MrMark
Well-known member
Lets try some math?
First the peak voltage for a 120VAC sine wave is actually 169.7V. Alternating current, or AC, is constantly changing direction and intensity. If you could periodically sample the voltage at sequential intervals in one cycle, you would find that in a 120V system the instantaneous voltage varies between -169.7V and +169.7V. Without doing any actual math, it turns out that the RMS value of a sine wave is approximately 0.707 times the peak value
In a single phase center tap application like is found in US Homes, the following two equations represent the voltage of each Leg:
Leg 1 Voltage = L1(t) = +169.7 * sin(120PI(t) + 0)
Leg 2 Voltage = L1(t) = -169.7 * sin(120PI(t) + 0)
At t=1/240, or 1/4 way through a cycle, the peek voltage you would get:
Leg 1 Voltage = L1(1/240) = +169.7 * sin(120PI(1/240) + 0) = +169.7
Leg 2 Voltage = L1(1/240) = -169.7 * sin(120PI(1/240) + 0) = -169.7
The differential between the Legs is now 339.4 Volts. Note I said differential, not the sum. Applying the RMS, the voltage between the two Legs is 240V.
Now is we looked at a two phase system with the sine waves perfectly out of sinc we would get:
Leg 1 Voltage = L1(t) = +169.7 * sin(120PI(t) + 0)
Leg 2 Voltage = L1(t) = -169.7 * sin(120PI(t) + PI)
At t=1/240, or 1/4 way through a cycle, the peek voltage you would get:
Leg 1 Voltage = L1(1/240) = +169.7 * sin(120PI(1/240) + 0) = +169.7
Leg 2 Voltage = L1(1/240) = -169.7 * sin(120PI(1/240) + PI) = +169.7
The differential between the Legs is now 0 Volts. Note I said differential, not the sum.
I believe I have all the math correct, it's been a few (many) years since I've done this, but there you go. Oh the reason the L2 voltage is a Negative is because of the transformer and the center tap.
Hope this helps.
I don't know about the math, it is probably right, but it is tangential to what we are discussing.
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