Impedance and Power Analysis

QUESTION ONE
a)
i. impedance of branch AB
Impedance, Z1=V/I
Where, I is the current through branch AB and
V is the voltage across the branch AB.
Z1=230<0/11.47<-4.49o
=20.05<4.49o
ii. Impedance of branch CD.
Impedance, Z2=230<0/ (3.88<32.50)
=59.28<-32.50
iii. Resistance of R1 and reactance of L1.
Impedance of branch AB, Z1=R1+Jx1,
from part (i), Z1=20.05<4.490=19.988+j1.5696
thus R1=19.988 ohms and X1=1.5696
iv. Resistance of R2 and reactance of C2.
Impedance of branch CD, Z2=R2+jX2,
From part (ii) Z2=59.28<-32.5o=50-j31.85
The real part of the equation above represent resistance and imaginary
part represent reactance
R2=50 ohms
X2=1/(2*phi*f*C2)
=31.85 ohms
v. X1=2*∏*f*L1
1.5696 = 2*3.14*50*L1
L1=5mH
b)
i. Relationship between VLINE and VPHASE
For star connection
VLINE= 3* VPHASE
ii. Phase voltages of the star connection
VPHASE=VLINE/3 and VLINE is 400V, thus
=400/3
=230.94 V
Taking as VAN reference phasor and considering voltages are out of
phase
VAN= 230.94<0
VBN=230.94<-1200

VCN=230.94<-2400
iii. phasor diagram for voltages
QUESTION TWO
a) 3 phase load connected in  and connected to 3- 400V 50Hz supply

ZAB=20+j5,
ZBC=15+j7
ZCA=20-j8
i. Phase voltages of the circuit
For delta connected load
VL=VP=400<0
Taking VAB as reference phase voltages are as follows
VAB=400<0
VBC=400<-1200
VCA=400<-2400
ii. Phase currents
The phase current, IAB=VAB/ZAB
=400<0/(20+j5)
=19.4<-14.040 A
The phase current, IBC=VBC/ZBC
=400<-120/(15+j7)
=24.16<-145.020 A
The phase current, ICA=VCA/ZCA
=400<+120/(20-j8)
=18.57<141.8o A
iii. Apparent, active and reactive power associated with each phase
Apparent power, SAB=VAB*IAB,
=400*19.4
=7.76KW
SBC=400*24.16
=9.664KW
SCA=400*18.57
=7.428KW
Active power, PAB=VAB*IAB*cos where cos is the phase angle
between phase voltage and phase current
=400*19.4*Cos (14.04)
=7.528kW
PBC=400*24.16*Cos (25.02)
=7.031Kw
PCA=400*18.57*Cos (21.8)
=6.896Kw
Reactive power, Q/phase=VP*IP*sin
QAB=400*19.4*Sin (14.04)
=1.882Kw
QBC=400*24.16*Sin (25.02)

=4.087Kw
QCA=400*18.57*Sin (21.8)
=2.758Kw
iv. Total active power, P= 7.528+7.031+6.896
=21.455kW
v. Total reactive power, Q=1.882+4.087+2.758
=24.852kW
vi. Total apparent power, S= 7.76+9.664+7.428
=24.852kw
b) One wattmeter method
This method applies only to balanced load, and the method uses one
wattmeter as shown in the figure below
Since its balanced load, I1=I2=I2=I and V1=V2=V3=V
When switch 1 is closed the wattmeter reads
P1= V13*I1*Cos(30-α)
=√3*V*I*Cos(30-α)
When switch 2 is closed the wattmeter reads
P2=V12*I2*Cos(30+α)
=√3*V*I*Cos(30+α)
Total power is thus, P=P1+P2
=3*V*I*cos α