Expressions Of Pj Problems
Three Phase Power

3 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:
3 Phase Power
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 Van, Vbn and Vcn in phasor form are:
Van = Van<0o
Vbn = Vbn<-(120)o
Vcn = Vcn<-(240)o = Vcn<120o
Sum of phase voltages equal zero. That is:
Van + Vbn + Vcn = 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', Vab is:
Vab = Van - Vbn
Vab = Van<0o - Vbn<(-120)o
So, Vab = V(cos0 + jsin0) - V[cos(-120) + jsin(-120)] = 3/2V + (√3/2)Vj = √3V<30o.
Line voltages between bb' and cc', Vbc is:
Vbc = Vbn - Vcn
= Vbn<(-120)o - Vcn<120o = √3V<(-90)o.
Line voltages between cc' and aa', Vca is:
Vca = Vcn - Van
= Vcn<120o - Van<0o = √3V<150o.
So, amplitudes of line voltages equal √3 times amplitude of phase voltages.

(b) In = current from node n to node n'
= Ia + Ib + Ic
= (Van + Vbn + Vcn)/Z = 0.
So, no need for a conducting wire between the nodes since In = 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 Vanis:
pa(t) = (V2/R)(1 + cos2ωt)
Instantaneous power due to source Vbn is:
pb(t)= (V2/R)[1 + cos(2ωt - 120o)]
Instantaneous power due to source Vcn is:
pc(t)= (V2/R)[1 + cos(2ωt + 120o)]
Total power P(t) = pa(t) + pb(t) + pc(t)
= 3V2/R + (V2/R)[cos2ωt + cos(2ωt - 120o) + cos(2ωt + 120o)] = 3V2/R + 0 = 3V2/R.
So, P(t) is a constant.


The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
Derivation Of The Area Of A Circle, A Sector Of A Circle And A Circular Ring
Derivation Of The Area Of A Trapezoid, A Rectangle And A Triangle
Derivation Of The Area Of An Ellipse
Derivation Of Volume Of A Cylinder
Derivation Of Volume Of A Sphere
Derivation Of Volume Of A Cone
Derivation Of Volume Of A Torus
Derivation Of Volume Of A Paraboloid
Volume Obtained By Revolving The Curve y = x2 About The X Axis
Single Variable Functions
Absolute Value Functions
Real Numbers
Vector Spaces
Equation Of The Ascent Path Of An Airplane
Calculating Capacity Of A Video Adapter Board Memory
Probability Density Functions
Boolean Algebra - Logic Functions
Ordinary Differential Equations (ODEs)
Infinite Sequences And Series
Introduction To Group Theory
Advanced Calculus - Partial Derivatives
Advanced Calculus - General Charateristics Of Partial Differential Equations
Advanced Calculus - Jacobians
Advanced Calculus - Solving PDEs By The Method Of Separation Of Variables
Advanced Calculus - Fourier Series
Advanced Calculus - Multiple Integrals
Production Schedule That Maximizes Profit Given Constraint Equation
Separation Of Variables As Solution Method For Homogeneous Heat Flow Equation
Newton And Fourier Cooling Laws Applied To Heat Flow Boundary Conditions
Fourier Series
Derivation Of Heat Equation For A One-Dimensional Heat Flow

The Universe is composed of matter and radiant energy. Matter is any kind of mass-energy that moves with velocities less than the velocity of light. Radiant energy is any kind of mass-energy that moves with the velocity of light.
Periodic Table
Composition And Structure Of Matter
How Matter Gets Composed
How Matter Gets Composed (2)
Molecular Structure Of Matter
Molecular Shapes: Bond Length, Bond Angle
Molecular Shapes: Valence Shell Electron Pair Repulsion
Molecular Shapes: Orbital Hybridization
Molecular Shapes: Sigma Bonds Pi Bonds
Molecular Shapes: Non ABn Molecules
Molecular Orbital Theory
More Pj Problem Strings

What is Time?
St Augustine On Time
Bergson On Time
Heidegger On Time
Kant On Time
Sagay On Time
What is Space?
Newton On Space
Space Governance
Imperfect Leaders
Essence Of Mathematics
Toolness Of Mathematics
The Number Line
The Windflower Saga
Who Am I?
Primordial Equilibrium
Primordial Care
Force Of Being

Blessed are they that have not seen, and yet have believed. John 20:29

TECTechnic Logo, Kimberlee J. Benart | © 2000-2021 | All rights reserved | Founder and Site Programmer, Peter O. Sagay.