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ECSE210
Winter 2011
Electric Circuits 1
51
Practice Set #5
P11.67
(Using all rms values)
(a)
(
)
1
220 7.6
1672 VA
1317
.788
1672
=cos
pf
38.0
sin
=1030 VAR
VI
P
pf
VI
Q
VI
!
"
°
=
=
=
=
=
=
#
=
(b)
To restore the pf to 1.0, a capacitor is required to eliminate
Q
by introducing –
Q
, then
2
2
c
(220)
1030 =
=
X = 47
1
1
=
=
= 56.5
F
(377)(47)
c
c
V
X
X
C
X
μ
"
#
$
(c)
*
cos
where
= 0
then 1317 = 220I
= 6.0Arms for corrected
Note
7.6Arms for uncorrected
.
P
VI
I
pf
I
pf
°
=
"
=
P11.68
First load:
( )
(
)
(
)
(
)
1
1
1
tan cos
.6
500 1
tan 53.1
500
667 kVA
P
jQ
P
j
j
j
!
=
+
=
+
=
+
°
=
+
S
Second load:
2
400
600 kVA
j
=
+
S
Total:
1
2
900
1267 kVA
j
=
+
S = S +S
1
desired
tan (cos
(.90))
900
436 VA
P
jP
j
!
=
+
=
+
S
desired
From the vector diagram:
jQ
=
+
S
S
. Therefore
900
436
900
1277
841 VAR
j
j
jQ
Q
+
=
+
+
!
=
"
2
2
2
*
*
(1000)
841
1189
1189
841
841
377
j
j
j
j
j
j
C
=
!
"
=
=
=
"
=
!
=
!
!
!
V
V
Z
Z
Z
Finally,
1
2.20
F
(1189)(377)
C
=
=
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View Full DocumentECSE210
Winter 2011
Electric Circuits 1
52
P11.613
Let
2
2
1
tan
X
R
j X
R
X
R
!
"
#
=
+
=
+
$
%
'
(
Z
be the impedance of the load.
Further, let
A
!
=
"
V
and
B
=
"
I
be the voltage across and current through the load. Then
(
)
2
2
1
tan
X
A
A
R
X
R
j X
R
B
B
"
#
$
%
+
$
=
+
=
=
=
$
#
'
(
$
)
*
Z
Equating angles gives
1
tan
X
R
#
$
%
=
#
'
(
)
Also, the complex power delivered to the load is
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This note was uploaded on 03/29/2011 for the course ECSE 291 taught by Professor N/a during the Spring '11 term at McGill.
 Spring '11
 N/A

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