Three-Phase Circuits
Demonstrative Video
VIDEO
Problem-1: Line and Phase Quantities
Determine the following quantities for the circuit shown
line currents
phase voltages of the load
line voltages of the load
phase voltages of the source
line voltages of the source
Solution-1
\[\begin{array}{l}
\mathbf{I}_{a A}=\frac{\mathbf{V}_{a^{\prime} n}}{Z_{g a}+Z_{1
a}+Z_{A}}=\frac{120 \angle 0^{\circ}}{40+j 30}=\left(2.4
\angle-36.87^{\circ}\right) \mathrm{A} \\
\mathbf{I}_{b B}=\mathbf{I}_{a A}<-120^{\circ}=\left(2.4
\angle-156.87^{\circ}\right) \mathrm{A} \\
\mathbf{I}_{c C}=\mathbf{I}_{a A}<+120^{\circ}=\left(2.4
\angle+83.13^{\circ}\right) \mathrm{A}
\end{array}\]
\[\begin{array}{l}
\mathbf{V}_{A N}=\mathbf{I}_{a A} Z_{A}=\left(2.4
\angle-36.87^{\circ}\right)(39+j 28)=\left(115.22
\angle-1.19^{\circ}\right) \mathrm{V}\\
\mathbf{V}_{B N}=\mathbf{V}_{A N} \angle-120^{\circ}=\left(115.22
\angle-121.19^{\circ}\right) \mathrm{V} \\
\mathbf{V}_{C N}=\mathbf{V}_{A N} \angle+120^{\circ}=\left(115.22
\angle+118.81^{\circ}\right) \mathrm{V}
\end{array}\]
The 3 line currents (of both load & source) are:
The 3 line voltages of the load are:
\[\begin{aligned}
\mathrm{V}_{A B} &=\left(\sqrt{3} \angle 30^{\circ}\right)
\mathbf{{V}_{A N}} \\
&=\left(\sqrt{3} \angle 30^{\circ}\right)\left(115.22
\angle-1.19^{\circ}\right) \\
&=\left(199.58<+28.81^{\circ}\right) \mathrm{V}
\end{aligned}\]
\[\begin{aligned}
\mathrm{V}_{B C} &=\mathbf{V}_{A B} \angle-120^{\circ} \\
&=\left(199.58 \angle-91.19^{\circ}\right) \mathrm{V} \\
\mathrm{{V}_{C A}} &=\mathbf{V}_{A B} \angle+120^{\circ} \\
&=\left(199.58 \angle+148.81^{\circ}\right) \mathrm{V}
\end{aligned}\]
\[\begin{aligned}
\mathbf{V}_{a n}=\mathbf{V}_{a^{\prime} \eta}-\mathbf{I}_{a A} Z_{g a}
&=120-\left(2.4 \angle-36.87^{\circ}\right)(0.2+j 0.5) \\
=&\left(118.9 \angle-0.32^{\circ}\right) \mathrm{V} \\
\mathbf{V}_{b n}=& \mathbf{V}_{a n} \angle-120^{\circ}=\left(118.9
\angle-120.32^{\circ}\right) \mathrm{V} \\
\mathbf{V}_{c n}=& \mathbf{V}_{a n} \angle+120^{\circ}=\left(118.9
\angle+119.68^{\circ}\right) \mathrm{V}
\end{aligned}\]
\[\begin{aligned}
\mathbf{V}_{a b} &=\left(\sqrt{3} \angle
30^{\circ}\right)\mathrm{V_{a n}}=\left(\sqrt{3} \angle
30^{\circ}\right)\left(118.9 \angle-0.32^{\circ}\right) \\
&=\left(205.94 \angle+29.68^{\circ}\right) \mathrm{V} \\
\mathbf{V}_{b c} &=\mathbf{V}_{a b} \angle-120^{\circ}=\left(205.94
\angle-90.32^{\circ}\right) \mathrm{V} \\
\mathbf{V}_{c a} &=\mathbf{V}_{a b} \angle+120^{\circ}=\left(205.94
\angle+149.68^{\circ}\right) \mathrm{V}
\end{aligned}\]
The 3 phase voltages of the source are:
Problem-2: Complex power and Power Dissipation
The equivalent Y-Y configuration of a circuit is shown. Determine the
complex powers provided by the source and dissipated by the line of a
a-phase?
Solution-2
\[\begin{aligned}
S_{\phi}& =\mathbf{V}_{\phi} \mathbf{I}_{\phi}^{*}\\
\Rightarrow (160+j 120) 10^{3}&=\frac{600}{\sqrt{3}} \mathbf{I}_{a
A}^{*} \\
\Rightarrow \mathbf{I}_{a A}& =\left(577.35
\angle-36.87^{\circ}\right) \mathrm{A}
\end{aligned}\]
\[\begin{aligned}
\mathbf{V}_{a n}& =\mathbf{V}_{A N}+\mathbf{I}_{a A} Z_{1 a} \\
&=600 / \sqrt{3}+\left(577.35
\angle-36.87^{\circ}\right)(0.005+j 0.025)\\
&=\left(357.51 \angle 1.57^{\circ}\right) \mathrm{V}
\end{aligned}\]
The line current of a-phase can be calculated by the complex power
is:
\[\begin{aligned}
S_{a n} &=\mathbf{V}_{a n} \mathbf{I}_{a A}^{*}=\left(357.51 \angle
1.57^{\circ}\right)\left(577.35 \angle 36.87^{\circ}\right) \\
&=\left(206.41 \angle 38.44^{\circ}\right) \mathrm{kV} \mathrm{A}
\end{aligned}\]
\[\begin{aligned}
S_{a A} &=\left|\mathbf{I}_{a A}\right|^{2} Z_{1
a}=(577.35)^{2}(0.005+j 0.025) \\
&=\left(8.50 \angle 78.66^{\circ}\right) \mathrm{kV} \mathrm{A}
\end{aligned}\]
The complex power provided by the source of aphase is:
Problem-3: Star and Delta Configurations
Each phase of a three-phase alternator, generates a voltage of 3810.5
V and can carry a maximum current of 30 A.
Find the following quantities:
line current,
line voltage
total kVA capacity,
if the alternator is connected in
star configuration
delta configuration
Solution-3
\(\text { Given data: }~~ E_{p h}=3810.5
\mathrm{V} ~~ I_{p h}=30 A\)
Star
Delta
$$\begin{aligned}
\Rightarrow & \text{Line current}~ \mathrm{I}_{t}
=\mathrm{I}_{\mathrm{ph}}=30 \mathrm{A} \\
\Rightarrow & \text{Line voltage} ~\mathrm{E}_{\ell} =\sqrt{3}
\mathrm{E}_{\mathrm{ph}}\\
& =6600 \mathrm{V} \\
\Rightarrow & \text{Total}~ \mathrm{kVA} =\sqrt{3} \mathrm{E}_{l}
\mathrm{I}_{l} \\
=&\sqrt{3} \times 6600 \times 30 \times 10^{-3} \\
&=342.95
\end{aligned}$$
$$\begin{aligned}
\Rightarrow & \mathrm{I}_{\ell}=\sqrt{3}
\mathrm{I}_{\mathrm{ph}}=51.96 \mathrm{A} \\
\Rightarrow & \mathrm{E}_{\ell}=3810.5 \mathrm{V} \\
\Rightarrow & \text { Total kVA }=\sqrt{3} \mathrm{E}_{l}
\mathrm{I}_{l} \\
=&\sqrt{3} \times 3810.5 \times 51.96 \times 10^{-3} \\
&=342.95
\end{aligned}$$
Problem-4: Balanced 3-phase System
A balanced \(3-\phi\) load connected
in star consists of \((6 + j
8)~\Omega\) impedance in each phase. It is connected to a \(3-\phi\) supply of 400 V, 50 Hz.
Find the following quantities:
magnitude of phase current
magnitude of line current
per phase power
total power.
Solution-4
\[\begin{aligned}
E_{ph}& =400 / \sqrt{3}=230.94 \mathrm{V}\\
Z&=(6+j 8) \Omega=10 \angle 53.13^{\circ} \Omega \\
\mathrm{I}_{\mathrm{ph}}&=230.94 / 10=23.094 \mathrm{A} \\
\end{aligned}\]
\[\begin{aligned}
\mathrm{I}_{\ell}&=\mathrm{I}_{\mathrm{ph}}=23.094 \mathrm{A}\\
P_{1-\phi}&=E_{p h} I_{p h} \cos \theta\\
=&230.94 \times 23.094 \times \cos 53.13^{\circ}\\
& =3200 \mathrm{W}
\end{aligned}\]
\[\begin{aligned}
\mathrm{P}_{\mathrm{T}}&=\sqrt{3} \mathrm{E}_{l} \mathrm{I}_{l}
\cos \theta \\
=&\sqrt{3} \times 400 \times 23.094 \times \cos 53.13^{\circ}\\
&=9600 \mathrm{W} \\
\text{or}~\mathrm{P}_{\mathrm{T}}&=3 \mathrm{P}=9600 \mathrm{W}
\end{aligned}\]
\(E_{p h}=400 ~\mathrm{V} \quad Z=(6+j
8) \Omega=10 \angle 53.13^{\circ} \Omega\)
Phase current, \(\mathrm{I}_{\mathrm{ph}}=400 / 10=40
\mathrm{A}\)
Line current, \(\mathrm{I}_{\ell}=\sqrt{3}
\mathrm{I}_{\mathrm{ph}}=69.282 \mathrm{A}\)
Per phase power, \(P=E_{p h} I_{p h}
\cos \theta=400 \times 40 \times \cos 53.13^{\circ}=9600
\mathrm{W}\)
Total power, \(\mathrm{P}_{T}=\sqrt{3}
\mathrm{E}_{\ell} \mathrm{I}_{\ell} \cos \theta=\sqrt{3} \times 400
\times 69.282 \times \cos 53.13^{\circ}=28800
\mathrm{W}\)
Alternatively, total power, \(\mathrm{P}_{\mathrm{T}}=3 \mathrm{P}=28800
\mathrm{W}\)
Problem-5: Phase Sequence
\[\begin{aligned}
\mathrm{V}_{\mathrm{AN}} & =400 \cos \left(\omega
\mathrm{t}-10^{\circ}\right) \mathrm{V} \\
\mathrm{v}_{\mathrm{BN}} & =400 \cos \left(\omega
\mathrm{t}-250^{\circ}\right) \mathrm{V} \\
\mathrm{v}_{\mathrm{CN}} & =400 \cos \left(\omega
\mathrm{t}-130^{\circ}\right) \mathrm{V}
\end{aligned}\]
Determine the phase sequence of the set of voltages given by
Solution-5
\[\begin{aligned}
\mathrm{V}_{\mathrm{AN}} & =400 \cos \left(\omega
\mathrm{t}-10^{\circ}\right) \mathrm{V} \\
\mathrm{v}_{\mathrm{BN}} & =400 \cos \left(\omega
\mathrm{t}-250^{\circ}\right) \mathrm{V} \\
\mathrm{v}_{\mathrm{CN}} & =400 \cos \left(\omega
\mathrm{t}-130^{\circ}\right) \mathrm{V}
\end{aligned}\]
Given voltage phasors are:
\[\begin{aligned}
\mathrm{V}_{\mathrm{AN}}=282.8 \angle-10^{\circ} \\
\mathrm{V}_{\mathrm{BN}}=282.8 \angle-250^{\circ}\\
\mathrm{V}_{\mathrm{CN}}=282.8 \angle-130^{\circ}
\end{aligned}\]
Note: \(\mathrm{V}_{\mathrm{C}}\) is lagging \(\mathrm{V}_{\mathrm{A}}\) . Hence phase
sequence is \(\mathrm{ACB}\)
Problem-6: Balanced System & Phase Sequence
In a three-phase balanced supply, voltage \(\mathrm{V}_{\mathrm{C}}=110 \angle 65^{\circ}
\mathrm{V}\) .
Taking the phase sequence as \(\mathrm{ABC}\) , find
the phase voltage
the line voltages
Solution-6
\[\begin{aligned}
\mathrm{V_A}& =\mathrm{Vc} \times 1 \angle-120^{\circ}=110
\angle-55^{\circ} \mathrm{V} \\
\mathrm{V_B}& =\mathrm{Vc} \times 1 \angle 120^{\circ}=110
\angle 185^{\circ} \mathrm{V} \\
\mathrm{V_C}& =110 \angle 65^{\circ} \mathrm{V}
\end{aligned}\]
\[\begin{aligned}
\mathrm{V_{AB}}& =\mathrm{V_A}-\mathrm{V_B}=190.5
\angle-25^{\circ} \mathrm{V} \\
\mathrm{V_{BC}}& =\mathrm{V_B}-\mathrm{V_C}=190.5 \angle-145^{\circ}
\mathrm{V} \\
\mathrm{V_{CA}}&=\mathrm{V_C}-\mathrm{VA}=190.5 \angle 95^{\circ}
\mathrm{V}
\end{aligned}\]
Problem-7: Unbalanced Delta System
\[\mathrm{Z}_{\mathrm{AB}}=10 \angle 0^{\circ}
\Omega ;~ \mathrm{Z}_{\mathrm{BC}}=10 \angle 30^{\circ} \Omega;~
\mathrm{Z}_{\mathrm{CA}}=15 \angle-30^{\circ} \Omega\]
\(\checkmark\)
line voltages
phase currents
line currents
the real power consumed by the load.
\(\mathrm{V_{BC}}\) ABC system has a
delta-connected load with impedances 3-wire A
Solution-7
$$\begin{aligned}
\mathrm{V}_{\mathrm{BC}}&=240 \angle 0^{\circ} \mathrm{V}\\
\mathrm{V} \mathrm{CA} & =240 \angle-120^{\circ} \mathrm{V}\\
\mathrm{V}_{\mathrm{AB}}&=240 \angle-240^{\circ} \mathrm{V}
\end{aligned}$$
$$\begin{aligned}
\mathrm{Z}_{\mathrm{AB}}&=10 \angle 0^{\circ} \Omega \\
\mathrm{Z}_{\mathrm{BC}}&=10 \angle 30^{\circ} \Omega\\
\mathrm{Z}_{\mathrm{CA}}&=15 \angle-30^{\circ} \Omega
\end{aligned}$$
\[\begin{aligned}
\mathrm{I_{BC}}&=\frac{\mathrm{V}_{\mathrm{BC}}}{\mathrm{Z}_{\mathrm{BC}}}=\frac{240
\angle 0^{\circ}}{10 \angle 30^{\circ}}=24 \angle-30^{\circ}
\mathrm{A}=(20.7846-\mathrm{j} 12) \mathrm{A}\\
\mathrm{I_{CA}}& =
\frac{\mathrm{V}_{\mathrm{CA}}}{\mathrm{Z}_{\mathrm{CA}}}=\frac{240
\angle-120^{\circ}}{15 \angle-30^{\circ}}=16 \angle-90^{\circ}
\mathrm{A}=(0-\mathrm{j} 16) \mathrm{A}\\
\mathrm{I_{AB}}&==\frac{\mathrm{V}_{\mathrm{AB}}}{\mathrm{Z}_{\mathrm{AB}}}=\frac{240
\angle-240^{\circ}}{10 \angle 0^{\circ}}=24 \angle-240^{\circ}
\mathrm{A}=(-12+\mathrm{j} 20.7846) \mathrm{A}
\end{aligned}\]
Phase currents are:
\[\begin{aligned}
\mathrm{I_A}&
=\mathrm{I}_{\mathrm{AB}}-\mathrm{I}_{\mathrm{CA}}=(-12+\mathrm{j}
20.7846)-(-\mathrm{j} 16)\\
& =(-12+\mathrm{j} 36.7846)=38.6925 \angle 108.07^{\circ}
\mathrm{A}\\
\mathrm{I_B}&=\mathrm{I}_{\mathrm{BC}}-\mathrm{I}_{\mathrm{AB}}=(32.7846-\mathrm{j}
32.7846)=46.3644 \angle-45^{\circ} \mathrm{A}\\
\mathrm{I_C}&=\mathrm{I_{CA}}-\mathrm{I_{BC}}=(-20.7846-\mathrm{j}
4)=21.166 \angle-169.11^{\circ} \mathrm{A}
\end{aligned}\]
. Thus and at junction Line currents are obtained by applying
\[\begin{aligned}
\mathrm{R}_{\mathrm{AB}}&=10 ~\Omega \\
\mathrm{R}_{\mathrm{BC}}&=10 \cos 30^{\circ}=8.6603 ~\Omega \\
\mathrm{R}_{\mathrm{CA}}&=15 \cos 30^{\circ}=12.9904 ~\Omega
\end{aligned}\]
\[\begin{aligned}
&= \left(24^{2} \times 10\right)+\left(24^{2} \times
8.6603\right)+\left(16^{2} \times 12.9904\right)\\
&=14073.8 \mathrm{W}=14.0738 \mathrm{kW}
\end{aligned}\]
Real power consumed by the load :
Problem-8: Unbalanced Star System
A \(3-\phi\) , 4 -wire, \(208~\mathrm{V}\) system has a star
connected load with impedances
\[\mathrm{Z}_{\mathrm{A}}=6 \angle
0^{\circ} \Omega ;~ \mathrm{Z}_{\mathrm{B}}=6 \angle 30^{\circ} \Omega;~
\mathrm{Z}_{\mathrm{C}}=5 \angle 45^{\circ} \Omega\]
Taking the voltage \(\mathrm{V_{AN}}\) as reference, determine
line and neutral currents. Draw the phasor diagram. Also calculate the
real power consumed by the load.
Solution-8
\(\mathrm{Z}_{\mathrm{A}}=6 \angle
\mathrm{0}^{\circ} \Omega ;~ \mathrm{Z}_{\mathrm{B}}=6 \angle 30^{\circ}
\Omega\) ; \(\mathrm{Z}_{\mathrm{C}}=5
\angle 45^{\circ} \Omega\)
Taking the phase sequence as ABC,
$$\begin{aligned}
\mathrm{V}_{\mathrm{AN}}&=\frac{208}{\sqrt{3}} \angle 0^{\circ} V\\
&=120.08 \angle 0^{\circ} \mathrm{V}\\
\mathrm{V}_{\mathrm{BN}}&=120.08 \angle-120^{\circ} \mathrm{V}\\
\mathrm{V}_{\mathrm{CN}}&=120.08 \angle-240^{\circ} \mathrm{V}
\end{aligned}$$
$$\begin{aligned}
\mathrm{I}_{\mathrm{A}}&=\frac{\mathrm{V}_{\mathrm{AN}}}{\mathrm{Z}_{\mathrm{A}}}=\frac{120.08}{6
\angle 0^{\circ}}\\
&=20.01 \angle 0^{\circ} \mathrm{A}\\
\mathrm{I}_{\mathrm{B}}&=\frac{\mathrm{V}_{\mathrm{BN}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{120.08
\angle-120^{\circ}}{6 \angle 30^{\circ}}\\
&=20.01 \angle-150^{\circ} \mathrm{A}\\
\mathrm{I}_{\mathrm{C}}&=\frac{V_{C N}}{Z_{C}}=\frac{120.08
\angle-240^{\circ}}{5 \angle 45^{\circ}}\\
&=24.01 \angle 75^{\circ} \mathrm{A}
\end{aligned}$$
$$\begin{aligned}
I_{N}&=-\left(I_{A}+I_{B}+I_{C}\right)\\
&=(-8.89-j 13.19) A\\
&=15.91 \angle-124^{\circ} A
\end{aligned}$$
Resistances in different phases are:
\(\mathrm{R}_{\mathrm{A}}=6 \Omega ~~
\mathrm{R}_{\mathrm{B}}=6 \cos 30^{\circ}=5.1962 \Omega ~~
\mathrm{R}_{\mathrm{C}}=3.53 \Omega\)
\[\begin{aligned}
=& \left(20.01^{2} \times 6\right)+\left(20.01^{2} \times
5.19\right)+\left(24.01^{2} \times 3.53\right) \\
=&6524.58~ \mathrm{W}= 6.5246 ~\mathrm{kW}
\end{aligned}\]
Real power consumed by the load
Problem-9: Unbalanced Systems
A \(3-\phi\) , 3-wire, 208 V, ACB
system has star connected load with impedances \(\mathrm{Z}_{\mathrm{A}}=6 \angle 0^{\circ} \Omega
; ~ \mathrm{Z}_{\mathrm{B}}=6 \angle 30^{\circ} \Omega\) and
\(\mathrm{Zc}=5 \angle 45^{\circ} \Omega
.\) Taking \(\mathrm{V}_{\mathrm{BC}}\) as reference,
Determine
line currents
the voltage across load impedances
Solution-9
$$\begin{aligned}
\mathrm{V}_{\mathrm{BC}}&=208 \angle \mathrm{0}^{\circ} \mathrm{V}
\\ \quad \mathrm{V}_{\mathrm{CA}}&=208 \angle 120^{\circ} \mathrm{V}
\\ \quad \mathrm{V}_{\mathrm{AB}}&=208 \angle-120^{\circ} \mathrm{V}
\end{aligned}$$
$$\begin{aligned}
Z_{A}&=6 \Omega \\
Z_{B}&=(5.1962+j 3) \Omega \\
Z_{C}&=(3.5355+j 3.5355) \Omega
\end{aligned}$$
\(\left[\begin{array}{cc}Z_{A}+Z_{B} &
-Z_{B} \\ -Z_{B} &
Z_{B}+Z_{c}\end{array}\right]\left[\begin{array}{l}I_{1} \\
I_{2}\end{array}\right]=\left[\begin{array}{c}V_{A B} \\ V_{B
C}\end{array}\right]\) \(\left[\begin{array}{cc}11.1962+j 3 & -5.1962-j
3 \\ -5.1962-j 3 & 8.7317+j
6.5355\end{array}\right]\left[\begin{array}{l}I_{1} \\
I_{2}\end{array}\right]=\left[\begin{array}{c}208 \angle-120^{\circ} \\
208 \angle 0^{\circ}\end{array}\right]\)
Determinant of the coefficient matrix \(=
90.9323 \angle 48.580\)
\(\left[\begin{array}{l}\mathrm{I}_{1} \\
\mathrm{I}_{2}\end{array}\right]=\frac{1}{90.9323 \angle
48.58^{\circ}}\left[\begin{array}{cc}8.7317+\mathrm{j} 6.5355 &
5.1962+\mathrm{j} 3 \\ 5.1962+\mathrm{j} 3 & 11.1962+\mathrm{j}
3\end{array}\right]\left[\begin{array}{c}208 \angle-120^{\circ} \\ 208
\angle 0^{\circ}\end{array}\right]\) \(=\left[\begin{array}{c}-3.6097-\mathrm{j} 22.9799
\\ 11.7974-\mathrm{j}
23.7446\end{array}\right]=\left[\begin{array}{l}23.2617
\angle-98.93^{\circ} \\ 26.5139
\angle-63.58^{\circ}\end{array}\right]\)
\(\left[\begin{array}{l}\mathrm{I}_{1} \\
\mathrm{I}_{2}\end{array}\right]=\left[\begin{array}{l}-3.6097-\mathrm{j}
22.9799 \\ 11.7974-\mathrm{j} 23.7446\end{array}\right]\)
From the mesh currents, line currents can be calculated:
\(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{1}=(-3.6109-\mathrm{j}
22.9797) \mathrm{A}=23.2617 \angle-98.93^{\circ} \mathrm{A}\)
\(\mathrm{I}_{B}=\mathrm{I}_{2}-\mathrm{I}_{1}=(15.4071-\mathrm{j}
0.7647) \mathrm{A}=15.426 \angle-2.84^{\circ} \mathrm{A}\) \(\mathrm{I_C}=-\mathrm{I}_{2}=(-11.7974+\mathrm{j}
23.7446) \mathrm{A}=26.5139 \angle 116.42^{\circ}
\mathrm{A}\)
Voltages across the loads can be calculated as:
\(\mathrm{V}_{\mathrm{AO}}=\mathrm{Z}_{\mathrm{A}}
\mathrm{I}_{\mathrm{A}}=(-21.6654-\mathrm{j} 137.8782)
\mathrm{A}=139.5702 \angle-98.93^{\circ} \mathrm{V}\) \(\mathrm{V}_{\mathrm{BO}}=\mathrm{Z}_{\mathrm{B}}
\mathrm{I}_{\mathrm{B}}=(82.3503+\mathrm{j} 42.2497) \mathrm{A}=92.556
\angle 27.16^{\circ} \mathrm{V}\) \(\mathrm{V}_{\mathrm{CO}}=\mathrm{Z}_{\mathrm{C}}
\mathrm{I_C}=(-125.6599+\mathrm{j} 42.2404) \mathrm{A}=132.5695 \angle
161.42^{\circ} \mathrm{V}\)
Problem-10: Power measurement by two-watt meter
The power input to a \(2000 ~V, 50
\mathrm{Hz},\) 3-phase motor is measured by two wattmeters which
indicate \(100 \mathrm{kW}\) and \(300 \mathrm{kW}\) respectively.
Calculate
the input power
the power factor
the line current.
Solution-10
Given : \(\quad \mathrm{W}_{1}=100
~\mathrm{kW}\) and \(\mathrm{W}_{2}=300~ \mathrm{kW}\)
Input power: \(\mathrm{P}=\mathrm{W}_{1}+\mathrm{W}_{2}=400~
\mathrm{kW}\)
Power factor angle
\[\begin{aligned}
\quad \tan \theta& =\sqrt{3}
\frac{\mathrm{W}_{2}-\mathrm{W}_{1}}{\mathrm{W}_{1}+\mathrm{W}_{2}}\\
&=\sqrt{3} \times \frac{200}{400}=0.8660\\
\theta& =40.89^{\circ} \\
\cos \theta&=0.7559
\end{aligned}\]
Power factor \(=0.7559\)
lagging
Three-phase power, \(P=\sqrt{3} V_{l}
I_{l} \cos \theta\)
Line current, \(\mathrm{I}_{l}=\frac{400 \times 10^{3}}{\sqrt{3}
\times 2000 \times 0.7559}=152.76 \mathrm{A}\)
Problem-11: Power measurement by two-watt meter
Two watt meters are connected to measure the power in a 3-phase
3-wire balanced load. Determine the total power and power factor, if the
two watt-meters read
\(1000 \mathrm{W}\) each, both
positive
\(1000 \mathrm{W}\) each of
opposite sign.
Solution-11
Case-1: Given data: \(\quad
\mathrm{W}_{1}=1000~ \mathrm{W} ; \quad \mathrm{W}_{2}=1000
\mathrm{W}\)
Total power \(\mathrm{P}=\mathrm{W}_{1}+\mathrm{W}_{2}=2000
\mathrm{W}\)
\(\tan \theta=\sqrt{3}
\frac{\mathrm{W}_{2}-\mathrm{W}_{1}}{\mathrm{W}_{1}+\mathrm{W}_{2}}=0\)
Power factor angle \(\theta=0\)
Power factor \(=\cos
0^{\circ}=1.0\)
Case-2: Given data: \(\quad
\mathrm{W}_{1}=-1000 \mathrm{W} ; \quad \mathrm{W}_{2}=1000 \mathrm{W} ;
\quad\)
Total power \(\mathrm{P}=\mathrm{W}_{1}+\mathrm{W}_{2}=0\)
\(\tan \theta=\sqrt{3}
\frac{W_{2}-W_{1}}{W_{1}+W_{2}}=\infty ;\)
Power factor angle \(\theta=90^{\circ}\)
Power factor = 0 lagging.
Problem-12: Power measurement by two-watt meter
Across \(400 \mathrm{V}, 3\) -phase
supply mains, a star-connected balanced load of \((16+\mathrm{j} 12) \Omega\) impedance is
connected.
Taking \(V_a\) as reference,
determine the line currents and the power absorbed by the load
If two wattmeters are used to measure the power, what will be the
readings of the wattmeters?
Solution-12
\(\quad V_{A}=\frac{400}{\sqrt{3}}
\angle 0^{\circ}=230.94 \angle 0^{\circ}~V ~~ Z=(16+j 12)=20 \angle
36.87^{\circ} \Omega\)
\[\begin{aligned}
\mathrm{I}_{\mathrm{A}}&=\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{Z}}=11.54
\angle-36.87^{\circ} \mathrm{A} \\
\mathrm{I}_{\mathrm{B}}&=11.54 \angle-156.87^{\circ}
\mathrm{A}\\ \mathrm{I}_{\mathrm{C}}&=11.54 \angle 83.13^{\circ}
\mathrm{A}
\end{aligned}\]
Line currents are:
\[P=\sqrt{3} V_{l} I_{l}
\cos \theta=\sqrt{3} \times 400 \times 11.54 \cos 36.87^{\circ}=6400
\mathrm{W}\]
Total power
\[\begin{aligned}
\quad \mathrm{W}_{1}&=\sqrt{3} \mathrm{V}_{\text {ph }}
\mathrm{I}_{\text {ph }} \cos \left(\theta+30^{\circ}\right)\\
&=\sqrt{3} \times 230.94 \times 11.54 \cos
66.87^{\circ}=1814.35 \mathrm{W}\\
\mathrm{W}_{2}&=\sqrt{3} \mathrm{V}_{\mathrm{ph}}
\mathrm{I}_{\mathrm{ph}} \cos \left(\theta-30^{\circ}\right)\\
& =\sqrt{3} \times 230.94 \times 11.54 \cos
6.87^{\circ}=4585.64 \mathrm{W}
\end{aligned}\]
Watt meter readings: