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( ∑

( ∑

Ry (10)

Q 1) 2020-Aug (2018 scheme ECE) 2020-Aug (2018 scheme EE) 2017-June Determine the Input resistance between PQ using star-delta transformation

6 Ω

Figure 2

Solution: In the given circuit 6 , 6 and 18 connected

between A,B and C are in delta connection, convert this into star network. The details are as shown in Figure 3

RA = RAB ×RAC∑

1.2Ω

3.6Ω

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 1

0.1. Star Delta Transformation

Figure 3

In the circuit 5 and 3.6 are in series, 6 and 1.2 are in series and 3.6 18 and 6 are in series.

7.2Ω 27.6Ω

8.6Ω P

Figure 4

7.2 27.6 are in parallel, which is in series with 8.6 . The net resistance is.

RPQ = 8.6 + 7.2 × 27.6

7.2 + 27.6 = 8.6 + 5.71

= 14.31

Q 2) For the circuit shown in Figure shown in Figure 5 determine the equivalent resistance between any two terminals.

Solution:

x

y

Solution:

In the given 4, 4, 4, and are in star network, convert this star network to delta network

Rxy = Rx + Ry + Rx ×Ry

Rz = 8 + 4 = 12

4 = 8 + 4 = 12

4 = 8 + 4 = 12

4 = 8 + 4 = 12

As shown in the Figure 6, by looking by any two terminals 12 is parallel with 4

= 12 × 4

Figure 6

By looking from XZ terminal the 3 resistance is in parallel with series connection of 3 and 3 resistance.

RXZ = 3 × 6

3 + 6 = 2

Q 3) For the circuit shown in Figure shown in Figure 7 determine the equivalent resistance between (1)A and B and (2) A and N.

C

A

B

N

Convert the internal star network to delta

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 2

0.1. Star Delta Transformation

4 = 32 + 15 = 47

2 = 34 + 60 = 94

30 = 6 + 0.266 = 6.266

Figure 8

In the circuit 10 is parallel with 47

= 10 × 47

= 94 × 15

= 6 × 6.266

C

A

B

12.93 Ω

Figure 9

While looking from terminal AB the resistances 12.93 and and 3.06 are in series and this resistance is parallel with 8.245

= 12.93 + 3.06 = 16

RAB = 8.245 × 16

8.245 + 16 = 5.44

C

A

B

N

Figure 10

From the Figure 10 by looking from the terminal resistance between A and N, there is a delta network between NCB which is replaced by star network.

Rx = RxyRxz

Figure 11

In the network of 6 and 1.225 are in series, 10 and 9.18 are in series. The details are as shown in Figure 12

6 Ω

1.225 Ω

10 Ω2Ω

2.5 Ω

The resistances 7.225 and 19.18 are in

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 3

0.1. Star Delta Transformation

7.75 Ω2Ω

Figure 14

The resistances 2.5 and 5.25 are in series. The resistances 2 7.75 are in in parallel and the net resistance between A and N is

= 2 × 7.75

2 + 7.75 = 1.59

Q 4) Determine the equivalent resistance between terminals A and B

4 Ω

15 Ω

20 Ω

= 1.2 + 7.446 + 3.3 = 11.94

Q 5) 2018-Dec (2010 scheme) Find the resistance of the circuit shown in Figure shown in Figure 5 between A and B

30Ω

2Ω

Solution:

In the given network 18 there is delta network between xyz and it is replaced by star network.

RX = 20 × 5

Figure 19

In the circuit 19, 10 2.5 are in series, similarly 5 7.5 are in series. The details are as shown in Figure 20

30Ω

Figure 20

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 4

0.1. Star Delta Transformation

In the circuit 20 at NPB 37.5 3.875 and 30 are in delta connection, convert into star network. The details are as shown in Figure 21

2.04Ω

Figure 21

In the circuit 21, 12.5 , 2.045 are in series, similarly 1.635 15 are in series. The details are as shown in Figure 22

14.54Ω

16.63Ω

15.76Ω

Figure 22

In the circuit 22 14.54 , 16.63 are in parallel, which is in series with similarly 15.76 . The net resistance between AB is.

RAB = 15.76 + 14.54 × 16.63

14.54 + 16.63 = 15.76 + 7.76

= 23.52

Q 6) For the circuit shown in Figure shown in Figure 5 determine the current i using star delta transformation.

Solution:

1 Ω

Figure 23

As shown in Figure 23 there is delta network between ABC, convert this to star network

RA = 1 × 1

0.333 Ω

Figure 24

The resistances 0.333 and 2 are in series and 0.333 and 1 are in series.

The resistances 2.333 and 1.333 are in parallel its equivalent resistance is

1 V

1 Ω

2.333 Ω

1.333 Ω

0.333 Ω

Figure 25

= 2.333 × 1.333

2.333 + 1.333 =

I = 1

2.181 = 0.458A

Q 7) Calculate the current in the 40 resistance of the circuit shown in Figure shown in Figure 5 using star delta transformation.

40 Ω 4 V

B

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 5

0.1. Star Delta Transformation

Solution: As shown in Figure 23 there is delta network

between ABC, convert this to star network

RA = 20 × 5

= 3.333 + 40.833 × 3.333

40.833 + 3.333 = 3.333

I1 = 4

6.41 = 0.624A

Using branch current division method the current in 40 is

I3 = 0.624A 3.333

44.163 = 0.047A

Q 8) Determine the currents supplied by the each battery using star-delta transformation

3 Ω

Solution: As shown in Figure 29 there is delta network

between ABC, convert this to star network

RA = 4 × 6

2 Ω

Figure 30

In the circuit 30, 3 and 1.6 are in series and also 2 and 1.3 are in series. The details are as shown in Figure 31

3 Ω 2 Ω1.6 Ω

15 V 20 V

15 − 6.6I1 − 2I2 = 0

20 − 5.3I2 − 2I1 = 0

2I1 + 5.3I2 = 20

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 6

0.1. Star Delta Transformation

=30.98

i1 =

30.98 = 3.292A

The current supplied by each battery are 1.275A and 3.292A respectively

9). Find the equivalent resistance between A and B as shown in Figure 33

30 40

60 50

6.94 × 3.55 = 38 + 24.3 + 80 = 142.3

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 7

0.2. Question Papers

0.2 Question Papers

Question Papers 2019-Jan (2017 scheme ECE) Find the equivalent resistance between a and b as shown in Figure 36 using star delta transformation

6 4

5 3

Solution: As shown in Figure 36 there is delta network

between 8 5 and 4 , convert this to star network

RA = 8 × 5

Figure 38

As shown in Figure 38 there is delta network between 6 5.17 and 5.35 , convert this to star

network

6.94 × 3.55 = 1.87 + 2.34 = 4.21

2019-DEC Determine the resistance between A and B of the network shown in Figure 41.

A

B

12.5

30

10

10

12.5

Solution:

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 8

0.2. Question Papers

Figure 42

As shown in Figure 42 there is delta network between 2 3 and 6 , convert this to star network

RA = 10 × 30

Figure 44

As shown in Figure 44 there is delta network between 6 12.5 and 24.5 , convert this to star network

RA = 6 × 12.5

= 8.1

Also by observation its a wheatstone bridge because the ratio is

12.5

12.5 =

6

RAB = 25 × 12

25 × 12 = 8.1

2019-DEC Determine the resistance between A and B of the network shown in Figure 41.

8

A

B

6

2

Solution:

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 9

0.2. Question Papers

5 E

Figure 48

As shown in Figure 48 there is star network between 2 3 and 5 , convert this to star network

RAB = 2 + 3 + 2 × 3

5 = 5 + 1.2 = 6.2

3 = 7 + 3.3 = 10.3

2 = 8 + 7.5 = 15.5

Figure 50

As shown in Figure 50 there is a delta network between 6.2 10.33 and 6.07 , convert this to star network

RA = 6.2 × 10.33

Figure 52

As shown in Figure 52 there is a delta network between 2.83 6 and 10.77, convert this to star network

RA = 2.83 × 6

= 3.1

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 10

0.2. Question Papers

2

5

6

Figure 56

As shown in Figure 56 there is delta network between 2 3 and 6 , convert this to star network

RA = 2 × 3

Figure 58

As shown in Figure 59 there is delta network between 1.1 5.636 and 5 , convert this to star

network

RMN = 1 + 1.06 + 0.468 = 2.528

2015-Dec Find the equivalent resistance between A and B of the network shown in Figure 61 using Star- Delta transformation.

7

A

8

8

5

10

10

9

3

9

Figure 62

As shown in Figure 62 there is delta network between 15 8 and 5 , and there is delta network

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 11

0.2. Question Papers

between 9 9 and 9 , convert this to star network

RA = 15 × 8

Figure 64

= 15.67 × 14.42

2013-DEC1 Using star/delta transformation deter- mine the resistance between M and N of the network shown in Figure 65.

30 30

2008-June Using star/delta transformation deter- mine the resistance between M and N of the network shown in Figure 70.

5 5

Solution:

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 12

0.2. Question Papers

Figure 71: 2008-June

As shown in Figure 71 there is delta network between A C D 5 5 and 10 , and there is delta network between B C E 5 5 and 10 , convert this to star network

RA = 5 × 5

2007-July Using star/delta transformation deter- mine the resistance between A and B of the network shown in Figure 75.

A

B

Solution:

As shown in Figure 75 there is star network between 6 6 and 6 , convert this to star to delta network

RAB = 6 + 6 + 6 × 6

6 = 12 + 6 = 18

6 = 12 + 6 = 18

6 = 12 + 6 = 18

= 18 × 18

36 = 9

Figure 77: 2007-July

9 and 9 are in series which is in parallel with 18

AB = 18 + 9 × 9

18 = 18 + 4.5 = 22.5

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 13

0.2. Question Papers

Using star/delta transformation determine the resistance between A and B of the network shown in Figure 78.

6kΩ

12kΩ

A

B

18kΩ

RPQ = 7 + 7.785 + 2 = 16.785k

( ∑

Ry (10)

Q 1) 2020-Aug (2018 scheme ECE) 2020-Aug (2018 scheme EE) 2017-June Determine the Input resistance between PQ using star-delta transformation

6 Ω

Figure 2

Solution: In the given circuit 6 , 6 and 18 connected

between A,B and C are in delta connection, convert this into star network. The details are as shown in Figure 3

RA = RAB ×RAC∑

1.2Ω

3.6Ω

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 1

0.1. Star Delta Transformation

Figure 3

In the circuit 5 and 3.6 are in series, 6 and 1.2 are in series and 3.6 18 and 6 are in series.

7.2Ω 27.6Ω

8.6Ω P

Figure 4

7.2 27.6 are in parallel, which is in series with 8.6 . The net resistance is.

RPQ = 8.6 + 7.2 × 27.6

7.2 + 27.6 = 8.6 + 5.71

= 14.31

Q 2) For the circuit shown in Figure shown in Figure 5 determine the equivalent resistance between any two terminals.

Solution:

x

y

Solution:

In the given 4, 4, 4, and are in star network, convert this star network to delta network

Rxy = Rx + Ry + Rx ×Ry

Rz = 8 + 4 = 12

4 = 8 + 4 = 12

4 = 8 + 4 = 12

4 = 8 + 4 = 12

As shown in the Figure 6, by looking by any two terminals 12 is parallel with 4

= 12 × 4

Figure 6

By looking from XZ terminal the 3 resistance is in parallel with series connection of 3 and 3 resistance.

RXZ = 3 × 6

3 + 6 = 2

Q 3) For the circuit shown in Figure shown in Figure 7 determine the equivalent resistance between (1)A and B and (2) A and N.

C

A

B

N

Convert the internal star network to delta

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 2

0.1. Star Delta Transformation

4 = 32 + 15 = 47

2 = 34 + 60 = 94

30 = 6 + 0.266 = 6.266

Figure 8

In the circuit 10 is parallel with 47

= 10 × 47

= 94 × 15

= 6 × 6.266

C

A

B

12.93 Ω

Figure 9

While looking from terminal AB the resistances 12.93 and and 3.06 are in series and this resistance is parallel with 8.245

= 12.93 + 3.06 = 16

RAB = 8.245 × 16

8.245 + 16 = 5.44

C

A

B

N

Figure 10

From the Figure 10 by looking from the terminal resistance between A and N, there is a delta network between NCB which is replaced by star network.

Rx = RxyRxz

Figure 11

In the network of 6 and 1.225 are in series, 10 and 9.18 are in series. The details are as shown in Figure 12

6 Ω

1.225 Ω

10 Ω2Ω

2.5 Ω

The resistances 7.225 and 19.18 are in

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 3

0.1. Star Delta Transformation

7.75 Ω2Ω

Figure 14

The resistances 2.5 and 5.25 are in series. The resistances 2 7.75 are in in parallel and the net resistance between A and N is

= 2 × 7.75

2 + 7.75 = 1.59

Q 4) Determine the equivalent resistance between terminals A and B

4 Ω

15 Ω

20 Ω

= 1.2 + 7.446 + 3.3 = 11.94

Q 5) 2018-Dec (2010 scheme) Find the resistance of the circuit shown in Figure shown in Figure 5 between A and B

30Ω

2Ω

Solution:

In the given network 18 there is delta network between xyz and it is replaced by star network.

RX = 20 × 5

Figure 19

In the circuit 19, 10 2.5 are in series, similarly 5 7.5 are in series. The details are as shown in Figure 20

30Ω

Figure 20

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 4

0.1. Star Delta Transformation

In the circuit 20 at NPB 37.5 3.875 and 30 are in delta connection, convert into star network. The details are as shown in Figure 21

2.04Ω

Figure 21

In the circuit 21, 12.5 , 2.045 are in series, similarly 1.635 15 are in series. The details are as shown in Figure 22

14.54Ω

16.63Ω

15.76Ω

Figure 22

In the circuit 22 14.54 , 16.63 are in parallel, which is in series with similarly 15.76 . The net resistance between AB is.

RAB = 15.76 + 14.54 × 16.63

14.54 + 16.63 = 15.76 + 7.76

= 23.52

Q 6) For the circuit shown in Figure shown in Figure 5 determine the current i using star delta transformation.

Solution:

1 Ω

Figure 23

As shown in Figure 23 there is delta network between ABC, convert this to star network

RA = 1 × 1

0.333 Ω

Figure 24

The resistances 0.333 and 2 are in series and 0.333 and 1 are in series.

The resistances 2.333 and 1.333 are in parallel its equivalent resistance is

1 V

1 Ω

2.333 Ω

1.333 Ω

0.333 Ω

Figure 25

= 2.333 × 1.333

2.333 + 1.333 =

I = 1

2.181 = 0.458A

Q 7) Calculate the current in the 40 resistance of the circuit shown in Figure shown in Figure 5 using star delta transformation.

40 Ω 4 V

B

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 5

0.1. Star Delta Transformation

Solution: As shown in Figure 23 there is delta network

between ABC, convert this to star network

RA = 20 × 5

= 3.333 + 40.833 × 3.333

40.833 + 3.333 = 3.333

I1 = 4

6.41 = 0.624A

Using branch current division method the current in 40 is

I3 = 0.624A 3.333

44.163 = 0.047A

Q 8) Determine the currents supplied by the each battery using star-delta transformation

3 Ω

Solution: As shown in Figure 29 there is delta network

between ABC, convert this to star network

RA = 4 × 6

2 Ω

Figure 30

In the circuit 30, 3 and 1.6 are in series and also 2 and 1.3 are in series. The details are as shown in Figure 31

3 Ω 2 Ω1.6 Ω

15 V 20 V

15 − 6.6I1 − 2I2 = 0

20 − 5.3I2 − 2I1 = 0

2I1 + 5.3I2 = 20

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 6

0.1. Star Delta Transformation

=30.98

i1 =

30.98 = 3.292A

The current supplied by each battery are 1.275A and 3.292A respectively

9). Find the equivalent resistance between A and B as shown in Figure 33

30 40

60 50

6.94 × 3.55 = 38 + 24.3 + 80 = 142.3

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 7

0.2. Question Papers

0.2 Question Papers

Question Papers 2019-Jan (2017 scheme ECE) Find the equivalent resistance between a and b as shown in Figure 36 using star delta transformation

6 4

5 3

Solution: As shown in Figure 36 there is delta network

between 8 5 and 4 , convert this to star network

RA = 8 × 5

Figure 38

As shown in Figure 38 there is delta network between 6 5.17 and 5.35 , convert this to star

network

6.94 × 3.55 = 1.87 + 2.34 = 4.21

2019-DEC Determine the resistance between A and B of the network shown in Figure 41.

A

B

12.5

30

10

10

12.5

Solution:

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 8

0.2. Question Papers

Figure 42

As shown in Figure 42 there is delta network between 2 3 and 6 , convert this to star network

RA = 10 × 30

Figure 44

As shown in Figure 44 there is delta network between 6 12.5 and 24.5 , convert this to star network

RA = 6 × 12.5

= 8.1

Also by observation its a wheatstone bridge because the ratio is

12.5

12.5 =

6

RAB = 25 × 12

25 × 12 = 8.1

2019-DEC Determine the resistance between A and B of the network shown in Figure 41.

8

A

B

6

2

Solution:

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 9

0.2. Question Papers

5 E

Figure 48

As shown in Figure 48 there is star network between 2 3 and 5 , convert this to star network

RAB = 2 + 3 + 2 × 3

5 = 5 + 1.2 = 6.2

3 = 7 + 3.3 = 10.3

2 = 8 + 7.5 = 15.5

Figure 50

As shown in Figure 50 there is a delta network between 6.2 10.33 and 6.07 , convert this to star network

RA = 6.2 × 10.33

Figure 52

As shown in Figure 52 there is a delta network between 2.83 6 and 10.77, convert this to star network

RA = 2.83 × 6

= 3.1

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 10

0.2. Question Papers

2

5

6

Figure 56

As shown in Figure 56 there is delta network between 2 3 and 6 , convert this to star network

RA = 2 × 3

Figure 58

As shown in Figure 59 there is delta network between 1.1 5.636 and 5 , convert this to star

network

RMN = 1 + 1.06 + 0.468 = 2.528

2015-Dec Find the equivalent resistance between A and B of the network shown in Figure 61 using Star- Delta transformation.

7

A

8

8

5

10

10

9

3

9

Figure 62

As shown in Figure 62 there is delta network between 15 8 and 5 , and there is delta network

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 11

0.2. Question Papers

between 9 9 and 9 , convert this to star network

RA = 15 × 8

Figure 64

= 15.67 × 14.42

2013-DEC1 Using star/delta transformation deter- mine the resistance between M and N of the network shown in Figure 65.

30 30

2008-June Using star/delta transformation deter- mine the resistance between M and N of the network shown in Figure 70.

5 5

Solution:

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 12

0.2. Question Papers

Figure 71: 2008-June

As shown in Figure 71 there is delta network between A C D 5 5 and 10 , and there is delta network between B C E 5 5 and 10 , convert this to star network

RA = 5 × 5

2007-July Using star/delta transformation deter- mine the resistance between A and B of the network shown in Figure 75.

A

B

Solution:

As shown in Figure 75 there is star network between 6 6 and 6 , convert this to star to delta network

RAB = 6 + 6 + 6 × 6

6 = 12 + 6 = 18

6 = 12 + 6 = 18

6 = 12 + 6 = 18

= 18 × 18

36 = 9

Figure 77: 2007-July

9 and 9 are in series which is in parallel with 18

AB = 18 + 9 × 9

18 = 18 + 4.5 = 22.5

Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga [email protected] 13

0.2. Question Papers

Using star/delta transformation determine the resistance between A and B of the network shown in Figure 78.

6kΩ

12kΩ

A

B

18kΩ

RPQ = 7 + 7.785 + 2 = 16.785k