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Three Phase Power
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Three Phase Power

Figure 8.56 is a Wye (or Y) configuration of a Balanced three phase AC circuit. Show that:
(a) The magnitude of the line voltages is equal to √3 times the magnitude of the phase voltages.
(b) No conducting wire is needed to connect nodes n and n^'.
(c) If the 3 balanced load impedances are replaced with 3 equal resistances R, the total instantaneous power delivered to the balanced load by the 3-phase generator is constant.

The string:
S₇P₃A₃₂ (Force - Push).
The math:

Pj Problem of interest is of type force. Power and energy problems are force problems
Three-phase power is a configuration of three sinusoidal voltages that are generated out of phase with each other. It came about as a result of the need to improve the efficiency of single phase power delivery. some of its advantages include:
(1) Delivery of steady constant supply of power. A single phase delivery is pulsating.
(2) Reduced transmission losses over long distances.
(3) Efficient use of conductors and circuit components
(4) Nonzero starting torque (moment of a force) for three-phase motors.
Balanced voltages implies that the voltages have equal amplitude and frequency and are out of phase by 120 degrees.
So, the phase voltages of V_an, V_bn and V_cn in phasor form are:
V_an = V_an<0^o
V_bn = V_bn<-(120)^o
V_cn = V_cn<-(240)^o = V_cn<120^o
Sum of phase voltages equal zero. That is:
V_an + V_bn + V_cn = 0.
Voltage amplitudes, V are rms values,
(a) The line voltages (or line-line-voltages) in figure 8.56 are:
Voltages between lines aa^' and bb^'
Voltages between lines aa^' and cc^'
Voltages between lines bb^' and cc^'.

Line voltage between aa^' and bb^', V_ab is:
V_ab = V_an - V_bn
V_ab = V_an<0^o - V_bn<(-120)^o
So, V_ab = V(cos0 + jsin0) - V[cos(-120) + jsin(-120)] = 3/2V + (√3/2)Vj = √3V<30^o.
Line voltages between bb^' and cc^', V_bc is:
V_bc = V_bn - V_cn
= V_bn<(-120)^o - V_cn<120^o = √3V<(-90)^o.
Line voltages between cc^' and aa^', V_ca is:
V_ca = V_cn - V_an
= V_cn<120^o - V_an<0^o = √3V<150^o.
So, amplitudes of line voltages equal √3 times amplitude of phase voltages.

(b) I_n = current from node n to node n^'
= I_a + I_b + I_c
= (V_an + V_bn + V_cn)/Z = 0.
So, no need for a conducting wire between the nodes since I_n = 0.

(c) General expression for instantaneous power p(t):
p(t) = (V²/R)[1 + cos(2ωt + θ)]
where V is rms value.
So, instantaneous power due to source V_anis:
p_a(t) = (V²/R)(1 + cos2ωt)
Instantaneous power due to source V_bn is:
p_b(t)= (V²/R)[1 + cos(2ωt - 120^o)]
Instantaneous power due to source V_cn is:
p_c(t)= (V²/R)[1 + cos(2ωt + 120^o)]
Total power P(t) = p_a(t) + p_b(t) + p_c(t)
= 3V²/R + (V²/R)[cos2ωt + cos(2ωt - 120^o) + cos(2ωt + 120^o)] = 3V²/R + 0 = 3V²/R.
So, P(t) is a constant.

Math

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