Page 453 - Basic Electrical Engineering

Let us consider the measurement of three-phase power of a star-connected
load using two single-phase wattmeters as has been shown in Fig. 4.17(a).
We will calculate the power measured by the two wattmeters separately. Let

W and W respectively be the two wattmeter readings. Current flowing
2
1
through the current coil of wattmeter W is I . The voltage appearing across
R
1
its pressure coil is V . The wattmeter reading will be equal to, W = V RB R
I .
RB
1
cos of angle between V RB and I . Similarly, the wattmeter reading W will be
R
2
equal to, W = V YB B YB and I . We will now draw
I cos of angle between V
2
B
the phasor diagram, and calculate W and W .
1
2
From the phasor diagram as shown in Fig. 4.17 (b),
W = V I cos (30 − ϕ) = V I cos (30 − ϕ) = V I cos (30 − ϕ)
L L
ph ph
1
RB L
(4.14)
And

W = V YB Y ph ph L L
I cos (30 + ϕ) = V I cos (30 + ϕ) = V I cos (30 + ϕ)
2
(4.15)

We know that the total power in a three-phase circuit is 3V I cos ϕ or
ph ph
equal to V I cos ϕ
L L

Figure 4.16 Two-wattmeter method of measuring power for star- and delta-connected load