EENG 457 Power System Analysis I
HOMEWORK 1
1. A single-phase voltage source is applied to an impedance Z  4.0  30  The source
current is measured as i(t )  5cos(t ) A . Determine
(a) the instantaneous power.
(b) the real and reactive power delivered by the source.
(c) the power factor at the terminals of the source.
2. A voltage source v(t )  100cos(100 t  60 ) V is connected to an impedance Z  5.030 .
(a) Determine the expressions for the current and the instantaneous power
delivered by the source as functions of time.
(b) Find the frequency and the average value of the instantaneous power.
3. The real power delivered by a source to two impedances Z1  4  j 5  and Z 2  8 
connected in parallel is 4000 W. Determine
(a) the real power absorbed by each impedance.
(b) the total source current Is.
4. A load consumes 60 kW at a power factor of 0.7 lagging from a 240-V 50-Hz source. The
power factor at the source terminals is to be raised to 0.95 lagging by placing a capacitor in
parallel with the load. Find the capacitance required in μF.
5. Two single-phase sources are connected by a line of
impedance Z L  0.7  j 2.4  as shown in the figure. The
source voltages are
E1  50016.26 V and E2  5850 V .
Find the complex power for each source and determine
whether they are supplying or receiving real and
reactive power. Also find the real and reactive power
loss on the line.
ZL=0.7+j2.4 Ω
+
+
I
E1
E2
6. A 380-V (line-to-line) three-phase source delivers power to a balanced delta-connected load
of Z  10  j5  per phase through a three-phase line as shown below. Find
(a) Current in phase ab of the load.
(b) Total complex power supplied by the source.
(c) Magnitude of the line-to-line voltage at the load terminals.
1+j2Ω
a
a'
V
L
1+j2Ω
b
1+j2Ω
c
Z
=380 V rms
b'
Z
Z
c'
SOLUTION
1.
(a) The source voltage in phasor form is
V  Z .I  50  4  30  20  30 V
In the time domain
v(t )  20cos(t  30 ) V
Instantaneous power
 p (t )  v(t ).i (t )  100 cos(t ).cos(t  30 ) W
 50 cos(30 ). 1  cos(2t )  +50sin(30 ).sin(2t )
(b)
 P  50.cos(30 )  43.3 W,
(c) Power factor = cos(30) = 0.866
2.
Q  50.sin(30 ) Var
Voltage phasor (rms) V  (100 / 2)60 V
V (100 / 2)60

 14.1430 A
Z
530
Instantaneous power
 I
 i(t )  14.14 2 cos(t  30 )
 p(t )  v(t ).i(t )  2000 cos(t  60 ).cos(t  30 ) W
 1000cos(30 ). 1  cos(2t  120 )   1000sin(30 ).sin(2t  120 )
Frequency of instantaneous power = 2
Average value = 1000cos(30 )  866 W
3-) (a) Real power absorbed by each impedance
V2
4
2
2
P1  Re 
  V 2 2  0.0976 V ,
 Z1 
4 5


P1  P2  4000 W  0.2226 V
2

V2
2 1
P2  Re 
 V
 0.125 V
 Z2 
8


V  17969 V 2  P1  1753.8 W, P2  2246.2 W
2
(b) Total source current,
I s  I1  I 2 
2
 1
V V
1

 134.05 
   29.83  j16.35 A
Z1 Z 2
 4  j5 8 
4-) Reactive power absorbed by the load
Q  P tan   60  tan(cos1 0.7)  60 1.0202  61.2122 kVar
When the power factor at source terminals is raised to 0.95, the source will supply
Q  P tan   60  tan(cos1 0.95)  19.721 kVar
Reactive power to be supplied by the capacitor
Qc  Q  Q  61.2122  19.721 kVar  41.4912 kVar  C V
2
41.4912 103
 C
 2300 F
2  50  2402
5-)
I
E1  E2 50016.26  5850

 42  j 56 A  7053.13 A
ZL
0.7  j 2.4
Complex power supplied by E1 :
S1  E1 I *  50016.26  70  53.13  28  j 21 kVA
Complex power absorbed by E2 : S2  E2 I *  5850  70  53.13  24.57  j 32.76 kVA
 E1 supplies 28 kW real power , and absorbs 21 kVar reactive power
 E2 absorbs 24.57 kW real power , and supplies 32.76 kVar reactive power
Complex power absorbed by the line impedance,
2
Sline  Z L I  (0.7  j 2.4)  702  3.43  j11.76 kVar
 Line absorbs 3.43 kW and 11.76 kVar
6-) Convert the  to Y: ZY  Z  / 3  (10  j5) / 3 .
Single-phase equivalent circuit:
1+j2Ω
a
a'
Ia
ZY
n
Van
380 / 30

1
13  j 11
1  j 2  3 (10  j5)
3
3
 29.5  j 24.97 A  38.65  40.23 A
Ia 
1
I a .  30  22.31  70.23 A
3
(b) Total complex power supplied by the source
I ab 
S  Van I a*  Vbn I b*  Vcn I c*  19.421  j16.43 kVA
(c)
5
Van  I a (10
3  j 3 )  139.96  j 34.044 V  144.04  13.67

Vab  3 144.04  249.48 V