n
Source connections
are shown soon.
Load connections
Summary of line and phase relationships:
Let VLL denote the magnitude of line voltages Vab , Vbc , and Vca .
Let IL denote the magnitude of line current.
Let Iφ denote the magnitude of phase current.
Let Vφ denote the magnitude of phase voltage.
We have the following relationships:
Y Connection
Voltage Magnitudes
VLL =
√
3Vφ
Current Magnitudes
IL = Iφ
For abc phase sequence
Vab leads Va by 30
Line voltage leads phase voltage by 30
∆ Connection
VLL = Vφ
IL =
0
√
3Iφ
Ia lags Iab by 30
0
0
Line current lags phase current by 30
0
Power in Three Phase Circuits
Let Vφ and Iφ , denote the magnitude of phase RMS values.
Similarly, let VL , and IL denote the magnitude of line RMS values.
Let θ be the phase angle of the load.
√
Note that in Y connection Iφ = IL , however VL is 3Vφ .
√
In ∆ connection, Vφ = VL , but IL is 3Iφ .
√
Real power = P = 3Vφ Iφ cos θ = 3VL IL cos θ
√
Reactive power = Q = 3Vφ Iφ sin θ = 3VL IL sin θ
√
Magnitude of Complex power = S = 3Vφ Iφ = 3VL IL
Power Factor = cos θ
Reactive Factor = sin θ
Power Triangle
Instantaneous power in three phase circuits
In each single phase, power varies as a function of time since both voltage
and current vary with respect to time. However, all three single phase instantaneous powers add up to a constant. That is, power delivered by a three
phase AC supply is the same at each instant of time as in the case of DC.
This is a remarkable property and is one of the reasons why three phase AC
power transmission became popular.
Let us compute next the three phase instantaneous power.
Consider a balanced Y connection with per phase impedance as Zφ = Z 6 θ.
Let V and I be per phase RMS voltage and RMS current.
√
2V cos ωt
van =
√
0
vbn =
2V cos(ωt − 120 )
√
0
2V cos(ωt − 240 )
vcn =
Magnitude of current I =
V
|Z| .
√
ia =
2I cos(ωt − θ)
√
0
ib =
2I cos(ωt − 120 − θ)
√
0
ic =
2I cos(ωt − 240 − θ)
Power supplied by Phase a is
pa (t) = van ia =
√
√
2V cos ωt 2I cos(ωt−θ) = 2V I cos ωt cos(ωt−θ) = V I cos θ−V I cos(2ωt−θ).
Note that 2 cos α cos β = cos(α − β) + cos(α + β)
Similarly,
0
pb (t) = vbn ib = V I cos θ − V I cos(2ωt − 240 − θ)
0
pc (t) = vcn ic = V I cos θ − V I cos(2ωt − 480 − θ)
Total power p(t) = pa (t) + pb (t) + pc (t)
= V I cos θ − V I cos(2ωt − θ)
0
+ V I cos θ − V I cos(2ωt − 240 − θ)
0
+ V I cos θ − V I cos(2ωt − 480 − θ)
= 3V I cos θ + 0
Total instantaneous power = 3V I cos θ =
Note that this does not vary with time t.
√
3VL IL cos θ