3-Phase AC Problem: UNIZOR.COM - Physics4Teens - Electromagnetism - Alte...

Notes to a video lecture on http://www.unizor.com



Problems on AC Induction



Problem A



Three-phase generator has four wires coming out from it connected to a "star" with three phase wires carrying sinusoidal EMF shifted by 120°=2π/3 from each other and one neutral wire.

The angular speed of a generator's rotor is ω.

Assume that the peak difference in electric potential between each phase wire and a neutral one is E.

Describe the difference in electric potential between each pair of phase wires as a function of time.



Solution



The difference in electric potential between phase wires and a neutral one can be described as

Phase 1: E₁(t)=E·sin(ωt)

Phase 2: E₂(t)=E·sin(ωt−2π/3)

Phase 3: E₃(t)=E·sin(ωt+2π/3)

The difference in electric potential between phase 1 wire and phase 2 wire can be represented as

E_1,2(t) = E₁(t) − E₂(t)

Similarly, the difference in electric potential between two other pairs of phase wires is

E_2,3(t) = E₂(t) − E₃(t)

E_3,1(t) = E₃(t) − E₁(t)

Let's calculate all these voltages.

E_1,2(t) = E·sin(ωt) − E·sin(ωt−2π/3) =

= E·sin((ωt−π/3)+π/3) − E·sin((ωt−π/3)−π/3) =

[substitute φ=ωt−π/3]

= E·sin(φ+π/3)−E·sin(φ−π/3) =

= E·[sin(φ+π/3)−sin(φ−π/3)]

Let's simplify the trigonometric expression.

sin(φ+π/3) − sin(φ−π/3) =

= sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) −

− sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) =

= 2cos(φ)·sin(π/3) =

= √3cos(φ) =

= √3cos(ωt−π/3)

Therefore,

E_1,2(t) = E√3cos(ωt−π/3)

As we see, the electric potential between phase 1 and phase 2 wires is also sinusoidal (since cos(x)=sin(x+π/2), cos(ωt−π/3) equals to sin(ωt+π/6)),
but shifted in time, and its peak voltage is greater than the peak
voltage between a phase wire and a neutral one by a factor of √3.

Similar factor difference of √3 is between effective voltages of these pairs of wires.

Analogous calculations for the other pairs of phase wires produce the following.

E_2,3(t) = E·sin(ωt−2π/3) − E·sin(ωt+2π/3) =

[substitute φ=ωt]

= sin(φ)·cos(2π/3) −

− cos(φ)·sin(2π/3) −

− sin(φ)·cos(2π/3) −

− cos(φ)·sin(2π/3) =

= −2cos(φ)·sin(2π/3) =

= −√3cos(φ) =

= −√3cos(ωt)

Therefore,

E_2,3(t) = −E√3cos(ωt)

Also the same factor difference of √3 relative to phase/neutral voltage.

Finally, the third phase/phase voltage calculations produce the following.

E_3,1(t) = E·sin(ωt+2π/3) − E·sin(ωt) =

= E·sin((ωt+π/3)+π/3) − E·sin((ωt+π/3)−π/3) =

[substitute φ=ωt+π/3]

= E·sin(φ+π/3)−E·sin(φ−π/3) =

= E·[sin(φ+π/3)−sin(φ−π/3)]

Let's simplify the trigonometric expression.

sin(φ+π/3) − sin(φ−π/3) =

= sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) −

− sin(φ)·cos(π/3) +

+ cos(φ)·sin(π/3) =

= 2cos(φ)·sin(π/3) =

= √3cos(φ) =

= √3cos(ωt+π/3)

Therefore,

E_3,1(t) = E√3cos(ωt+π/3)

Again, the same factor difference of √3 relative to phase/neutral voltage.

CONCLUSION

The voltage between any two phase wires is in magnitude greater than phase to neutral voltage by √3. Both are sinusoidal, but one is shifted in time relatively to another.

The phase to phase effective voltage (which is by √2 less then peak voltage) also is greater than phase to neutral by the same √3.

EXAMPLES

If phase to neutral effective voltage is 127V, the phase to phase effective voltage is 220V.

If phase to neutral effective voltage is 220V, the phase to phase effective voltage is 380V.