In this slide I have explained how two watt meters can be used to measure 3 phase power. Some of the added advantage of this method is that we can calculate 3 phase reactive power and power factor of load as well.

1. ELECTRICAL
ENGINEERING
by - Mohammed Waris Senan

2. Mohammed Waris Senan 2
THREE PHASE SYSTEM
Measurement of 3 phase power
by two wattmeter method

3. Mohammed Waris Senan 3
Wattmeter Terminals & Symbol
M
VC
L
Current Coil (CC)
Potential Coil (PC)

4. Mohammed Waris Senan 4
Wattmeter Connection
P = Vpc Icc cos (angle between Vpc & Icc )
Connection Diagram
CC
PC
voltage
supply
L
O
A
D
LI
LV
M
C V
L

5. Mohammed Waris Senan 5
Measurement of 3 phase Power
Connection Diagram
Reading of wattmeter 1 :
P1 = VAB IA cos (angle between VAB & IA)
Reading of wattmeter 2 :
P2 = VCB IC cos (angle between VCB & IC)
For a balanced three phase STAR connected Load,
Let, Vphase = V, & Iline = Iphase = I
∴ Vline = 𝟑 V & Load pf is cos𝝓 lag
P1 = 𝟑 VI cos (angle between VAB & IA)
P2 = 𝟑 VI cos (angle between VCB & IC)

6. Mohammed Waris Senan 6
Measurement of 3 phase Power
𝝓
𝝓
𝝓
BNV
ANV
CNV
CBV
ABV
AI
CI
BI
BNV
𝟑𝟎°
𝟑𝟎°
Angle between VAB & IA = 30° + 𝝓
Angle between VCB & IC = 30° - 𝝓
P1 = 𝟑 VI cos (30° + 𝝓)
P2 = 𝟑 VI cos (30° - 𝝓)
Total 3 phase power
P3-𝜙 = P1 + P2 = 3VI cos𝝓

7. Mohammed Waris Senan 7
Measurement of 3 phase Power
P1 = 𝟑 VI cos (30°)
P2 = 𝟑 VI cos (30°)
Some Important Cases:
1. If Power factor of load is UNITY
i.e. cos𝝓 = 1, 𝝓 = 0°
Reading of both wattmeter is EQUAL
2. If Power factor of load is 1/2
i.e. cos𝝓 = 1/2, 𝝓 = 60°
P1 = 𝟑 VI cos (30° + 60°) = 𝟑 VI cos (90°) = 0
P2 = 𝟑 VI cos (30° - 60°) = 𝟑 VI cos (30°) = 𝟑
𝟐 VI
Reading of one wattmeter is NEGATIVE
3. If Power factor of load is 0 and ½
i.e. 0 ≤ cos𝝓 < ½ , 60° < 𝝓 ≤ 90°
Reading is taken by reversing the
connection of Current Coil (CC)or
Potential Coil (PC)
Generally connection of Current Coil
is reversed
Reading of one wattmeter is ZERO

8. Mohammed Waris Senan 8
Measurement of Power Factor (cos 𝝓)
P1 – P2 = 3 VI cos (30° + 𝜙) - 3 VI cos (30° - 𝜙)
= 3 VI sin𝜙
So, Three Phase Reactive Power, Q3-𝝓 = 3 (P1 – P2 )


3VIcos
VIsin3
PP
PP
21
21



3
tan









 
21
211
PP
PP
3tan
Now,

9. If power factor is UNITY, cos𝜙 = 1 ⇒ 𝜙 = 0°
P3-𝜙 = 3VLILcos𝜙 = 3 × 400 × 115.4 × 1 = 80 kW
So, Reading of watt-meters will be:
P1 = 3 VI cos (30° + 𝜙) = 3
𝑉𝐿
3
I cos (30° + 𝜙) = VLIL cos (30° + 𝜙)
P1 = 400 × 115.4 × cos (30° + 0°) = 40 kW
P2 = 3 VI cos (30° - 𝜙) = 3
𝑉𝐿
3
I cos (30° + 𝜙) = VLIL cos (30° - 𝜙)
P2 = 400 × 115.4 × cos (30° - 0°) = 40 kW
Mohammed Waris Senan 9
Numerical Problems
In a balanced 3-phase 400 V circuit, Iline = 115.4 A. When power is measured by two watt-meter method, one of the meter
reads 40 kW and another zero. What is the power factor of the load? If power factor is unity and line current is same, what
would be the readings of each watt-meter? [AKU, 2019 (odd sem)] {8 marks}
Question:
Solution:
Given data:
VL = 400 V
IL = 115.4 A
P1 = 40 kW
P2 = 0
power factor, cos𝜙 = ?
Alternately:
We know that if one of the wattmeter reads zero then
pf of the load will be ½ .
Total power, P3-𝜙 = P1 + P2 = 40 + 0 = 40 kW
As, P3-𝜙 = 3VLILcos𝜙
⇒ 3 × 400 × 115.4 × cos𝜙 = 40000
∴ cos𝜙 = 0.5

10. Mohammed Waris Senan 10
Numerical Problems
In a balanced 3-phase star connected load, two watt-meters are
connected to measure the 3-phase power as shown in figure. If
impedance of each phase is 10∠45° Ω and line-to-line voltage is
220 V (rms), what will be readings of two watt-meter?
Question:
Solution:
Reading of wattmeter 1 :
P1 = VAC IA cos (angle between VAC & IA)
Reading of wattmeter 2 :
P2 = VBC IB cos (angle between VBC & IB)
V01270
3
V
V,.V
L
ANAN

Soceas referenTake
A457.12
4510
0127
Z
V
I,
AN
A






Now
Let phase sequence be ABC

11. P1 = 220×12.7×cos (15°) = 2698.8 W
P2 = 220×12.7×cos (75°) = 723.14 W
Mohammed Waris Senan 11
Numerical Problems
Solution:
A1657.12)120-45(7.12I
A457.12I
B
A




V30220V
V90220V
AC
BC





1530-45I&V AAC betweenAngle

7590-165I&V BBC betweenAngle

12. Mohammed Waris Senan 12

30V3VAC 
ANV
CNV
BNV CNV
BCV
9𝟎°
ACV
𝟑𝟎°

90V3VBC 

13. Mohammed Waris Senan 13



120VVor240VV
120VV
0VV
cnCN
BN
AN






150VVor210V3V
90V3V
30V3V
cnCA
BC
AB



ANV
CNV
BNV
CAV
ANV
𝟑𝟎°
𝟑𝟎°
CNV
BNV
ABV
ANV
𝟑𝟎°
CNV
𝟑𝟎°

30V3VAC 