Equivalent Single Load for Complex Delta and Y Connected Loads

  • Context: Engineering 
  • Thread starter Thread starter billyray
  • Start date Start date
  • Tags Tags
    Circuit Diagram
Click For Summary
SUMMARY

The discussion focuses on calculating the equivalent single Δ-connected load from complex delta and Y-connected loads. Participants emphasize the necessity of transforming Y loads to Δ and vice versa to achieve accurate load summation. The solution involves transforming the Y load to a Δ, summing the two parallel Δs into a single Δ load, and then converting that result back to a Y load. The conclusion is that the approach does not affect the final result, as both methods yield the same equivalent load.

PREREQUISITES
  • Understanding of complex load calculations in electrical circuits
  • Knowledge of Δ (Delta) and Y (Wye) load transformations
  • Familiarity with impedance and power calculations in AC circuits
  • Basic principles of parallel and series circuit analysis
NEXT STEPS
  • Study the process of transforming Y loads to Δ loads in detail
  • Learn about calculating equivalent loads in AC circuits
  • Explore complex impedance calculations for both Δ and Y configurations
  • Research real power dissipation in complex loads
USEFUL FOR

Electrical engineers, students studying circuit theory, and professionals involved in load analysis and power distribution will benefit from this discussion.

billyray
Messages
58
Reaction score
1

Homework Statement


a) the equivalent single Δ-connected load
I have uploaded circuit

Homework Equations


none yet

The Attempt at a Solution


I am not sure if the question is asking me to find the resistance of both delta and y connected loads then work out them both in parallel. After I have the answer find a delta load for a circuit that equals this or simply find the delta loads impedance. That seems a bit easy from what i have done on my last question.
The questions after the question number a make me think i should do it the first way i said but i am a bit unsure.
also question b says i need to transform delta to y in question a but i don't see why.
Any advice please
 

Attachments

Physics news on Phys.org
Note that the loads are complex.
Approach load summation from the Δ side.
3a. Transform the Y load to a Δ, then sum the two parallel Δs into a single Δ load.
3b. Transform that parallel sum Δ from 3a, into a single Y load.
Now approach load summation from the Y side.
3c. Transform the Δ load of fig 3, to a Y, then sum it in parallel with the existing Y load to make one Y load.
3d. Find the real power dissipated in the complex Δ load.
billyray said:
also question b says i need to transform delta to y in question a but i don't see why.
So you can see that it does not matter how you approach the result. The Yb solution should be the same as the Yc solution.
 
  • Like
Likes   Reactions: scottdave

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 43 ·
2
Replies
43
Views
17K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K