Three-phase power is a popular method of electric power transmission. One major benefit is smooth power quality. Overhead power lines as well as step-down transformers are fed with 3-phase current because of its many other desirable properties. Systems that employ 3-phase current can be wired either in star or in delta formations. Star systems offer higher voltages and less amperes while delta systems have more amperes and lower voltages. Interestingly, most of the electric power in the world is 3 phase.
- Skill level:
- Moderately Challenging
Determine the three transmission lines (variables) have the same amplitude. Use an oscilloscope. Connect the lines to an oscilloscope and use the screen to check if they are the same height from the line labelled 0.
Determine the three variables (transmission lines) have the same frequency. Use an oscilloscope to determine frequency by checking that all occupy the same number of boxes horizontally.
Determine that the three variables are in phase at 120 degrees. Use an oscilloscope to check this by ensuring that all the lines intersect line 0 at exactly the same points.
Use the formula below to determine the Root Mean Square (RMS) of the wave forms:
Va = Vm < ø m Vb = Vm < ø -120 degrees Vc = Vm < -240 degrees = Vm < ø+120 degrees
Vb = Va ( 1< -120 degrees), and Vc = Va ( 1< +120 degrees)
Let ø = 0 degrees
Va = Vm < 0 degrees, and Vb = Vm <-120 degrees
Use the formula below to determine the wave forms:
VL1 = A sin x VL2 = A sin (x -- 2 / 3 ∏) VL3 = A sin (x -- 4 / 3 ∏)
A = Peak Voltage Voltages L1, L2 and L3 are relative to the neutral
Determine the power to a resistive load using the formula:
P = VI = V*V / R
Use R =1 to calculate power on each load as follows:
PL1 = VVL1 / R = VVL1
PL2 = VVL2 / R = VVL2
PL3 = VVL3 / R = VVL3
Ptot = PL1 + PL2 + PL3
Ptot = SinSin x + SinSin ( x -- 2 / 3∏) + Sin*Sin ( x -- 4 / 3 ∏)
Use the angle subtraction formulae to finish the calculation below:
Ptot = Sin*Sinx + (Sinx cos (2 /3 ∏) -- cos x sin (2/3∏))(Sinx cos (2 /3 ∏) -- cos x sin (2/3∏)) + (Sinx cos ( 4/3 ∏) -- cos x sin ( 4 / 3 ∏)(Sinx cos ( 4/3 ∏) -- cos x sin ( 4 / 3 ∏)
Ptot = 6 / 4 SinSin x + 6 / 4 CosCos x
Ptot = 3 / 2 ( SinSin x + CosCos x)
Use Pythagorean trigonometric identity to solve as shown below:
Ptot = 3 / 2
Tips and warnings
- Where there is no neutral current, the neutral current is calculated as the sum of the phase currents.
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