Solution of ECE 202 Test 4 S13
1.
In a positive-phase-sequence, three-phase balanced source Van = 250∠50° Vrms. Find the numerical values of
the magnitude and phase (in degrees between -180° and +180°) of Vcn .
Vcn = ____________ ∠ ____________° Vrms
Vcn = 250∠170°
With reference to the circuit diagram below, find the numerical magnitudes and angles (in the range from -180°
to +180°) of these voltages, currents and impedances. The source is balanced and has a positive phase sequence.
(Remember magnitudes are never negative.) (Be sure to carefully observe the sequence of the subscripts.)
(a)
Vab = ____________ ∠ ____________° Vrms
(b)
Van = ____________ ∠ ____________° Vrms
(c)
Vac = ____________ ∠ ____________° Vrms
(d)
Z AN = ____________ ∠ ____________° Ω
(e)
I NB = ____________ ∠ ____________° Arms
(f)
Find the numerical power factor of the load and indicate whether it is leading or lagging by circling the
correct word.
PF = ____________
leading
lagging
I aA = 4∠ − 150° Arms , Vba = 180∠40° Vrms
Z AN = Z BN = ZCN
c
C
cn
CN
V
V
Z
2.
an
n
b
Vab = −Vba = 180∠ − 140° Vrms
Van + Vnb = Van − Vbn = Van − Van (1∠ − 120° ) = Vab = 180∠ − 140°
180∠ − 140°
Van =
= 103.923∠ − 170° Vrms
1 − 1∠ − 120°
Vac = Van + Vnc = Van − Vcn = Van − Van (1∠120° ) = 103.923∠ − 170° (1 − 1∠120° )
Vac = 180∠160° Vrms
V
103.923∠ − 170°
Z AN = an =
= 25.981∠ − 20°
I aA
4∠ − 150°
V
V
103.923∠70°
I NB = −I BN = − bn = − bn = −
= 4∠ − 90°
Z BN
Z AN
25.981∠ − 20°
PF = cos ( Z AN ) = cos ( −20° ) = 0.9396 leading
A
ZAN
N
Z BN
Vbn
a
B
Solution of ECE 202 Test 4 S13
1.
In a positive-phase-sequence, three-phase balanced source Van = 250∠40° Vrms. Find the numerical values of
the magnitude and phase (in degrees between -180° and +180°) of Vcn .
Vcn = ____________ ∠ ____________° Vrms
Vcn = 250∠160°
With reference to the circuit diagram below, find the numerical magnitudes and angles (in the range from -180°
to +180°) of these voltages, currents and impedances. The source is balanced and has a positive phase sequence.
(Remember magnitudes are never negative.) (Be sure to carefully observe the sequence of the subscripts.)
(a)
Vab = ____________ ∠ ____________° Vrms
(b)
Van = ____________ ∠ ____________° Vrms
(c)
Vac = ____________ ∠ ____________° Vrms
(d)
Z AN = ____________ ∠ ____________° Ω
(e)
I NB = ____________ ∠ ____________° Arms
(f)
Find the numerical power factor of the load and indicate whether it is leading or lagging by circling the
correct word.
PF = ____________
leading
lagging
I aA = 5∠ − 110° Arms , Vba = 140∠30° Vrms
Z AN = Z BN = ZCN
c
C
cn
CN
V
V
Z
2.
an
n
b
Vab = −Vba = 140∠ − 150° Vrms
Van + Vnb = Van − Vbn = Van − Van (1∠ − 120° ) = Vab = 140∠ − 150°
140∠ − 150°
Van =
= 80.829∠ ± 180° Vrms
1 − 1∠ − 120°
Vac = Van + Vnc = Van − Vcn = Van − Van (1∠120° ) = 80.829∠ ± 180° (1 − 1∠120° )
Vac = 140∠150° Vrms
V
80.829∠ ± 180°
Z AN = an =
= 16.1658∠ − 70°
I aA
5∠ − 110°
V
V
80.829∠60°
I NB = −I BN = − bn = − bn = −
= 5∠ − 50°
Z BN
Z AN
16.1658∠ − 70°
PF = cos ( Z AN ) = cos ( −70° ) = 0.342 leading
A
ZAN
N
Z BN
Vbn
a
B
Solution of ECE 202 Test 4 S13
1.
In a positive-phase-sequence, three-phase balanced source Van = 250∠ − 10° Vrms. Find the numerical values of
the magnitude and phase (in degrees between -180° and +180°) of Vcn .
Vcn = ____________ ∠ ____________° Vrms
Vcn = 250∠110°
With reference to the circuit diagram below, find the numerical magnitudes and angles (in the range from -180°
to +180°) of these voltages, currents and impedances. The source is balanced and has a positive phase sequence.
(Remember magnitudes are never negative.) (Be sure to carefully observe the sequence of the subscripts.)
(a)
Vab = ____________ ∠ ____________° Vrms
(b)
Van = ____________ ∠ ____________° Vrms
(c)
Vac = ____________ ∠ ____________° Vrms
(d)
Z AN = ____________ ∠ ____________° Ω
(e)
I NB = ____________ ∠ ____________° Arms
(f)
Find the numerical power factor of the load and indicate whether it is leading or lagging by circling the
correct word.
PF = ____________
leading
lagging
I aA = 3∠ − 120° Arms , Vba = 200∠50° Vrms
Z AN = Z BN = ZCN
c
C
cn
CN
V
V
Z
2.
an
n
b
Vab = −Vba = 200∠ − 130° Vrms
Van + Vnb = Van − Vbn = Van − Van (1∠ − 120° ) = Vab = 200∠ − 130°
200∠ − 130°
Van =
= 115.47∠ − 160° Vrms
1 − 1∠ − 120°
Vac = Van + Vnc = Van − Vcn = Van − Van (1∠120° ) = (115.47∠ − 160° ) (1 − 1∠120° )
Vac = 200∠170° Vrms
V
115.47∠ − 160°
Z AN = an =
= 34.89∠ − 40°
I aA
3∠ − 120°
V
V
115.47∠80°
I NB = −I BN = − bn = − bn = −
= 3∠ − 60°
Z BN
Z AN
34.89∠ − 40°
PF = cos ( Z AN ) = cos ( −40° ) = 0.766 leading
A
ZAN
N
Z BN
Vbn
a
B