L5 Examples
Electric Power System 2014
1(2)
Voltages, currents and power in AC circuits
IL
I12
Zth
Vth
Z
ZL
V
V1
V2
Figure 1 The Thévenin equivalent can Figure 2 The voltage sources can represent
represent any point in the electricity generators in a power system. Z represents
network. ZL represents an impedance load impedance of power lines and transformers
connected to this point.
between the two sources.
1. One voltage source
Use the circuit in Figure 1 and for each case below

Determine numerical values of complex current IL, complex voltage V and magnitude of
V in % of Vth.

Draw Vth (phase angle = 0), V and IL in the complex plane. Adjust angles for clarity.

Determine the complex power S=P+jQ leaving Zth to the right.

Describe how angle and magnitude of V relates to those of Vth.
a. Wall outlet, Vth=230 V, Zth=1Heater ZL=51.
b. Transformer station, Vth=145/3 kV, transformers and cables Zth=j50eating in 1/3
city (we analyze only one of three phases) ZL=1000.
c. Transformer station, Vth=145/3 kV, transformers and cables Zth=j50Shunt
inductance in one phase ZL=j1400.
d. Transformer station, Vth=145/3 kV, transformers and cables Zth=j50Shunt
capacitor in one phase ZL=-j1400.
2. Two voltage sources
Use the circuit in Figure 2 and for each case below

Determine numerical values of the complex current I12.

Draw V1, V2 and I12 in the complex plane. Adjust angles for clarity.

Determine the complex power S=P+jQ leaving Z to the right.

Describe how P and Q relate to magnitude difference between V1 and V2 and angle
difference between V1 and V2.
a. V1= V2=145/3 kV0, High voltage lines and transformers Z=j50.
b. V1=145/3 kV0, V2=145/3 kV-30, High voltage lines and transformers Z=j50.
c. V1=145/3 kV0, V2=140/3 kV0, High voltage lines and transformers Z=j50.
d. V1=145/3 kV0, V2=150/3 kV0, High voltage lines and transformers Z=j50.
OS
IEA
L5 Examples
Electric Power System 2014
1. Solution
Vth=230;
Vth=145e3/sqrt(3);
Vth=145e3/sqrt(3);
Vth=145e3/sqrt(3);
IL=Vth/(ZL+Zth)
145e3/sqrt(3)=83.7e3
Zth=1;
Zth=j*50;
Zth=j*50;
Zth=j*50;
V=ZL*IL
IL (A)
1a
4.42
V (V)
98
%1a
%1b
%1c
%1d
S (VA)
Zth ZL
998
83507 - 4175i, 100 6.99e6
0.0008 – 57.7i 80829
1d 0.0 +62.0i
97
86816
1a.
V and Vth
RR
V<Vth, in phase
XL R
V<Vth, lags Vth
0.00 +4.67e6i XL XL
V<Vth, in phase
104 0.00 - 5.38e6i XL XC
V>Vth, in phase
1b.
Vth
Vth
V
I
I
1c.
1d.
I
Vth
V
Vth
V
V
I
2. Solution
V1=145e3/sqrt(3);
V1=145e3/sqrt(3);
V1=145e3/sqrt(3);
V1=145e3/sqrt(3);
I12=(V1-V2)/Z,
V2=145e3/sqrt(3);
V2=145e3/sqrt(3)*exp(-j*pi*3/180);
V2=140e3/sqrt(3);
V2=150e3/sqrt(3);
S=V2*conj(I12)
I12 (A)
2a
V%
226
1b 83.5 - 4.17i
1c
ZL=51;
ZL=1000;
ZL=1400;
ZL=-j*1400;
S=V*conj(IL)
2(2)
S (VA)
%2a
%2b
%2c
%2d
P
Q
0
=0 Arg(V1)=Arg(V2) 0 |V1|=|V2|
2b 87.6269 - 2.2946i 7.3358e+06 - 1.9209e+05i
>0 Arg(V1)>Arg(V2) 0 |V1||V2|
2c
0
Z=j*50;
Z=j*50;
Z=j*50;
Z=j*50;
0.0000 -57.7350i
0.0000e+00 + 4.6667e+06i =0 Arg(V1)=Arg(V2) >0 |V1|>|V2|
2d 0.0000 +57.7350i 0.0000e+00 - 5.0000e+06i
=0 Arg(V1)=Arg(V2) <0 |V1|<|V2|
2a.
2b.
V1=V2
I12=0
V1
I12
V2
2c.
I12
OS
I12
V1
2d.
V1
V2
V2
IEA