THREE-PHASE CIRCUITS PART II

POWER IN A BALANCED SYSTEM

• For a Y-connected load with θ∠= ZZY , the phase voltages and currents are:

,cos2 tVv pAN ω= ),120cos(2 o−= tVv pBN ω )120cos(2 o+= tVv pCN ω ),cos(2 θω −= tIi pAN ),120cos(2 o−−= θωtIi pBN )120cos(2 o+−= θωtIi pCN where pV and pI are the rms voltage and current, respectively.

• The total instantaneous power absorbed by the load is

)()()()( tptptptp CBA ++= CNCNBNBNANAN iviviv ++= )120cos()120cos()cos([cos2 oo −−−+−= θωωθωω ttttIV pp

)]120cos()120cos( oo +−++ θωω tt

• Applying )]cos()[cos(21coscos BABABA −++= gives

)240cos(coscos3[)( o−++= ααθppIVtp )]240cos( o++ α

where θωα −= t2

• Applying BSBABA sinsincoscos)cos( −=+ yields

oo 240sinsin240coscoscoscos3[)( αααθ +++= ppIVtp ]240sinsin240coscos oo αα −+

]cos)21(2coscos3[ ααθ −++= ppIV

θcos3 ppV I=

→ The total instantaneous power in a balanced three-phase system is constant regardless of whether the load is Y- or Δ-connected while that of each phase is still time varying.

→ The average power consumed by each phase of the load is one- ).( third of tp

θcosppp IVP =

→ The reactive power consumed by each phase of the load is

θsinppp IVQ = → The apparent power consumed by each phase of the load is

ppp IVS = → The complex power consumed by each phase of the load is

∗=+= ppppp IVjQPS

Note that pV is the phasor rms voltage whose magnitude is pV pI is the phasor rms current whose magnitude is pI

• The total complex power is

*

22 3

333p

pppppp

ZV

ZIIVSS ==== ∗

where θ∠= Yp ZZ or θ∠ΔZ (the load impedance per phase)

• Alternatively, since pLpL VVII 3, == for a Y-connected load or pLpL VVII == ,3 for a Δ-connected load,

θθ cos3cos3 LLpp IVIVP == θθ sin3sin3 LLpp IVIVQ ==

θθ ∠=∠=+= LLpp IVIVjQPS 33

Ex. Practice problem 12.6

For the Y-Y circuit of Practice Prob.12.2, calculate the complex power at the source and at the load.

Ex. Practice problem 12.7

Calculate the line current (magnitude only) required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 440 V.

UNBALANCED THREE-PHASE SYSTEMS

• An unbalanced system is due to unbalanced voltage sources or an unbalanced load.

• Solved by direct application of mesh and nodal analysis or ohm’s law

• For a four-wire Y-Y system, the neutral line current is not zero

)( cban IIII ++−=

• Each phase of the load consumes unequal power. Thus, the total power is the sum of the powers in the three phases.

Ex. Practice problem 12.9

The unbalanced Δ-load shown is supplied by balanced line-to-line voltages of 240 V in the positive sequence. Find the line currents using Vab as reference.

Ex. Practice problem 12.9

For the unbalanced circuit shown, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source.

RESIDENTIAL WIRING

SINGLE-PHASE THREE-WIRE RESIDENTIAL WIRING

GROUNDING & SAFETY

GROUND FAULT CIRCUIT INTERRUPTER (GFCI)

GENERATION & DISTRIBUTION OF AC POWER