Delta-Wye Connection Problem

joIp_rkAs, ECE, NEHU, Shillong - 793 022

North-Eastern Hill University

December 27, 2017

Delta-Wye Connection Problem

Q. For the following circuit in Fig. 4, find the equivalent resistance between (i) A and
B; (ii) A and N:
Soln . (i)
A Since the equivalent resistance is required between vertices A and
B, we now convert the star connection inside the network in to a
delta connection as A

4Ω 6×4+4×5+5×6
10

RA = = 18.50 Ω
5Ω

4Ω
4 RB RC
Ω

6×4+4×5+5×6
RB = = 14.80 Ω 5Ω
5Ω 5 6Ω N
6Ω N 6×4+4×5+5×6
RC = = 12.33 Ω C B
6 RA
C B
25 Ω The required resistance between vertices A and B is given by
Fig. 4 A
RAB = {(RB ||5) + (RA ||25)} ||(RC ||10)
= {(14.8||5) + (18.5||25)}||(12.33||10)
10

= (3.74 + 10.63)||(5.52)
5Ω

= 14.37||5.52 RB RC
RA
= 3.99 Ω
C B
25 Ω
Delta-Wye Connection Problem
Identical Issues

Q. For the following circuit in Fig. 4, find the equivalent resistance between (i) A and
B; (ii) A and N:
Soln . (i)
A This time for the equivalent resistance, resistors connected to ver-
A
tex C have been used to obtain a delta connection as
5 × 6 + 6 × 25 + 25 × 5
RA = = 12.20 Ω
4Ω 25 5Ω
RA
RB
10
5Ω

5 × 6 + 6 × 25 + 25 × 5
Ω
RB = = 50.83 Ω
6 RC
6Ω N
5 × 6 + 6 × 25 + 25 × 5
5Ω RC = = 61.00 Ω
6Ω N 5 C B
25 Ω
C B The required resistance between vertices A and B is given by
A
25 Ω
RAB = {(RA ||4) + (RC ||10)}||(RB ||5)
Fig. 4
= {(12.2||4) + (61||5)}||(50.83||10) RB
RA 4Ω

10
= (3.09 + 4.62)||(8.36)

Ω
= 7.71||8.36 5Ω
Verification of earlier result! N
= 4.01 Ω
B
RC

Delta-Wye Connection Problem

Q. For the following circuit in Fig. 4, find the equivalent resistance between (i) A and
B; (ii) A and N:
Soln . (ii)
A Star connection at the vertex C have been converted into a delta
A
connection as
5 × 6 + 6 × 25 + 25 × 5
RA = = 12.20 Ω
4Ω 25 5Ω
RA
RB
10
5Ω

5 × 6 + 6 × 25 + 25 × 5
Ω

RB = = 50.83 Ω
6 RC
6Ω N
5 × 6 + 6 × 25 + 25 × 5
5Ω RC = = 61.00 Ω
6Ω N 5 C B
25 Ω
C B The required resistance between vertices A and N is given by
A
25 Ω
RAN = (RA ||4)||{(RB ||10) + (RC ||5)}
Fig. 4
= (12.2||4)||{(50.83||10) + (61||5)} RB
RA 4Ω
10

= (3.09)||(8.36 + 4.62)
Ω

= 3.09||12.98 5Ω
Similar to earlier solution! N
= 2.5 Ω
B
RC