EE 212 – Assignment 2 Solutions
Problem # 1
Figure 1
Voltage
=
120
∠
0
0
Frequency =
60 Hz
Angular frequency,
ϖ
=
2
×π×
f
=
377 rad/s
X
L1
=
ϖ
L
1
=
377 * 50 * 10
3
=
18.85
Ω
X
L2
=
ϖ
L
2
=
377 *100 *10
3
=
37.70
Ω
Z
=
(10 + j18.85) + (100  j37.7)
=
(10 + j18.85) + (35.3
∠
69.3
0
)
=
(10 + j18.85) + (12.4 + j33.0)
=
22.4 + j51.9
=
56.5
∠
66.6
0
ohms
i)
Calculate the current drawn from the source.
Ι
= V/Z
= 120
∠
0
0
V/
56.5
∠
66.6
0
Ω
= 2.12
∠
66.6
0
A
ii)
Value of the capacitor required to correct power factor to unity.
Current p.f. =cos 66.6
0
=
0.397 lagging
S
=
V
Ι
*
=
(120
∠
0
0
) (2.12
∠
66.6
0
)* VA
=
(120
∠
0
0
) (2.12
∠
66.6
0
) VA
=
255
∠
66.6
0
VA
=
(101 + j234) VA
120 V
~
j18.85
Ω
100
Ω
10
Ω
+

A
B
j37.7
Ω
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Q
=
234 VAr
Capacitor C must generate the reactive power, Q
c
= 234 VAr
Figure 1a
Q
c
= V
2
/ X
c
and
X
c
= 1 /
ϖ
C
C
= Q /
ϖ
V
2
= 234 / (377)
*
(120)
2
= 43.1 µF
iii) With the power factor corrected to unity, the current drawn from the source.
Current drawn from source,
Ι
new
S
=
V
Ι
*
new
With unity power factor, S
=
101 + j0
(i.e. no reactive power)
Ι
*
new
=
S / V
=
101
∠
0
0
/ 120
∠
0
0
=
0.843
∠
0
0
A
Alternate Method:
X
c
=
(120)
2
V
2
/ 234 VAr = 61.6
∠
90
0
Ω
since X
c
lags V by 90
0
Z
new
=
Z  X
c
=
(
56.507
∠
66.5972
0
)Ω  (61.5724
∠
90
0
)Ω
=
142
∠
0
0
Ω
Ι
*
new
=
V / Z
new
=
(120
∠
0
0
) V/ (142
∠
0
0
) Ω
=
0.843
∠
0
0
A
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 Spring '10
 Gander
 Alternating Current, Frequency, Volt, Power factor, resonant frequency, Hz Angular frequency

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