Answer Posted / thowsif_emerson

The power in watts is equal to volts times amps, but
we must use "phase" volts and "phase" amps to calculate
power.
Calculating these phase relationships requires a little bit
of
trigonometry.

In a "Y" connected circuit, the voltage measured line to
line is not the
true "phase" voltage [phase to neutral voltage], but the
combination of
two voltages that are out of phase by 120 degrees.

Assuming a balanced load, the current in any of the three
phases is the
same as that measured in each line because the line will be
attached to
one end of the phase so there cannot be any difference. We
need the

phase voltage times the phase current to get the phase
power.

There are two phases connected between each pair of lines
in a "Y"
circuit, and since the voltages are not in phase, they do
not add together
to make the line voltage twice the phase voltage. It turns
out through
some basic trigonometry, that the line voltage is equal to
each of the two
phase voltages times the sine of 120 degrees, and the sine
of 120 degrees
is "one-half" the square root of 3. Adding those two halves
together
gives the LINE voltage as the square root of 3 times the
phase voltage.
Or conversely the phase voltage is the line voltage DIVIDED
by the square
root of 3.

So the power in any phase, assuming, again, a balanced load
is
the line voltage times the line current divided by the
square root of 3.

For the three phases, then, the total power is three times
the power in
any phase. 3 X the line voltage X the line current divided
by the square
root of 3. 3 divided by the square root of 3 simplifies to
just the
square root of 3. Multiplying that, as you noted, by the
power factor
converts volt-amps to watts assuming the power factor is
other than one.

In a delta connected circuit, the same problem exists
except with the current measurement. Here, the line to line
voltage
measurement is the phase voltage, but the line current is
composed of two
currents that are out of phase by 120 degrees, and you
guessed it, the
"vector sum" of the two phase current components is the
square root of
three times the current in any phase. And, the phase
current is the line
current divided by the square root of 3, so the power
equation works
exactly the same for both the "Y" and the Delta connected
circuits.

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