
kVAR at 0.
8
n
 phasor d
o
h
ase angle 
b
r
cuits in Fi
g
t
ransfer an
d
n
t capacito
r
i
8
5 lagging p
f
,
o
main and 
S
b
etween (a) 
g
. Q10.   W
h
d
 what is th
i
s added ac
r
f
 to unity pf.
      (
4
S
ketch the 
p
i
1
(t)
 and 
i
2
(
   (4 
m
h
atshould
b
e
 value of t
h
(5 m
a
r
oss the 
4
marks) 
p
hasor
(
t)
 and  
m
arks)
b
e the 
h
is
a
rks)

 

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