Electrical circuit Analysis

ECE 2224 ELECTRIC CIRCUIT ANALYSIS II
ASSIGNMENT II

1. A star connected 3-phase load has an impedance of ( 7.07 + j 7.07 )  in each
phase. It is fed from a 173.2 V 3-phase balanced supply. Find
a) Line current.
b) Total active power
c) Power factor
d) Draw phasor diagram showing phase and line voltages and all currents.
[9 marks]
Solution:

Ic
c
300 V
(14.5 + j14.5) 

)  n 300 V
(14




4.5
.5




j1
+
+
j1




5
4.
4.




(1
5
)




b
Ib
a
Ia

a) VL = 300 V
V 300
Vp = L = = 173.2 V [½ mark]
3 3

Z = 14.5 + j14.5 = 20.51450 

Vp 173.2
IL = I p = = = 8.44 A [½ mark]
Z 20.51

b) P = 3 VL I L cos ( ) = 3 ( 300  8.44 ) cos ( −450 − 300 ) = 1135.06 W
[1 mark]
c) power factor

(
p. f . = cos  = cos −750 = 0.2588 (lagging ) ) [1 mark]




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, d) Draw phasor diagram showing phase and line voltages and all currents.
Let Van = 173.200 V

Vbn = 173.2 − 1200 V

Vcn = 173.21200 V

Vab = Van − Vbn = 173.200 − 173.2 − 1200 = 300300 V [1 mark]

Vbc = Vbn − Vcn = 173.2 − 1200 − 173.21200 = 300 − 900 V [1 mark]

Vca = Vcn − Van = 173.21200 − 173.200 = 3001500 V [1 mark]

Van 173.200
Ia = = = 8.45 − 450 A [½ mark]
Z 20.51450



Vbn 173.2 − 1200
Ib = = = 8.45 − 1650 A [½ mark]
Z 20.51450

Vcn 173.21200
Ic = = = 8.45750 A [½ mark]
Z 20.51450


The phasor diagram is as shown in below.
Vcn = 173.2 V
Vca = 300 V Vab = 300 V
Ic = 8.45 A




Van = 173.2 V
Ib = 8.45 A


Ia = 8.45 A




Vbn = 173.2 V




Vbc = 300 V

[1½ marks]
2. A 3-phase balanced star connected load is connected to a 240 V 3 phase
supply. The total volt amperes are 6000 VA and total active power is 4800 W.
Find

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, a) Power factor
b) Line current.
c) Phase current
d) Resistance and reactance of load in each phase [6 marks]
Solution:
Given VL = 240 V , S = 6000 VA and P = 4800 W

a) Power factor
Since P = S cos ( ) = S  p. f

P 4800
Hence, Power factor, p. f = = = 0.8 [1 mark]
S 6000
b) Line current
Since S = 3 VL I L or

S 6000
IL = = = 14.43 A [1 mark]
3 VL 3  240

c) Phase current
For a star connected 3-phase load, the line and phase currents are equal
i.e.

I p = I L = 14.43 A [1 mark]

d) Resistance and reactance of load in each phase
V
Since, I L = I p = p , then
Z

Vp VL 240
Z = = = 9.6  [1 mark]
Ip 3I p 3  14.43

The angle of impedance is the angle that produces the power factor. Hence,
the angle of impedance is

 = cos −1 ( 0.8 ) = 36.870 [½ mark]

Therefore, Z = 9.6 − 36.870  = ( 7.68 − j 5.76 )  [½ mark]

The resistance of load is R = 7.68  [½ mark]

The reactance of the load is XC = − j 5.76  [½ mark]




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