3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Steps to be followed while solving a problem on
three-phase circuits:
(i) Assume supply voltage as line voltage.
(ii) Indentify the type of load, i.e., whether star or delta connected
and determine the phase voltage.
(iii) Determine the phase current as, Iph =
V ph
Z ph
(iv) Determine the line current depending on whether the load is
star or delta connected.
(v) The phase angle,  is the angle between Vph and Iph.
.
Calculate its value from Zph.
Q.1) Three coils, each having resistance and inductance of 8 Ω
and 0.02 H respectively, are connected in star across a threephase, 230 V, 50 Hz supply. Find the line current, power factor,
power, reactive volt-amperes and total volt-amperes.
Data :
R=8Ω
L = 0.02 H
VL = 230 V
𝑓 = 50 Hz
For a star-connected load,
𝑉𝑝𝑕 =
𝑉𝐿
3
=
230
3
= 132.79 V
𝑋𝐿 = 2𝜋𝑓𝐿 = 2𝜋 × 50 × 0.02 = 6.28 Ω
1
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
𝑍𝑝𝑕 = 𝑅 + 𝑗𝑋𝐿
= 8 + 𝑗6.28 = 10.17 < 38.13°Ω
𝑍𝑝𝑕 = 10.17 Ω
∅
= 38.13 °
Power factor = 𝑐𝑜𝑠 (38.13°) = 0.786(𝑙𝑎𝑔𝑔𝑖𝑛𝑔)
𝐼𝑝𝑕 =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
132.79
10.17
= 13.05 𝐴
𝐼𝐿 = 𝐼𝑝𝑕 = 13.05 𝐴
𝑃=
3 𝑉𝐿 𝐼𝐿 𝑐𝑜𝑠∅ = 3 × 230 × 13.05 × 0.786 = 𝟒. 𝟎𝟖𝟖 𝒌𝑾
𝑄=
3 𝑉𝐿 𝐼𝐿 𝑠𝑖𝑛∅
𝑄=
3 × 230 × 13.05 × sin(38.13) = 𝟑. 𝟐𝟏 𝑲𝑽𝑨𝑹
𝑆=
3 𝑈𝐿 𝐼𝐿 =
3 × 230 × 13.05 = 𝟓. 𝟏𝟗𝟖 𝑲𝑽𝑨
Q. 2) Three similar coils A, B and C are available. Each coil has a
9  resistance and a 12  reactance. They are connected in delta
to a three-phase, 440 V, 50 Hz supply. Calculate for this load the
(i) phase current, (ii) line current, (iii) power factor, (iv) total kVA,
(v) active power, and (vi) reactive power. If these coils are
connected in star across the same supply, calculate all the above
quantities.
2
3 PHASE CIRCUITS
Data :
INTERCONNECTION
R=9
f = 50 Hz
XL = 12 
VL = 440 V
SOLVED
For a delta-connected load,
VL = Vph = 440 V
𝑍ph =9+j12 = 1553.13°
Zph = 15 
 = 53.13°
Iph =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
400
= 29.33 A
15
IL= 3 Iph= 3 x 29.33 = 50,8 A
Power factor = cos = cos (53.13°) = 0.6 (lagging)
S = 3 V L IL
= 3 x 440 x 50.8 = 38.71 kVA
P = 3 VLIL cos
= 3 x 440 x 50.8 x 0.6 = 23.23 kW
Q = 3 VLIL sin
= 3 x 440 x 50.8 x sin(53.13°)= 30.97 kVAR
If these coils are connected in star across the same supply,
Vph =
𝑉𝐿
3
=
440
3
= 254.03 V
Zph = 15 
Iph =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
254.03
15
= 16.94 A
3
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
IL = Iph = 16.94 A
Power factor = 0.6 (lagging)
S = 3 V L IL =
3 x 440 x 16.94 = 12.91 kVA
P = 3 V L IL =
3 x 440 x 16.94 x 0.6 = 7.74 kW
Q = 3 VLIL sin 
=
3 x 440 x 16.94 x 0.8 = 12.33 kVAR
Q. 3) A 415 V, 50 Hz, three-phase voltage is applied to three starconnected identical impedances. Each impedance consists of a
resistance of 15  a capacitance of 177 F and an inductance of
0.1 henry in series. Find the (i) phase current, (ii) line current, (iii)
power factor, (iv) active power, (v) reactive power, and (vi) total
VA. Draw a neat phasor diagram. If the same impedances are
connected in delta, find the (i) line current, and (ii) power
consumed.
Data :
VL = 50Hz
f = 50 Hz
R = 15
C = 177F
L = 0.1 H
For a star-connected load,
Vph =
𝑉𝐿
3
=
415
3
= 239.6 V
XL = 2fL = 2 x 50 x 0.1 =31.42 
4
3 PHASE CIRCUITS
INTERCONNECTION
XC =
1
=
2fL
1
2 x 50 x 177 x 10−6
SOLVED
= 17.98 
𝑍ph = R + jXL - jXC = 15 + j31.42 - j17.98
= 15 + j13.44 = 20.14  41.860 
Zph = 20.14 

=41.68°
Power factor = cos  = cos (41.86°) = 0.744 (lagging)
Iph =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
239.6
20.14
= 11.9 A
IL = Iph = 11.9 A
P =
Q =
3 VL IL cos= 3 x 415 x 11.9 x .744 = 6.36 kw
3 VL IL sin= 3 x 415 x 11.9 x sin (41.860)
Q = 5.71 KVAR
S =
S=
3 V L IL
3 x 415 x 11.9 = 8.55 Kva
Phasor Diagram:
5
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
If the same impedances are connecte in delta
VL = Vph = 415 V
Zph = 20.14
→ Iph =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
415
20.14
= 20.61 A
→ IL =
3 Iph =
→ P=
3 VL IL cos =
3 x 20.61 = 35.69 A
3 x 415 x 35.69 x .744
P = 19.09 kW
Q.4) Three similar choke coils are connected in star to a threephase supply. If the line current is 15A, the total power consumed
is 11 kW and the volt-ampere input is 15 kVA, find the line and
phase voltages, the VAR input and the reactance arid resistance
6
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
of each coil. If these coils are now connected in delta to the same
supply, calculate phase and line currents, active and reactive
power.
Data:
IL = 15 A
P = 11 kW
S = 15 kVA
For a star-connected load,
S = 3 V L IL
15 x 103 = 3 x VL x 15
VL = 577.35 V
𝑉𝐿
Vph =
3
cos =
𝑃
𝑆
=
577.35
=
3
= 333.33V
11 𝑋 10 3
15 𝑋 10 3
= 0.733
 = 42.860
Q = 3 VL IL sin  = 3 x 577.35 x 15 x sin (42.860) = 10.2 kVAR
Iph = IL = 15A
Zph =
𝑉𝑝 𝑕
𝐼𝑝 𝑕
=
333.33
15
= 22.22 
R = Zph cos  = 22.22 X 0.733 = 16.29 
XL = Zph sin 
= 22.22 x sin (42.86°) = 15.11 
If these coils are now connected in delta,
Vph =VL = 577.35V
Zph = 22.22 
7
3 PHASE CIRCUITS
INTERCONNECTION
Iph =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
577.35
22.22
SOLVED
= 25.98 A
IL = 3 Iph
= 3 x 25.98 = 45 A
P = 3 VL IL cos
= 3 x 577.35 x 45 x 0.733 = 32.98 kW
Q = 3 VL IL sin
= 3 x 577.35 x 45 x sin (42.860) = 30.61 kVAR
Q. 5) A 3, star connected source feeds 1500 kw at 0.85 pf lag to
a balanced delta connected load. Calculate the current, its active
& reactive components in each phase of source & load. The given
line voltage is 2.2kV.
𝑝𝑓 = 0.85(𝑙𝑎𝑔𝑔𝑖𝑛𝑔)
Data : P = 1500 kW
VL = 2.2 kV
For a delta-connected load,
P = 3 VL IL cos∅
1500 × 103 =
3 × 2.2 × 103 × I𝐿 × 0.85
IL = 463.12A
Iph =
IL
3
=
463.12
3
= 267.38A
Active component in each phase of the load
8
= 𝐼𝑝𝑕 cos ∅
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
= 267.38 × 0.85 = 227.27𝐴
Reactive component in each phase of the load = 𝐼𝑝𝑕 sin ∅
= 267.38 × sin 𝑐𝑜𝑠 −1 0.85
= 267.38 × 0.526 = 140.85 𝐴
For a star-connected source, the phase current in the source will
be the same as the line current drawn by load.
Active component of this current in each phase of the source
= 463.12 × 0.85 = 393.65 𝐴
Reactive component of this current in each phase of the source
= 463.12 × 0.526 = 243.6 𝐴
9
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Q. 6) For a balanced, three-phase Wye-connected load, the phase
voltage VR is 10045° V and it draws a line current Iy of 5
180°A. (i) Find the complex impedance per phase, (ii) Draw a
power triangle and identify all its sides with magnitudes and
appropriate units. Assume phase sequence R-Y-B.
VR = 𝑉𝑝𝑕 = 100 - 450 V
Data:
Iy = 5180°A
For a Wye-connected load,
𝑉𝑅 = 100 -45°V
The current IR leads current Iy by angle 120°.
𝐼𝑅
= 5  - 60°A
𝑍𝑝𝑕 =
𝑉𝑅
𝐼𝑅
=
100  − 45°
5 − 60°
= 20 15°  = 19.32 +j5.18 
10
3 PHASE CIRCUITS
INTERCONNECTION
Active power
SOLVED
P = 3 Vph Iph cos
= 3x 100 x 5 x cos (15°) = 1.45 kW
Q = 3 Vph Iph sin 
Reactive power
= 3 x 100 x 5 x sin (150) = 0.39 kVAR
Reactive power
S = 3 Vph Iph
.5 K
1
=
S
VA

Q=0.39K VA R
= 3 x 100 x 5 = 1.5 KVA
P=1.45K W
Q.7) Each leg of a balanced, delta-connected load consists of a
7 resistance m series with a 4 inductive reactance. The lineto-line voltages are
Eab = 2360 00
E bc = 2360 120° V
Eca = 2360 120°V
Determine
(i) phase current Iab, Ibc and Ica(both magnitude and phase),
(ii) each line current and its associated phase angle.
(iii) the load power factor, and
11
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
(iv) find the impedance per phase that draws the same power at
the same power factor.
Data :
R =7
XL = 4 
VL = 2360 V
For a delta-connected load,
Vph = VL = 2360 V
𝑍ph = 7 + j 4 = 8.06  29.740
Phase current Iab =
=
𝐼𝑏𝑐 =
𝐼𝑐𝑎 =
𝐸𝑎𝑏
𝑍𝑝 𝑕
2360  0°
8.06  29.74°
2360  120°
8.06  29.74°
2360  120°
8.06  29.74°
= 292.8  29.740A
= 292.8  149.710A
= 292.8 90.260A
In a delta-connected, three-phase system, line currents lag
behind respective phase currents by 300.
IL = 3 Iph
= 3 x 292.8 = 507.14 A
ILa = 507.14 59.710A
ILb = 507.14 179.710A
ILc = 507.14 60.260A
Load power factor = cos (29.740) = 0.868 (lagging)
12
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Assuming that impedances are now connected in star, the power
per phase and power factor remains the one.
For a delta-connected load,
Power per phase = Vph Iph cos  = 2360 x 292.8 x 0.868 =
599.79 kW
For a star-connected load,
Vph =
𝑉𝐿
3
=
2360
3
= 1362.55 V
Power per phase = Vph Iph cos 
599.79 x 103 = 1362.55 x Iph x 0.868
Iph = 507.14 A
Zph =
𝑉𝑝 𝑕
𝐼𝑝 𝑕
=
1362 .55
507.14
= 2.69 
Q.8) A three-phase, 200 kW, 50 Hz, delta-connected induction
motor is supplied from a three-phase, 440 V, 50 Hz supply
system. The efficiency and power factor of the three-phase
induction motor are 91 % and 0.86 respectively. Calculate (i) line
currents, (ii) currents in each phase of the motor, and (Hi) active
and reactive components of phase current.
Data:
P0 = 200 kW
f
= 50Hz
VL = 440 V
 = 91%
13
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
pf = 0.86
For a delta-connected load (induction motor),
Vph = VL = 440 V
Efficiency
𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
 =
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
0.91 =
Input power
200 × 10 3
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
Pi = 219.78 kW
3 𝑉𝐿 𝐼𝐿 cos
Pi =
219.78 X 103 =
𝐼𝐿
=
3 x 440 x 𝐼𝐿 x 0.86
335.3A
Iph =
𝐼𝐿
3
=
335.3
3
= 193.6A
Active component of phase current = Iphcos=193.6x0.86
=166.5 A
Reactive component of phase current = Iph sin  = 193.6 x 0.51
= 98.7 A
Q.9) A 3 400 V star connected alternator supplies a 3, 112 kw
mesh connected induction motor of effieiency & pf of 0.88 and
0.86 respectively. Find (i) current in each motor phase (ii) currents
in each alternator phase (iii) active and reactive components of
current in each case.
14
3 PHASE CIRCUITS
Data:
INTERCONNECTION
SOLVED
VL = 400V
P0 = 112 Kw
pf = 0.86 V
 = 0.88
For mesh connected load (induction motor),
Vph = VL = 440 V
Efficiency
0.88 =
Input power
𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
 =
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
112 𝑥 10 3
𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
Pi = input power = 127.27 kw
Now Pi = 3 𝑉𝐿 𝐼𝐿 cos
 127.27 x 103 =
3 x 400 x 𝐼𝐿
X 0.86
 𝑰𝑳 = 213.6A
𝑰𝒑𝒉 =
→
𝑰𝑳
𝟑
=213.6/ 3 =123.32A
Current in star connected alternator phase will be same as
line current drawn by motor.
 current in each alternator phase = 213.6 A
→
Active component of current in each phase of motor
=𝐼𝑝𝑕 cos 
= 123.32 x 0.86 = 105A
→
Reactive component = 𝐼𝑝𝑕 sin  = 123.32 x 0.51 = 62.89 A of
current in each motor .
15
3 PHASE CIRCUITS
→
INTERCONNECTION
Active component
SOLVED
= 213 x 0.86
of current in each alternator  = 183. 7 A
→
Reactive component of
= 213 x 0.51 = 108.9A
current in each alternator 
Q. 10)Three identical impedances of 10 30°  each are
connected in star and another set of three identical impedances of
18 60°  are connected in delta. If both the sets of impedances
are connected across a balanced, three-phase 400 V supply, find
the line current, total voltamperes, active power and reactive
power.
Data :
𝑍𝑌 = 10  30° 
𝑍𝐴 = 10  30° 
𝑉𝐿 = 400 𝑉
Three identical delta impedances can be converted into
equivalent star impedances.
𝑍′𝑌 =
𝑍𝐴
3
=
18 600
3
= 6 600 
Now two star-connected impedances 10  30° and 6  60°
are in parallel across a three-phase supply.
𝑍𝑒𝑞 =
( 10  30°) ( 6  60°)
10  30°+ 6  60°
16
= 3.87 48.830 
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
For a star-connected load,
𝑉𝑝𝑕 =
𝐼𝑝𝑕 =
𝑉𝐿
3
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
400
=
3
= 230.94 V
𝑉𝑝 𝑕
𝑍𝑒𝑞
=
230.94
3.87
= 59.67 𝐴
𝐼𝐿 = 𝐼𝑝𝑕 = 59.67 𝐴
S = 3 𝑉𝐿 𝐼𝐿 = 3 x 400 x 59.67 = 41.34 kVA
𝑃 = 3 𝑉𝐿 𝐼𝐿 cos ∅ = 3 x 400 x 59.67 x cos (48.830) = 27.21kW
Q = 3 𝑉𝐿 𝐼𝐿 sin ∅ = 3 x 400 x 59.67 x sin (48.830) = 31.12kVAR
Q. 11) Three 100 𝜴, non- inductive resistances are connected in
(a) star, and (b) delta across a 400 V, 50 Hz, three- phase supply.
Calculate the power taken from the supply in each case. If one of
the resistances is open circuited, what would be the value of total
power taken from the mains in each of the two cases?
𝑉𝐿 = 400 𝑉
DATA:
𝑍𝑝𝑕 = 100 Ω,
For a star-connected load,
𝑉𝑝𝑕 =
𝑉𝐿
3
𝐼𝑝𝑕 =
=
400
𝑉𝑝 𝑕
𝑍𝑝 𝑕
3
=
= 230.94 𝑉
230.94
100
= 2.31 𝐴
𝐼𝐿 = 𝐼𝑝𝑕 = 2.31 𝐴
17
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
cos ∅ = 1
P = 𝟑 𝑽𝑳 𝑰𝑳 𝐜𝐨𝐬 ∅
= 3 × 400 × 2.31 × 1 = 1600.41 𝑊
For a delta-connected load,
𝑉𝑝𝑕 = 𝑉𝐿 = 400 V
𝐼𝑝𝑕 =
𝑉𝑝 𝑕
𝑍𝑝 𝑕
=
400
100
4𝐴
𝐼𝐿 = 3 𝐼𝑝𝑕
= 3 × 4 = 6.93 A
𝑃 = 3 𝑉𝑙 𝐼𝑙 cos ∅
= 3 × 400 × 6.93 × 1 = 4801.24
When one of the resistors is open circuited
Stat connection
The circuit consists of two 100 𝜴 resistors in
series across a 400 V supply.
Currents in lines A and C =
400
200
= 2A
Power taken from the mains = 400× 2 = 800 𝑊
18
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Hence, when one resistor is open circuited, the power
consumption is reduced by half.
Delta connection In this case, currents in A and C remain as
usual out of phase with each other.
∴ current in each phase =
400
100
= 4A
→ power taken from mains = 2× 4 × 400 = 320000
∴ when one resistor is open circuited, power consumption
is
reduced by one third.
Q.12) A balanced delta-connected load having an impedance ZL
= (300+j210) ohm in each phase is supplied from a 400 V, threephase supply through a three-phase line having an impedance of
Zs=(4+j8) ohm in each phase. Find current and voltage in each of
the load.
Data : 𝑍𝐿 = 300 + 𝑗210 𝜴
𝑍𝑆 = 4 + 𝑗8 𝜴
𝑉𝐿 = 400𝑉
19
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Three identical delta impedance can be converted into equivalent
star impedances.
𝑍′𝐿 =
𝑍𝐿
3
=
300+𝑗 210
3
= 100 + 𝑗70 𝛀
The circuit can be drawn as
𝑍𝐴𝑁 = 𝑍𝑆 + 𝑍 ′𝐿
= 4 + 𝑗8 + 100 + 𝑗70
= 104 + 𝑗78 𝛺 = 13036.87°𝛺
For a star-connected load,
𝑉𝐴𝑁 =
𝐼𝐴𝑁 =
Voltage across
400
230.940°
130 36.87°
𝑍𝑆 = 𝐼𝐴𝑁 . 𝑍𝑆
20
3
= 230.94V
=1.78 − 36.87°A
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
= 1.78 − 36.87° × 4 + 𝑗8 =
15.9226.56°
𝑍′𝐿 = 230.94  0° − 15.9226.56°
Voltage across
= 216.82  − 1.88°𝑉
Voltage in each phase of the load = 216.82 𝑉
Current in each phase of the load = 1.78 A
PROBLEMS BASED ON CONVERSION
Q.1. Find RAB
4
4.5
4.5
R4
R2
A
R1
R3
3
3
R6
R5
7.5
7.5
3
Converting the two delta networks to star.
21
B
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
4
R2
R4
R6
R1
B
A
R5
R3
3
R1
=
RC
=
R2
=
R5
=
R3
=
R4
=
4.5 𝑥 7.5
4.5 + 7.5 + 3
7.5 𝑥 3
4.5 + 7.5 + 3
4.5 𝑥 3
4.5 + 7.5 + 3
0.9
A
2.25
4
= 2.25
= 1. 5
= 0.9
0.9
2.25
5.8
6
1.5
3
1.5
Now 5.8 || 6 = 2.95
A
2.25
2.95
7.45
22
2.25
B
B
3 PHASE CIRCUITS
 RAB
=
INTERCONNECTION
SOLVED
7.5
Q. 2
6
9
1.5
4
3
1
Note
:
for such a network convert internal star to delta
network, do not do vice versa until explicitly mentioned.
 Converting star n/w to delta n/w we have
A
9
1.5
R2
R1
R3
B
C
1
we have 9 and R1 in parallel
23
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
we have 1.5 and R2 in parallel
we have 9 and R1 in parallel
R1 = 6 + 4 +
9
6𝑥4
= 18
3
R2 = 6 + 3 +
6𝑥3
6
18
=
4
13
.5
9
1.3 1.5
5
0.9
13.5
R3 = 4 + 3 +
4𝑥3
6
1
= 9
→
The network can be simplified as in
1)
Between terminals A and B
9 || 18 = 6
2)
Between terminals B and C
1.5 || 13.5 = 1.35

Redrawing the network as simplified
A
below.
RAB = 6 || (1.35 + 0.9)
1.35
6
= 6 || 2.25
= 1.64 
C
B
0.9
24
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Q.3
15
20
A
30
45
35
B
40
Drawing the resistance of 30 from outside
20
15
A
25
45
35
30
B
40
Converting delta network formed by resistance 20, 25
and 35 into an equivalent star network.
25
3 PHASE CIRCUITS
INTERCONNECTION
15
C
A
R2
45
R1
30
E
R3
B
D
R1 =
R2 =
R3 =
SOLVED
20 𝑥 35
20 + 35 + 25
20 𝑥 25
20 + 35 + 25
35 𝑥 25
20 + 35 + 25
F
40
= 8.75
= 6.25
= 10.95
Redrawing the circuit, where B & D are equipotential.
A
15
C
6.25
8.75
E
F
10.95
30
45
40
B
D
D
By Series - Parallel Reduction
26
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
6.25
15
A
45
38.75
50.94
B
Now 50.94  || 38.75  = 22.01  = 2.2 
 22  in series with 6.25
= 22 + 6.25 = 28.25
15
A
45
28.25
B
parallel
45 || 28.25 = 17.35 
A
A
17.35
32.35
=
B
B
 RAB = 32.35Ω
27
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Q.4 Find RAB
A
6
4
3
5
5
4
8
B
Hint convert : 3Ω , 5Ω and 8Ω star to delta
A
6
4
R1
R2
5
parallel
R3
4
l
lle
a
r
pa
B
Now (R2 and 5Ω in parallel) and (R3 and 4Ω in parallel)
A
6
A
4
n
parallel
y
y
z
Convert
X1Y1Z to
star
B
z
series
series
B
28
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
 RAB = 4.23Ω
Q.5
Find equivalent resistance between A and B.
→ converting star n/w formed by 3Ω, 4Ω and 5Ω into delta
n/w.
R3
5
11.75
3
15.67
4
R1 = 5 + 4 +
R2 = 3 + 4 +
R1
5𝑥4
3
5𝑥4
5
= 15.67
= 9.4
29
9.4
R2
3 PHASE CIRCUITS
INTERCONNECTION
R3 = 5 + 3 +
5𝑥3
4
SOLVED
= 11.75
Similarly, Converting star n/w formed by resistors 4, 6,
8 into  network.
R6
6
13
4
26
8
R4 = 6 + 8 +
R5 = 4 + 8 +
R6 = 6 + 4 +
→
R4
6𝑥8
4
4𝑥8
6
6𝑥4
8
17.33
R5
= 26
= 17.33
= 13
connecting the 2 delta n/w's in parallel between A & B
30
3 PHASE CIRCUITS
→
INTERCONNECTION
SOLVED
 Resistances of 9.4Ω and 17.33 Ω are in parallel with a
short. its equivalent
R=0
 RAB = 6.17 || 9.78 = 3.78Ω
31
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Q.6 Find equivalent resistance between terminals x and y in
the n/w shown.
→
 8 Ω and 4 Ω are in series as well as 17Ω and 13Ω are in
series as well as 17Ω and 13Ω as shown in above fig. we
have.
34
N
12
12
O
P
30
30
30
RL
M
12
X
Y
32
6
Series
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
converting XMN from  →  Similarly converting YoP
from  → 
34
N
R1
R3
10
4
4
10
RX
A
RY
N
R2
M
P
O
R2
4
X
10
y
∆
³ Delta Resistances are equal  = 
3
R1 = R2 = R3 =
12 𝑥 12
12 + 12 + 12
=
Rx = Ry = Rz =
= 4 on
30 𝑥 30
30 + 30 + 30
33
12
3
∆
3
=
= 4
= 10 on
30
3
= 10
3 PHASE CIRCUITS
INTERCONNECTION
48
SOLVED
14
48
A
A
10
4
14
Now 48 || 14 = 10.84
10.84
4
10
X
Y
 Rxy = 4 + 10 + 10.84 = 24.84
34
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
Q.17 Find RAB
A
6
6
B
12
12
12
C
D
3
E
F
3
Converting internal delta n/w →  n/w
A
B
6
R1
R2
6
4
Series
R3
4
4
3
3
Series
35
3 PHASE CIRCUITS
INTERCONNECTION
SOLVED
A
6
6
B
C
S
7
F
7
C and F are equipotential  SC and SF are in parallel, ll'y AC and
AF are parallel.
A
A
A
B
4
4
6
6
3
S
6
7
10.5
7
6
3.5
C,F
C,F
 RAB = 10.5 Ω
36
B