A 3-phase, 4-wire system which has a system voltage of 254-V L-N with loads connected in wye; Za = 1020° ohm; Zb=10437° ohm and Zc = 102-53° ohm. Calculate the power dissipated on each load and the neutral current when phase sequence is a) abc and b) acb You can use the sample problem below as guide answering the problem, please use 3 decimal points for the final answer!
A 3-phase, 4-wire system which has a system voltage of 254-V L-N with loads connected in wye; Za = 1020° ohm; Zb=10437° ohm and Zc = 102-53° ohm. Calculate the power dissipated on each load and the neutral current when phase sequence is a) abc and b) acb You can use the sample problem below as guide answering the problem, please use 3 decimal points for the final answer!
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![A 3-phase, 4-wire system which has a system voltage of 254-V L-N with loads connected in wye;
Za = 10/0° ohm; Zb= 10/37° ohm and Zc = 102-53° ohm.
Calculate the power dissipated on each load and the neutral current when phase sequence is
a) abc and b) acb
You can use the sample problem below as guide answering the problem, please use 3 decimal points
for the final answer!
Sample Problem solution for guide (the given from this problem is different from the problem above)
Solution. (i) Phase sequence RYB (Fig. 19.80)
254 0: VYN
RN
254
120; Vav 254 120
VRN 254 0
IR IRN
25.4 0.
RR
10 0
2542-120°
.=25.42-157= 25.4(–0.9205 – j0.3907)=–2338– 19.95
YN
10237°
-IN
VAN
ly
(a)
Fig. 19.80
2542120°
AN
= 25.42173°= 25.4(-0.9925+ 0.1219)=-252+j31
102-53°
(1+1+1)--[25.4+(-23.38-9.55) + (−25.21 +£3.1)] = 23.49 +j6.85
24.46/16°15'
Now
R = 102; R₂ = 10 cos 37°= 82; R₂ = 10 cos 53° = 652
25.4² 8 5,162 W
25.4² 10 6.452 W: Wy
W₂ 25.4² 6 3,871 W
(ii) Phase sequence RBY [Fig. 19.81]
254 0:VYN 254 120
R.N
BN
254
120
254 0
10 0
25.4 0
120°
254 120
120°
25.4 83
10 37
- (3.1 +25.2)
120
254
120
10 53
IN
25.4 67 (9.95 j23.4)
(Ig Iy 1₂) (38.45 1.8)
--38.45-1.8-3852-177.3°.
254 V
VIN
Fig. 19.81](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd79262ba-431b-4ea2-8202-e105b35e4016%2F5e822205-0015-4b68-83ac-dbd4169fedb6%2Fu0h4jge_processed.png&w=3840&q=75)
Transcribed Image Text:A 3-phase, 4-wire system which has a system voltage of 254-V L-N with loads connected in wye;
Za = 10/0° ohm; Zb= 10/37° ohm and Zc = 102-53° ohm.
Calculate the power dissipated on each load and the neutral current when phase sequence is
a) abc and b) acb
You can use the sample problem below as guide answering the problem, please use 3 decimal points
for the final answer!
Sample Problem solution for guide (the given from this problem is different from the problem above)
Solution. (i) Phase sequence RYB (Fig. 19.80)
254 0: VYN
RN
254
120; Vav 254 120
VRN 254 0
IR IRN
25.4 0.
RR
10 0
2542-120°
.=25.42-157= 25.4(–0.9205 – j0.3907)=–2338– 19.95
YN
10237°
-IN
VAN
ly
(a)
Fig. 19.80
2542120°
AN
= 25.42173°= 25.4(-0.9925+ 0.1219)=-252+j31
102-53°
(1+1+1)--[25.4+(-23.38-9.55) + (−25.21 +£3.1)] = 23.49 +j6.85
24.46/16°15'
Now
R = 102; R₂ = 10 cos 37°= 82; R₂ = 10 cos 53° = 652
25.4² 8 5,162 W
25.4² 10 6.452 W: Wy
W₂ 25.4² 6 3,871 W
(ii) Phase sequence RBY [Fig. 19.81]
254 0:VYN 254 120
R.N
BN
254
120
254 0
10 0
25.4 0
120°
254 120
120°
25.4 83
10 37
- (3.1 +25.2)
120
254
120
10 53
IN
25.4 67 (9.95 j23.4)
(Ig Iy 1₂) (38.45 1.8)
--38.45-1.8-3852-177.3°.
254 V
VIN
Fig. 19.81
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