MAXIMUM POWER TRANSFER IN AC CIRCUITS  OHM'S LAW 
The alternating current networks that we have studied so far are widely used to model AC mains electric power networks in homes. However, for industrial use and also for electric power generation, a network of AC generators is more effective. This is realized by polyphase networks consisting of a number of identical sinusoidal generators with a phase angle difference. The most common polyphase networks are two or threephase networks. We will limit our discussion here to threephase networks.
Note that TINA provides special tools for drawing threephase networks in the Special component toolbar, under the Stars and Y buttons.
A threephase network can be seen as a special connection of three single phase or simple AC circuits. Threephase networks consist of three simple networks, each having the same amplitude and frequency, and a 120Â° phase difference between adjacent networks. The time diagram of the voltages in a 120V_{eff }system is shown in the diagram below.
We can also represent these voltages with phasors using TINAâ€™s Phasor Diagram.
Compared to singlephase systems, three phase networks are superior because both the power stations and the transmission lines require thinner conductors for transmitting the same power. Due to the fact that one of the three voltages is always nonzero, threephase equipment has better characteristics, and threephase motors are selfstarting without any additional circuitry. It is also much easier to convert threephase voltages into DC (rectification), due to the reduced fluctuation in the rectified voltage.
The frequency of threephase electric power networks is 60 Hz in United States and 50 Hz in Europe. The single phase home network is simply one of the voltages from a threephase network.
In practice, the three phases are connected in one of two ways.
1) The Wye or Yconnection, where the negative terminals of each generator or load are connected to form the neutral terminal. This results in a threewire system, or if a neutral wire is provided, a fourwire system.
The V_{p1},V_{p2},V_{p3 }voltages of the generators are called phase voltages, while the voltages V_{L1},V_{L2},V_{L3 }between any two connecting lines (but excluding the neutral wire) are called line voltages. Similarly, the I_{p1},I_{p2},I_{p3 }currents of the generators are called phase currents while the currents I_{L1},I_{L2},I_{L3 }in the connecting lines (excluding the neutral wire) are called line currents.
In Yconnection, the phase and line currents are obviously the same, but the line
voltages are greater than the phase voltages. In the balanced case:
Let â€˜s demonstrate this by a phasor diagram:
Letâ€™s calculate V_{L }for the phasor diagram above using the cosine rule of trigonometry:
Now letâ€™s calculate the same quantity using complex peak values:
V_{p1} = 169.7 e^{j 0 } ^{ Â° } = 169.7
V_{p2} = 169.7 e^{j 120 } ^{ Â° } = 84.85+j146.96
V_{L} = V_{p2}  V_{p1 }= 254.55+j146.96^{ = }293.9 e^{ j150 } ^{ Â° } ^{ }
The same result with the TINA Interpreter:
Vp1:=169.7
Vp2:=169.7 *exp(j*degtorad(120))
Vp2=[84.85+146.9645*j]
VL:=Vp2Vp1
VL=[254.55+146.9645*j]
radtodeg(arc(VL))=[150]
abs(VL)=[293.929]
Similarly the complex peak values of the line voltages
V_{L21} = 293.9 e^{j 150} ^{ Â° } V,
V_{L23} = 293.9 e^{j 270} ^{ Â° } V,
V_{L13} = 293.9 e^{j 30} ^{ Â° } V.
The complex effective values:
V_{L21eff} = 207.85 e^{j 150} ^{ Â° } V,
V_{L23eff} = 207.85 e^{j 270} ^{ Â° } V,
V_{L13eff} = 207.85 e^{j 30} ^{ Â° } V.
Finally letâ€™s check the same results using TINA for a circuit with
120 V_{eff} ; V_{P1} = V_{P2} = V_{P3} =169.7 V and Z_{1}= Z_{2} =Z_{3 }= 1 ohms
2) The delta or Dconnection of three phases is achieved by connecting the three loads in series forming a closed loop. This is only used for threewire systems.
As opposed to a Yconnection, in D connection
the phase and line voltages are obviously the same, but the line currents are
greater than the phase currents. In the balanced case:
Letâ€™s demonstrate this with TINA for a network with 120 V_{eff }Z=10 ohms.
Result:
Since either the generator or the load can be connected in D or in Y, there are four possible interconnections: YY, Y D, DY and D D. If the load impedances of the different phases are equal, the threephase network is balanced.
Some further important definitions and facts:
The phase difference between the phase voltage or current and the nearest line voltage and current (if they are not the same) is 30 Â°.
If the load is balanced (i.e. all the loads have the same impedance), each phaseâ€™s voltages and currents are equal. Furthermore, in the Yconnection, there is no neutral current even if there is a neutral wire.
If the load is unbalanced, the phase voltages and currents are different Also, in the Yâ€“Yconnection with no neutral wire, the common nodes (star points) are not at the same potential. In this case we can solve for node potential V_{0} (the common node of the loads) using a node equation. Calculating V_{0} allows you to solve for the phase voltages of the load, current in the neutral wire, etc. Yconnected generators always incorporate a neutral wire.
The power in a balanced three phase system is P_{T} = 3 V_{p}I_{p }cos J = V_{L}I_{L} cos J
where J is the phase angle between the voltage and the current of the load.
The total apparent power in a balanced three phase system: S_{T} = V_{L}I_{L }
The total reactive power in a balanced three phase system: Q_{T} = V_{L }I_{L }sin J
Example 1
The rms value of the phase voltages of a threephase balanced Yconnected generator is 220 V; its frequency is 50 Hz.
a/ Find the time function of the phase currents of the load!
b/ Calculate all the average and reactive powers of the load!
Both the generator and the load are balanced, so we need to calculate only one phase and can get the other voltages or currents by changing the phase angles. In the schematic above we did not draw the neutral wire, but instead assigned â€˜earthâ€™ at both sides. This can serve as a neutral wire; however, because the circuit is balanced the neutral wire is not needed.
The load is connected in Y, so the phase currents are equal to the line currents: the peak values:
I_{P1} = V_{P}/(R+j w L) = 311/(100+j314*0.3) = 311/(100+j94.2) = 1.65j1.55 = 2.26 e^{j43.3} ^{Â° }A
V_{P1} = 311 V
I_{P2} = I_{P1} e^{ j 120} ^{Â° } =2.26 e^{j76.7} ^{Â° } A
I_{P3} = I_{P2} e^{ j 120} ^{Â° } =2.26 e^{j163.3} ^{Â° } A
i_{P1} = 2.26 cos ( wÃ—t â€“ 44.3 Â° ) A
i_{P2} = 2.26 cos ( wÃ— t + 76.7 Â°) A
i_{P3} = 2.26 cos ( wÃ— t â€“ 163.3 Â° ) AThe powers are also equal: P_{1} = P_{2} = P_{3} = = 2.26^{2}*100/2 = 256.1 W
{Solution by TINA's Interpreter} {Since
both the generator and the load are balanced

This is the same as calculated results by hand and TINAâ€™s Interpreter.
Example 2
A threephase balanced Yconnected generator is loaded by a deltaconnected threepole load with equal impedances. f=50 Hz.
Find the time functions of a/ the phase voltages of the load,
b/ the phase currents of the load,
c/ the line currents!
The phase voltage of the load equals the line voltage of the generator:
V_{L} =
The phase currents of the load: I_{1} = V_{L}/R_{1}+V_{L}j w C = 1.228 + j1.337 = 1.815 e^{j 47.46} ^{Â° } A
I_{2} = I_{1} * e^{j120} ^{Â° } = 1.815 e^{j72.54} ^{Â° } A = 0.543 â€“ j1.73 A
I_{3} = I_{1} * e^{j120} ^{Â° } = 1.815 e^{j167.46} ^{Â° } = 1.772 + j0.394
Seeing the directions: I_{a} = I_{1} â€“ I_{3} = 3+j0.933 A = 3.14 e^{j17.26} ^{Â° } A.
i_{a}(t) = 3.14 cos ( wÃ— t + 17.3 Â° ) AAccording to the results calculated by hand and TINAâ€™s Interpreter.
{Solution
by TINA's Interpreter. Since the symmetry we

Finally an example with an unbalanced load:
Example 3
The rms value of the phase voltages of a threephase balanced
Yconnected generator is 220 V; its frequency is 50 Hz.
a/ Find the phasor of the voltage V_{0} !
b/ Find the amplitudes and initial phase angles of the phase currents !
Now the load is an asymmetrical one and we have no neutral wire, so we can expect a potential difference between the neutral points. Use an equation for the node potential V_{0}:
hence V_{0} = 192.71+ j39.54 V = 196.7 e^{j11.6} ^{Â° } V
and: I_{1} = (V_{1}V_{0})*j w C = 0.125 e^{j71.5} ^{Â° } A; I_{2} = (V_{2}V_{0})*j w C = 0.465 e^{j48.43} ^{Â° }
and I_{3} =(V_{3}V_{0})/R = 0.417 e^{j 146.6} ^{Â° } A
v_{0}(t) = 196.7 cos ( wÃ— t + 11.6 Â° ) V;
i_{1}(t) = 0.125 cos ( wÃ— t + 71.5 Â° ) A;
i_{2}(t) = 0.465 cos ( wÃ— t  48.4 Â° ) A;
i_{3}(t) = 0.417 cos ( wÃ— t + 146.6 Â° ) A;{Solution by TINA's Interpreter}
{Because of nonsymmetry we have to
calculate all phases individually}
om:=314;
V1:=311;
V2:=311*exp(j*4*pi/3);
V3:=311*exp(j*2*pi/3);
Sys V0
(V0V1)*j*om*C+(V0V2)*j*om*C+(V0V3)/R=0
end;
V0=[192.7123+39.5329*j]
abs(V0)=[196.7254]
I1:=(V1V0)*j*om*C;
abs(I1)=[124.6519m]
radtodeg(arc(I1))=[71.5199]
I2:=(V2V0)*j*om*C;
abs(I2)=[465.2069m]
radtodeg(arc(I2))=[48.4267]
I3:=(V3V0)/R;
abs(I3)=[417.2054m]
radtodeg(arc(I3))=[146.5774]
And, finally, the results calculated by TINA agree with the results calculated by the other techniques.
MAXIMUM POWER TRANSFER IN AC CIRCUITS  OHM'S LAW 