Solution of ECE 202 Test 5 S13
With reference to the circuit diagram below, find the following numerical quantities. (All magnitudes of complex
numbers should be non-negative and all angles of complex numbers should be in the range −180° to +180° .)
(a)
Vbn = ____________ ∠ ____________° V rms
Vbn = Van × 1∠ − 120° = 120∠ − 120° V rms
(b)
Vac = ____________ ∠ ____________° V rms
Vac = Van + Vnc , Vnc = −Vcn = − (Van × 1∠120° ) = 120∠ − 60° ⇒ Vac = 120∠0° + 120∠ − 60° = 207.85∠ − 30°
(c)
I AC = ____________ ∠ ____________° A rms
I AC =
(d)
VAC Vac 207.85∠ − 30°
=
=
= 5.823∠ − 41.3°
ZCA ZCA
35 + j7
I cC = ____________ ∠ ____________° A rms
I cC = I CA + I CB = −I AC +
Vcb
V + Vnb
120∠120° − Vbn
= − ( 5.823∠ − 41.3° ) + cn
= − ( 5.823∠ − 41.3° ) +
Z BC
35 + j7
35 + j7
I cC = − ( 5.823∠ − 41.3° ) +
(e)
120∠120° − 120∠ − 120°
= 10.0853∠108.69°
35 + j7
Power Factor of the load = ____________
⎛
⎛ 7 ⎞⎞
PF = cos ⎜ tan −1 ⎜ ⎟ ⎟ = 0.9806
⎝ 35 ⎠ ⎠
⎝
(f)
Total real average power delivered to the delta-connected load = ____________ W
The average real power delivered to ZCA is
PCA = VCA I CA cos ( VCA − I CA ) = 207.85 × 5.823 × cos (150° − 138.7° ) = 1186.85 W
Therefore the total power delivered to all three delta-connected loads is P = 3560.5 W .
(g)
Total complex power delivered to the delta-connected load = ____________ VA
The complex power delivered to ZCA is
*
SCA = VCA I CA
= 207.85∠150° × 5.823∠ − 138.7° = 1210.3∠11.3°
Therefore the total complex power delivered to all three delta-connected loads is S = 3630.9∠11.3° .
(h)
Total reactive power delivered to the delta-connected load = ____________ VAR
The total reactive power delivered to all three delta-connected loads is Q = S sin (11.3° ) = 711.46 VAR .
Van = 120∠0° Vrms
Z AB = Z BC = ZCA = 35 + j7 Ω
Positive Phase Sequence
c
n
Vbn
b
Van
Z CA
a
A
Z
ZBC
C
Vcn
AB
B
Solution of ECE 202 Test 5 S13
With reference to the circuit diagram below, find the following numerical quantities. (All magnitudes of complex
numbers should be non-negative and all angles of complex numbers should be in the range −180° to +180° .)
(a)
Vbn = ____________ ∠ ____________° V rms
Vbn = Van × 1∠ − 120° = 150∠ − 120° V rms
(b)
Vac = ____________ ∠ ____________° V rms
Vac = Van + Vnc , Vnc = −Vcn = − (Van × 1∠120° ) = 150∠ − 60° ⇒ Vac = 150∠0° + 150∠ − 60° = 259.81∠ − 30°
(c)
I AC = ____________ ∠ ____________° A rms
I AC =
(d)
VAC Vac 259.81∠ − 30°
=
=
= 10.008∠ − 45.642°
ZCA ZCA
25 + j7
I cC = ____________ ∠ ____________° A rms
I cC = I CA + I CB = −I AC +
Vcb
V + Vnb
150∠120° − Vbn
= − (10.008∠ − 45.642° ) + cn
= − (10.008∠ − 45.642° ) +
Z BC
25 + j7
25 + j7
I cC = − (10.008∠ − 45.642° ) +
(e)
150∠120° − 150∠ − 120°
= 17.334∠104.36°
25 + j7
Power Factor of the load = ____________
⎛
⎛ 7 ⎞⎞
PF = cos ⎜ tan −1 ⎜ ⎟ ⎟ = 0.963
⎝ 25 ⎠ ⎠
⎝
(f)
Total real average power delivered to the delta-connected load = ____________ W
The average real power delivered to ZCA is
PCA = VCA I CA cos ( VCA − I CA ) = 259.81 × 10.008 × cos (150° − 134.36° ) = 2503.91 W
Therefore the total power delivered to all three delta-connected loads is P = 7511.72 W .
(g)
Total complex power delivered to the delta-connected load = ____________ VA
The complex power delivered to ZCA is
*
SCA = VCA I CA
= 259.81∠150° × 10.008∠ − 134.36° = 2600.18∠15.64°
Therefore the total complex power delivered to all three delta-connected loads is S = 7800.54∠15.64° .
(h)
Total reactive power delivered to the delta-connected load = ____________ VAR
The total reactive power delivered to all three delta-connected loads is Q = S sin (11.3° ) = 2102.96 VAR .
Van = 150∠0° Vrms
Z AB = Z BC = ZCA = 25 + j7 Ω
Positive Phase Sequence
c
C
cn
n
Vbn
b
Van
Z CA
a
A
Z
ZBC
V
AB
B
Solution of ECE 202 Test 5 S13
With reference to the circuit diagram below, find the following numerical quantities. (All magnitudes of complex
numbers should be non-negative and all angles of complex numbers should be in the range −180° to +180° .)
(a)
Vbn = ____________ ∠ ____________° V rms
Vbn = Van × 1∠ − 120° = 180∠ − 120° V rms
(b)
Vac = ____________ ∠ ____________° V rms
Vac = Van + Vnc , Vnc = −Vcn = − (Van × 1∠120° ) = 180∠ − 60° ⇒ Vac = 180∠0° + 180∠ − 60° = 311.78∠ − 30°
(c)
I AC = ____________ ∠ ____________° A rms
I AC =
(d)
VAC Vac 311.78∠ − 30°
=
=
= 6.8461∠ − 38.84°
ZCA ZCA
45 + j7
I cC = ____________ ∠ ____________° A rms
I cC = I CA + I CB = −I AC +
Vcb
V + Vnb
180∠120° − Vbn
= − ( 6.8461∠ − 38.84° ) + cn
= − ( 6.8461∠ − 38.84° ) +
Z BC
45 + j7
45 + j7
I cC = − ( 6.8461∠ − 38.84° ) +
(e)
180∠120° − 180∠ − 120°
= 11.858∠111.16°
45 + j7
Power Factor of the load = ____________
⎛
⎛ 7 ⎞⎞
PF = cos ⎜ tan −1 ⎜ ⎟ ⎟ = 0.9882
⎝ 35 ⎠ ⎠
⎝
(f)
Total real average power delivered to the delta-connected load = ____________ W
The average real power delivered to ZCA is
PCA = VCA I CA cos ( VCA − I CA ) = 311.78 × 6.8461 × cos (150° − 141.16° ) = 2109.12 W
Therefore the total power delivered to all three delta-connected loads is P = 6327.4 W .
(g)
Total complex power delivered to the delta-connected load = ____________ VA
The complex power delivered to ZCA is
*
SCA = VCA I CA
= 311.78∠150° × 6.8461∠ − 141.16° = 2134.5∠8.84°
Therefore the total complex power delivered to all three delta-connected loads is S = 6403.5∠8.84° .
(h)
Total reactive power delivered to the delta-connected load = ____________ VAR
The total reactive power delivered to all three delta-connected loads is Q = S sin (11.3° ) = 984.12 VAR .
Van = 180∠0° Vrms
Z AB = Z BC = ZCA = 45 + j7 Ω
Positive Phase Sequence
c
C
cn
n
Vbn
b
Van
Z CA
a
A
Z
ZBC
V
AB
B