Star Delta Problems

1. V
3

5
.
2

10

10
C

20
A B
2. Find the current supplied by the source to the circuit shown in fig.2.
20Ω and 60Ω resistor are in parallel















15
10
150
15
10
150
60
5
.
2
150
5
.
2
25
25
100
5
.
2
10
*
5
.
2
5
.
2
*
10
10
*
10
CA
BC
AB
R
R
R
V
3

15

20
C

60
A
B

15
Fig.2
In ABC three star connected resistor is
need to be converted into delta 


 15
80
60
*
20
Req
V
3

15
C

15
A
B

15

2. Now 15Ω and 15Ω resistor are in series
Ω
Req 30
15
15 



Now 15Ω and 30Ω resistor are connected in parallel
V
3

15 
30
Ω
*
Req 10
45
30
15



V
3

10
Now by ohm’s law the current
supplied can be calculated
A
.
R
V
I 3
0
10
3




3. 3. Find the equivalent resistance across A and B for
the circuit shown in fig.5.
Fig.5.

6

3

5
.
8

9

7
B
A
a
b c
In a, b and c 3Ω, 6Ω and 9Ω resistor are delta
Connected and it is need to be converted to star


















3
18
54
9
6
3
6
*
9
5
.
1
18
27
9
6
3
3
*
9
1
18
18
9
6
3
6
*
3
c
b
a
R
R
R

5
.
1 
3

5
.
8

1

7
B
A
Now 1.5Ω and 8.5Ω resistor are in series and
3Ω and 7Ω resistor are in series










10
3
7
R
10
5
.
8
5
.
1
R
eq2
eq1

4. 
10

1
B
A

10
Now two 10Ω resistors are connected in
parallel



 5
20
10
*
10
Req

5

1
B
A
Now 5Ω and 1Ω resistors are
connected in series




 6
5
1
RAB

5. 4. Obtain the equivalent resistance between the points A and B as
shown in fig.1
A
B

2

3

4 
8

5

6
In the given circuit the 4Ω, 8Ω and 3Ω resistors are connected in the
star connection this can be converted into the delta connection
A
R
B
R
C
R
A
B

2

3

4 
8

5

6
Fig.1

6. 





















5
.
8
8
3
4
3
4
67
22
3
4
8
4
8
17
4
8
3
8
3
*
R
.R
R
R
R
R
Ω
.
*
R
.R
R
R
R
R
Ω
*
R
.R
R
R
R
R
B
C
A
C
A
CA
A
C
B
C
B
BC
C
B
A
B
A
AB
The reduced diagram is shown in
below (after converting from star
to delta)

5
CA
R AB
R
BC
R
A
B

2

5
.
8

67
.
22

17

6
The above diagram is re arranged
as shown in below
A
B

2

5
.
8

67
.
22

17

6

5

7. A
B

2

5
.
8

67
.
22

17

6

5
The above circuit 5Ω and 17Ω are connected in parallel, similarly 6Ω and 22.67Ω resistance
are connected in parallel, the equivalent value for this parallel connection is given in below
Ω
.
.
*
.
Ω
Ω||
.
Similarly
Ω
.
*
Ω
Ω||
74
4
6
67
22
6
67
22
6
67
22
86
3
5
17
5
17
17
5








Reduced after calculating the value for
parallel connection is shown in below
A
B

2

5
.
8

74
.
4

86
.
3

8. In the equivalent circuit
4.74Ω and 3.86Ω are in
series so the circuit again
reduced in the below format
A
B

2

5
.
8

74
.
4

86
.
3
B
A

2

5
.
8 
86
.
3 Ω
.
Ω
.
Ω
. 6
8
74
4
86
3 

Ω
.
.
.
.
*
.
Ω
.
Ω||
.
Now
27
4
6
8
5
8
6
8
5
8
6
8
5
8




4.27Ω and 2Ω are in series so the equivalent resistance
between terminals A and B is 6.7Ω
A

2

27
.
4
B

9. A
5. Obtain the equivalent resistance between the points A and B as
shown in fig.2
35Ω, 25Ω and 15Ω resistance are connected in delta connection, this
can be converted into star with the equivalent value RA, RB and RC
CA
R
AB
R
BC
R
B

15

35

25

20

40 
30
A
B

15

35

25

20

40 
30
Fig.2

10. Ω
*
R
R
R
.R
R
R
Ω
.
*
R
R
R
.R
R
R
Ω
*
R
R
R
.R
R
R
CA
BC
AB
CA
BC
C
CA
BC
AB
BA
AB
B
CA
BC
AB
CA
AB
A
5
25
15
35
15
25
67
11
25
15
35
25
35
7
25
15
35
15
35






















20
B

67
.
11

7

5

40 
30
A
In the above equivalent circuit 5Ω and 30Ω are
connected in series, similarly 11.67Ω and 40Ω
are connected in series so the above circuit
reduced as shown in below
B

7

67
.
51 
35
A

20

11. B

7

67
.
51 
35
A

20
Ω
.
.
*
.
Ω
Ω||
.
86
20
35
67
51
35
67
51
35
67
51




The equivalent value for the 51.67||35 is 20.86Ω and the 7Ω
and 20Ω resistance are connected in series so the equivalent
value is 27Ω, finally the resistance between terminal A and B
is given by
47.86 Ω