Exam 12: Three-Phase Systems

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In a balanced ΔΔ\Delta - \Delta circuit shown below, the source voltage Vab=6550V\mathrm { V } _ { \mathrm { ab } } = 655 \angle 0 ^ { \circ } \mathrm { V } (rms).The wire impedance is per phase. Find the ZW=15+j12Z _ { W } = 15 + j 12 and the phase impedance is ZΔ=54+j48ΩZ _ { \Delta } = 54 + j 48 \Omega per phase. Find the phase currents and line currents.  In a balanced  \Delta - \Delta  circuit shown below, the source voltage    \mathrm { V } _ { \mathrm { ab } } = 655 \angle 0 ^ { \circ } \mathrm { V }  (rms).The wire impedance is per phase. Find the  Z _ { W } = 15 + j 12     and the phase impedance is  Z _ { \Delta } = 54 + j 48 \Omega  per phase. Find the phase currents and line currents.

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Mesh equations: Vab+ZwI1+ZΔ(I1I3)+Zw(I1I2)=0Vbc+Zw(I2I1)+ZΔ(I2I3)+ZwI2=0ZΔ(I3I2)+ZΔ(I3I1)+ZΔI3=0IA=I1=8.73870.3141AIB=I2I1=8.738169.6859AIC=I2=8.73849.6859AIAB=I1I3=5.044940.3141AIBC=I2I3=5.0449160.3141AICA=I3=5.044979.6859A\begin{array} { l } - V _ { a b } + Z _ { w } I _ { 1 } + Z _ { \Delta } \left( I _ { 1 } - I _ { 3 } \right) + Z _ { w } \left( I _ { 1 } - I _ { 2 } \right) = 0 \\- V _ { b c } + Z _ { w } \left( I _ { 2 } - I _ { 1 } \right) + Z _ { \Delta } \left( I _ { 2 } - I _ { 3 } \right) + Z _ { w } I _ { 2 } = 0 \\Z _ { \Delta } \left( I _ { 3 } - I _ { 2 } \right) + Z _ { \Delta } \left( I _ { 3 } - I _ { 1 } \right) + Z _ { \Delta } I _ { 3 } = 0 \\I _ { A } = I _ { 1 } = 8.738 \angle - 70.3141 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { B } = I _ { 2 } - I _ { 1 } = 8.738 \angle 169.6859 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { C } = - \mathrm { I } _ { 2 } = 8.738 \angle 49.6859 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { AB } } = \mathrm { I } _ { 1 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 40.3141 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { BC } } = \mathrm { I } _ { 2 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 160.3141 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { CA } } = - \mathrm { I } _ { 3 } = 5.0449 \angle 79.6859 ^ { \circ } \mathrm { A }\end{array}  Mesh equations:  \begin{array} { l }  - V _ { a b } + Z _ { w } I _ { 1 } + Z _ { \Delta } \left( I _ { 1 } - I _ { 3 } \right) + Z _ { w } \left( I _ { 1 } - I _ { 2 } \right) = 0 \\ - V _ { b c } + Z _ { w } \left( I _ { 2 } - I _ { 1 } \right) + Z _ { \Delta } \left( I _ { 2 } - I _ { 3 } \right) + Z _ { w } I _ { 2 } = 0 \\ Z _ { \Delta } \left( I _ { 3 } - I _ { 2 } \right) + Z _ { \Delta } \left( I _ { 3 } - I _ { 1 } \right) + Z _ { \Delta } I _ { 3 } = 0 \\ I _ { A } = I _ { 1 } = 8.738 \angle - 70.3141 ^ { \circ } \mathrm { A } \\ \mathrm { I } _ { B } = I _ { 2 } - I _ { 1 } = 8.738 \angle 169.6859 ^ { \circ } \mathrm { A } \\  \mathrm { I } _ { C } = - \mathrm { I } _ { 2 } = 8.738 \angle 49.6859 ^ { \circ } \mathrm { A } \\ \mathrm { I } _ { \mathrm { AB } } = \mathrm { I } _ { 1 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 40.3141 ^ { \circ } \mathrm { A } \\ \mathrm { I } _ { \mathrm { BC } } = \mathrm { I } _ { 2 } - \mathrm { I } _ { 3 } = 5.0449 \angle - 160.3141 ^ { \circ } \mathrm { A } \\ \mathrm { I } _ { \mathrm { CA } } = - \mathrm { I } _ { 3 } = 5.0449 \angle 79.6859 ^ { \circ } \mathrm { A } \end{array}     \text { ICAbp=vpa (ICAbp, 7) }  ICAbp=vpa (ICAbp, 7) \text { ICAbp=vpa (ICAbp, 7) }

In a balanced Y -Δ circuit shown below, the magnitude of the phase voltage is VP=635 VV _ { P } = 635 \mathrm {~V} (rms). The phase impedance of the load is ZΔ=66+j39ΩZ _ { \Delta } = 66 + \mathrm { j } 39 \Omega per phase. The wire impedance is Zw\mathrm { Z } _ { \mathrm { w } } =11+j16= 11 + \mathrm { j } 16 per wire. Find line current and phase current assuming Van =0\angle \mathrm { V } _ { \text {an } } = 0 ^ { \circ } .  In a balanced  Y -Δ circuit shown below, the magnitude of the phase voltage is   V _ { P } = 635 \mathrm {~V}  (rms). The phase impedance of the load is  Z _ { \Delta } = 66 + \mathrm { j } 39 \Omega  per phase. The wire impedance is  \mathrm { Z } _ { \mathrm { w } }   = 11 + \mathrm { j } 16  per wire. Find line current and phase current assuming  \angle \mathrm { V } _ { \text {an } } = 0 ^ { \circ } .

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ZY=ZΔ/3=22+j13=25.553930.5792ΩZt=ZW+ZY=33+j29 Van=6350V,Vbn=635120V,Vcn=635120VIA=Van/Zt=14.454241.3086AIB=Vbn/Zt=14.4542161.3086AIC=Vcn/Zt=14.454278.6914A\begin{array} { l } \mathrm { Z } _ { \mathrm { Y } } = \mathrm { Z } _ { \Delta } / 3 = 22 + \mathrm { j } 13 = 25.5539 \angle 30.5792 ^ { \circ } \Omega \\\mathrm { Z } _ { \mathrm { t } } = \mathrm { Z } _ { \mathrm { W } } + \mathrm { Z } _ { \mathrm { Y } } = 33 + \mathrm { j } 29 \\\mathrm {~V} _ { \mathrm { an } } = 635 \angle 0 ^ { \circ } \mathrm { V } , \mathrm { V } _ { \mathrm { bn } } = 635 \angle - 120 ^ { \circ } \mathrm { V } , \mathrm { V } _ { \mathrm { cn } } = 635 \angle 120 ^ { \circ } \mathrm { V } \\\mathrm { I } _ { \mathrm { A } } = \mathrm { V } _ { \mathrm { an } } / \mathrm { Z } _ { \mathrm { t } } = 14.4542 \angle - 41.3086 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { B } } = \mathrm { V } _ { \mathrm { bn } } / \mathrm { Z } _ { \mathrm { t } } = 14.4542 \angle - 161.3086 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { C } } = \mathrm { V } _ { \mathrm { cn } } / \mathrm { Z } _ { \mathrm { t } } = 14.4542 \angle 78.6914 ^ { \circ } \mathrm { A }\end{array} VA=Van ZW×IA=369.361510.7294VVB=VbnZW×IB=369.3615130.7294VVC=VcnZW×IC=369.3615109.2706VVAB=VAVB=639.752919.2706VVBC=VBVC=639.7529100.7294VVCA=VCVA=639.7529139.2706VIAB=VAB/ZΔ=8.345211.3086AIBC=VBC/ZΔ=8.3452131.3086AICA=VCA/ZΔ=8.3452108.6914A\begin{array} { l } V _ { A } = V _ { \text {an } } - Z _ { W } \times I _ { A } = 369.3615 \angle - 10.7294 ^ { \circ } \mathrm { V } \\V _ { B } = V _ { b n } - Z _ { W } \times I _ { B } = 369.3615 \angle - 130.7294 ^ { \circ } \mathrm { V } \\V _ { C } = V _ { c n } - Z _ { W } \times I _ { C } = 369.3615 \angle 109.2706 ^ { \circ } \mathrm { V } \\V _ { A B } = V _ { A } - V _ { B } = 639.7529 \angle 19.2706 ^ { \circ } \mathrm { V } \\V _ { B C } = V _ { B } - V _ { C } = 639.7529 \angle - 100.7294 ^ { \circ } \mathrm { V } \\V _ { C A } = V _ { C } - V _ { A } = 639.7529 \angle 139.2706 ^ { \circ } \mathrm { V } \\I _ { A B } = V _ { A B } / Z _ { \Delta } = 8.3452 \angle - 11.3086 ^ { \circ } \mathrm { A } \\I _ { B C } = V _ { B C } / Z _ { \Delta } = 8.3452 \angle - 131.3086 ^ { \circ } \mathrm { A } \\I _ { C A } = V _ { C A } / Z _ { \Delta } = 8.3452 \angle 108.6914 ^ { \circ } \mathrm { A }\end{array} ;
1+16j; clear ali; format long
Vp=635;ZD=66+39j;Zw=1\mathrm { Vp } = 635 ; \mathrm { ZD } = 66 + 39 \mathrm { j } ; \mathrm { Zw } = 1
ZDp=R2P(ZD)Z D p = R 2 P ( Z D )
theta-2Dp (3)
thetad=ZDp (2)
ZY=ZD/3\mathrm { ZY } = \mathrm { ZD } / 3
ZYp=R2P(ZY)\mathrm { ZYp } = \mathrm { R } 2 \mathrm { P } ( \mathrm { ZY } )
Zt=ZW+ZY\mathrm { Zt } = \mathrm { ZW } + \mathrm { ZY }
Van=P2Rd(Vp,O)V a n = \mathrm { P } 2 \mathrm { Rd } ( \mathrm { Vp } , \mathrm { O } )
Vanp=R2P{VanV a n p = R 2 P \{ V a n \rangle
Vbn=P2Rd(Vp,120)V b n = P 2 R d ( V p , - 120 )
Vbnp=R2P (Vbn\}
Vcn=P2Rd{VP,240}V c n = P 2 R d \{ V P , - 240 \}
VCnp-R2P \{Vcn \rangle
IA=Van/ZtI A = V a n / Z t
IAP =R2P= R 2 P (IA)
IB=Vbn/Zt\mathrm { I } B = V \mathrm { bn } / \mathrm { Zt }
IBp=R2PI B p = R 2 P (IB)
IC=Vcn/Zt\mathrm { IC } = \mathrm { Vcn } / \mathrm { Zt }
ICP=R2P(IC)I C P = R 2 P ( I C )
VA=Van ZWIA- Z W ^ { * } I A
VAP=R2P(VA)V A P = R 2 P ( V A )
VB=VbnZWIBV B = V b n - Z W ^ { * } I B
VBP=R2P(VB)\mathrm { VBP } = \mathrm { R } 2 \mathrm { P } ( \mathrm { VB } )
VC=VCn2WICV C = V C n - 2 W ^ { \star } I C
VCP=R2P(VC)\mathrm { VCP } = \mathrm { R } 2 \mathrm { P } ( \mathrm { VC } )
VAB=VAVBV A B = V A - V B
VABP=R2P(VAB)V A B P = R 2 P ( V A B )
VBC=VBVC\mathrm { VBC } = \mathrm { VB } - \mathrm { VC }
BCP=R2P{VBC}\vee B C P = R 2 P \{ V B C \}
VCA=VCVA\mathrm { VCA } = \mathrm { VC } - \mathrm { VA }
VCAP=R2PV C A P = R 2 \mathrm { P } (VCA)
IAB=VAB/ZDI A B = V A B / Z D
IABP=R2PI A B P = R 2 P (IAB)
IBC=VBC/ZDI B C = V B C / Z D
IBCP=R2P\mathrm { IBCP } = \mathrm { R } 2 \mathrm { P } (IBC)
ICA=VCA/ZDI C A = V C A / Z D
ICAP=R2P{ICA}I C A P = R 2 P \{ I C A \}

In a balanced Y-Y circuit shown below, the magnitude of the line voltage is VL=535 V\mathrm { V } _ { \mathrm { L } } = 535 \mathrm {~V} (rms). The phase impedance is per phase, the wire ipedance is ZY=75+i42Ω\mathrm { Z } _ { \mathrm { Y } } = 75 + \mathrm { i } 42 \Omega ZW=15+i11Z W = 15 + i 11 Ω per phase, the source impedance is assuing ZS=12+j10ΩZ _ { S } = 12 + j 10 \Omega  Find IA,VAN and VAB\text { Find } \mathrm { I } _ { \mathrm { A } } , V _ { \mathrm { AN } } \text { and } \mathrm { V } _ { \mathrm { AB } } Van =0\angle \mathrm { V } _ { \text {an } } = 0 ^ { \circ }  In a balanced Y-Y circuit shown below, the magnitude of the line voltage is  \mathrm { V } _ { \mathrm { L } } = 535 \mathrm {~V}  (rms). The phase impedance is per phase, the wire ipedance is  \mathrm { Z } _ { \mathrm { Y } } = 75 + \mathrm { i } 42 \Omega   Z W = 15 + i 11  Ω per phase, the source impedance is assuing  Z _ { S } = 12 + j 10 \Omega   \text { Find } \mathrm { I } _ { \mathrm { A } } , V _ { \mathrm { AN } } \text { and } \mathrm { V } _ { \mathrm { AB } }   \angle \mathrm { V } _ { \text {an } } = 0 ^ { \circ }

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Vp=VL/3=308.8824 V Van=308.88240VVbn=308.8824120VVcn=308.8824120VZt=Zs+ZwW+ZY=102+j63\begin{array} { l } \mathrm { V } _ { \mathrm { p } } = \mathrm { V } _ { \mathrm { L } } / \sqrt { 3 } = 308.8824 \mathrm {~V} \\\mathrm {~V} _ { \mathrm { an } } = 308.8824 \angle 0 ^ { \circ } \mathrm { V } \\\mathrm { V } _ { \mathrm { bn } } = 308.8824 \angle - 120 ^ { \circ } \mathrm { V } \\\mathrm { V } _ { \mathrm { cn } } = 308.8824 \angle 120 ^ { \circ } \mathrm { V } \\\mathrm { Z } _ { \mathrm { t } } = \mathrm { Z } _ { \mathrm { s } } + \mathrm { Zw } _ { \mathrm { W } } + \mathrm { Z } _ { \mathrm { Y } } = 102 + \mathrm { j } 63\end{array} IA=Van/Zt=2.576431.7014AIB=Vbn/Zt=2.5764151.7014AIC=Vcn/Zt=2.576488.2986AVAN=ZY×IA=221.46872.4526VVBN=ZY×IB=221.4687122.4526VVCN=ZY×IC=221.4687117.5474VVAB=VANVBN=383.59527.5474V\begin{array} { l } \mathrm { I } _ { \mathrm { A } } = \mathrm { V } _ { \mathrm { an } } / \mathrm { Z } _ { \mathrm { t } } = 2.5764 \angle - 31.7014 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { B } } = \mathrm { V } _ { \mathrm { bn } } / \mathrm { Z } _ { \mathrm { t } } = 2.5764 \angle - 151.7014 ^ { \circ } \mathrm { A } \\\mathrm { I } _ { \mathrm { C } } = \mathrm { V } _ { \mathrm { cn } } / \mathrm { Z } _ { \mathrm { t } } = 2.5764 \angle 88.2986 ^ { \circ } \mathrm { A } \\\mathrm { V } _ { \mathrm { AN } } = \mathrm { Z } _ { \mathrm { Y } } \times \mathrm { I } _ { \mathrm { A } } = 221.4687 \angle - 2.4526 ^ { \circ } \mathrm { V } \\\mathrm { V } _ { \mathrm { BN } } = \mathrm { Z } _ { \mathrm { Y } } \times \mathrm { I } _ { \mathrm { B } } = 221.4687 \angle - 122.4526 ^ { \circ } \mathrm { V } \\\mathrm { V } _ { \mathrm { CN } } = \mathrm { Z } _ { \mathrm { Y } } \times \mathrm { I } _ { \mathrm { C } } = 221.4687 \angle 117.5474 ^ { \circ } \mathrm { V } \\\mathrm { V } _ { \mathrm { AB } } = \mathrm { V } _ { \mathrm { AN } } - \mathrm { V } _ { \mathrm { BN } } = 383.595 \angle 27.5474 ^ { \circ } \mathrm { V }\end{array} clear all; format long;
f=60;w=2piff = 60 ; w = 2 * p i \star f
VL=535;2y75+42j;2w=15+11j;zs=12+10j;V L = 535 ; 2 y - 75 + 42 j ; 2 w = 15 + 11 j ; z s = 12 + 10 j ;
Vp=VL/sqrt\mathrm { Vp } = \mathrm { VL } / \mathrm { sqrt } (3)
phi=0
Van=P2Rd (Vp, phi)
Vanp=R2P (Van)
Vbn=P2Rd (Vp, phi-120)
Vbnp=R2P (Vbn)
Ven=P2Rd (Vp, phi 240)- 240 )
Venp-R2P (Ven)
Zt=Zw+Zy+Zs\mathrm { Zt } = \mathrm { Zw } + \mathrm { Zy } + \mathrm { Zs }
IA=Van/Zt\mathrm { IA } = \operatorname { Van } / \mathrm { Zt }
IAP=I A P = R2P (IA)
IB=Vbn/Zt\mathrm { IB } = \mathrm { Vbn } / \mathrm { Zt }
IBp =R2P= \mathrm { R } 2 \mathrm { P } (IB)
IC=Vcn/2tI C = V c n / 2 t
ICp=R2P(IC)I C p = R 2 P ( I C )
VAN =2yIA= 2 y ^ { * } I A
VANp=R2P\mathrm { VANp } = \mathrm { R } 2 \mathrm { P } (VAN)
VBN =ZYIB= Z Y ^ { \star } I B
VBNp=R2P (VBN)
VCN=ZYICV C N = Z Y ^ { \star } I C
VCNp-R2P (VCN)
VAB=VANVBN\mathrm { VAB } = \mathrm { VAN } - \mathrm { VBN }
VABp=R2P(VAB)\mathrm { VABp } = \mathrm { R } 2 \mathrm { P } ( \mathrm { VAB } )

 If Vbc=295115V, what are Vab,Vca and Van,Vbn,Vcn for the balanced three-phase \text { If } \mathrm { V } _ { \mathrm { bc } } = 295 \angle -115 ^ { \circ } \mathrm { V } \text {, what are } \mathrm { V } _ { \mathrm { ab } } , \mathrm { V } _ { \mathrm { ca } } \text { and } \mathrm { V } _ { \mathrm { an } } , \mathrm { V } _ { \mathrm { bn } } , \mathrm { V } _ { \mathrm { cn } } \text { for the balanced three-phase } sources?

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If Vcn=275105V, what are Van,Vbn and Vab,Vbc,Vca for the balanced three-phase V _ { c n } = 275 \angle 105 ^ { \circ } V \text {, what are } V _ { a n } , V _ { b n } \text { and } V _ { a b } , V _ { b c } , V _ { c a } \text { for the balanced three-phase } sources?

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In a balanced ΔY\Delta - Y circuit shown below, the agnitude of the phase voltage is VP = 565V (rms). The wire Impedance is per Zw=17+j11\mathrm { Zw } = 17 + \mathrm { j } 11 , and the phase impedance is ZY=47+j49ΩZ _ { Y } = 47 + j 49 \Omega per phase. Find line currents and voltages of the equivalent Y-connected source assuming Vab=0\angle \mathrm { V } _ { \mathrm { ab } } = 0 ^ { \circ }  In a balanced  \Delta - Y  circuit shown below, the agnitude of the phase voltage is V<sub>P</sub> = 565V  (rms). The wire Impedance is per  \mathrm { Zw } = 17 + \mathrm { j } 11  , and the phase impedance is  Z _ { Y } = 47 + j 49 \Omega  per phase. Find line currents and voltages of the equivalent Y-connected source assuming  \angle \mathrm { V } _ { \mathrm { ab } } = 0 ^ { \circ }

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