118 Three-phase a.c. circuits
cos ~b so that the phase currents are lagging the phase voltages by ~b. Wattmeter
W~ reads
VAclA cos (30 ~ - ~b) (5.10)
where VAC is a line voltage and IA is a line current.
Wattmeter W2 reads
VBClB cos (30 ~ + 6) (5.11)
where VBc is a line voltage and IB is a line current.
For a leading power factor of cos ~b the sign of ~b will change in Equations
(5.10) and (5.11). The power represented by the reading on Wl is:
P1 = VEIL (COS 30 ~ COS 6 + sin 30 ~ sin 6) - VLIL[(~V/3/2) COS 6 + (1/2)sin 6]
The power represented by the reading on W2 is:
P1 = VEIL (COS 30 ~ COS ~b - sin 30 ~ sin ~b) = VLIL[(V3/2) COS ~b - (1/2)sin th]
The power represented by the readings on WI and W2 is thus
P, + P2 = V3 VEIL COS 6 (5.12)
which is the total power in a balanced three-phase circuit.
Therefore the sum of the readings on the two wattmeters gives the total
power in the three-phase circuit. If the phase angle is greater than 60 ~ (leading
or lagging) one of the wattmeters will read negative because cos (30 ~ + ~b) is
then negative and the reading must be subtracted from the other to give the
total power.
Now P~ - P2 = VEIL sin ~b, which is 1/V'3 of the total reactive power, so if we
multiply (P~ - P2) by V'3 we obtain the total reactive power in the three-phase
circuit. Thus
Var = N/3(Pl- P2) = X/3 VLIL sin 6 (5.13)Figure 5.17
! b--_~_ r ..... ....
I I // "/
i I /
I I //~1 VAC
./~/'/ / /./ I
VAN /IA I I. I V cN
r ""\'-. VBc
I~.~ ..... ..~> ..... J-/-
Vc, r ,"t" ~ //
IB \\\ \ / /r
\ ,,. \ / J
\\ //
\