A wattmeter is connected as shown in the figure. The wattmeter reads
Concept:
Wattmeter reads the power and it is given by
P = VPC ICC cos ϕ
VPC is the voltage across pressure coil
ICC is current flows through the current coil
ϕ is the phase angle between VPC and ICC
Application:
The circuit representation of the given question is as shown below.
Concept:
In a two-wattmeter method, for lagging load
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)
Power factor = cos ϕ
\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)
Calculation:
Given that, W1 = 2 W2
\(\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} - {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}\)
⇒ ϕ = 30°
Power factor = cos 30° = 0.866
Important Point:
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
In the measurement of power of balanced load by two wattmeter method in a 3-phase circuit, The readings of the wattmeters are 4 kW and 2 kW respectively, the later is being taken after reversing the connections of current coil. the power factor and reactive power of the load is
Concept:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Calculation:
Given that,
Reversing the connections
W1 = 4 kW, W2 = - 2 kW
Total reactive power (Q) = √3[4 - (- 2)] = 6 √3 kVAR
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{4} - {(-2)}} \right)}}{{{4} - {2}}}} \right)\)
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{6\sqrt 3}}{{{2}}}} \right)\)
ϕ = 79.10^{o}
cosϕ = 0.2 (approx)
Important Points
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I_L}}{2}\) |
VLIL |
1.5 VL IL |
W1= 2W2 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
According to Blondell’s theorem, the no. of watt meters to be required to measure the total power in n-phase system is either N (or) N–1.
When a separate neutral wire is available in the system the no. of watt meters to be required is N.
When the neutral wire is not available in the system, then the no. of watt meters to be required is (N–1).
Power is measured according to Blondel's theorem
Phase/conductor |
Wattmeter required |
n |
n-1 |
n with neutral wire |
n |
One line is acting as a common line for the return path. Hence the minimum number of wattmeters required is 2.
Additional Information
Measurement of polyphase power: 3 - ϕ
1) For 3ϕ – 4 wire → neutral is present
2) As per Blondel’s theorem, if neutral is not present for the ‘n’ wire system, then the ‘n-1’ number of wattmeter’s are required for the ‘n’ wire system.
Type of Load |
Star 3 - ϕ 3 - wire |
Star 3 - ϕ 4 – wire |
Delta 3 - ϕ 3 – wire |
Balanced |
2 |
1, 2, 3 |
2 |
Unbalanced |
2 |
2, 3 |
2 |
Concept:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)
Total power in the circuit (P) = W1 + W2 = 3 Vph Iph cos ϕ
CALCULATION:
Given: VL = 400 V, I_{L} = 5 A, ϕ = \(30^\circ\)
W_{1} = 400 × 5 × cos (30 + 30) = 1000 W
W_{2} = 400 × 5 × cos (30 - 30) = 2000 W
Additional Information
In two wattmeter method, the power factor is given by
\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
Here \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
When both the wattmeter’s show the same positive reading, ϕ becomes zero and hence power factor is unity.
In two wattmeter method, the wattmeter readings are given by
W1 = VLIL cos (30 – ϕ)
W2 = VLIL cos (30 + ϕ)
Total power = W1 + W2
Reactive power = √3 (W1 – W2)
Power factor = cos ϕ
Where \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}} \right)\)
Important Point:
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
According to Blondel’s theorem:
When two wattmeters are used to measure power of a 3 - φ balanced circuit and one wattmeter reads negative, it means the angle of lag is
Concept:
The two watt-meter method used to measure three-phase power. Readings of both watt-meters are
\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30 - \phi } \right)\)
\({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {30 + \phi } \right)\)
In a two watt-meter method, the Power factor can be calculated as
\(\cos \phi = \cos \left[ {{{\tan }^{ - 1}}\sqrt 3 \left( {\frac{{{W_1} - {W_2}}}{{{W_1} + {W_2}}}} \right)} \right]\)
In a two watt-meter method, the total power can be calculated as
P = W1 + W2
Where, Vp = Voltage across pressure coil
Ip = Current through the current coil
ϕ = Phase angle
Case 1st
Case 2nd
Case 3rd
Case 4th
Note:
Concept:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I_L}}{2}\) |
VLIL |
1.5 VL IL |
W1= 2W2 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
Explanation:
W_{1} = -ve, W_{2} = +ve
|W_{1}| ≠ |W_{2}|, then the powere factor range is given as
The range of ϕ is, 60 < ϕ < 90
Now the range of power factor is, 0.5 > cos ϕ > 0.
Concept:
Two-wattmeter method:
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)
Power factor = cos ϕ
\(ϕ =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)
If,
W_{1} > W_{2}: The power factor consider as lagging
W_{2} > W_{1}: The power factor consider as leading
Hence, When the power factor is changed from lagging to leading, the reading of W1 and W2 get interchanged.
The wattmeter reads
Wattmeter:
A wattmeter is an electrical instrument that is used to measure the electric power of any electrical circuit.
The internal construction of a wattmeter is such that it consists of two coils. One of the coils is in series and the other is connected in parallel.
The coil that is connected in series with the circuit is known as the current coil and the one that is connected in parallel with the circuit is known as the voltage coil
Wattmeter measures the average power.
The reading of wattmeter will be,
Average power = V_{pc} I_{cc} cos ϕ
Where,
V_{pc} is the voltage across pressure coil
I_{cc} is current flows through the current coil
ϕ is the phase angle between V_{pc} and Icc
So, wattmeter is a device capable of detecting voltage, current and the angle between the voltage and the current to provide power readings.
Concept:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Calculation:
Given that,
W_{1} = 20 kW, W_{2} = 10 kW
Total reactive power (Q) = √3(20 – 10) = 17.42 kvar
Electro dynamo type wattmeter:
Pressure coil Inductance error in Electrodynometer wattmeter
NOTE:
In two wattmeter method, the power factor is given by
\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
Here \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
When both the wattmeter’s show the same positive reading, ϕ becomes zero and hence power factor is unity.
Important Point:
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
Wattmeter:
Wattmeter works on the power formula given below
P = VI cos ϕ
Where, P = Power, V = Voltage, I = current, cosϕ = power factor.
3-phase power and power factor can be measured by using a single-phase wattmeter as given below
Blondel’s Theorem: According to this theorem for the n-wire system, (n – 1) single-phase wattmeters are used.
Concept:
In a two-wattmeter method, for lagging load
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)
Power factor = cos ϕ
\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)
Calculation:
Given that, one of the wattmeter reads zero.
W2 = 0
\(\tan \phi = \frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}\)
\(\tan \phi = \frac{{\sqrt 3 \left( {{W_1} - 0} \right)}}{{\left( {{W_1} + 0} \right)}}\)
⇒ tan ϕ = √3
⇒ ϕ = tan-1 (√3) = 60°
Power factor = \(\cos \phi = \cos 60^\circ = \frac{1}{2} = 0.5\)Concept:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Calculation:
Given VL = 400 V, IL = 5 A
Phase angle, ϕ = 30°
W1 = 40 × 5 × cos (30 – 30) = 2000 W
W2 = 40 × 5 × cos (30 + 30) = 1000 WIn 3-phase power measurement by two-wattmeter method, the two wattmeters read as W_{1} = 300 W and W_{2} = 300 W. Then the load is said to be operating at _________.
Concept:
In two wattmeter method, the power factor is given by
\(cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
Here \(ϕ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
W_{1} = reading of 1^{st} wattmeter
W_{2} = reading of 2^{nd} wattmeter
When both the wattmeter’s show the same positive reading, ϕ becomes zero and hence power factor is unity.
Calculation:
Given that
W_{1} = 300 Watt and W_{2} = 300 Watt
Now power factor
\(\cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
cos ϕ = 1 (Unity power factor)
Additional Information
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
Concept:
In a two-wattmeter method, for lagging load
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Calculation:
Given that, W2 = 2W1
\(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - 2{W_1}} \right)}}{{\left( {{W_1} + 2{W_2}} \right)}}} \right) = - 30^\circ \)
Power factor = cos ϕ = 0.866
Blondel’s Theorem:
Application:
Given reading of wattmeter A (W_{A}) = 500 W
Reading of wattmeter B (W_{B}) = 1000 W
According to Blondel’s theorem, the total power of the circuit (W) will be,
W = W_{A} + W_{B}
W = 500 + 1000 = 1500 W