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## Homework Statement

Based on Fig. 4,

Z

_{Δ}=6+8j Ω

Z

_{Y}=4+3j Ω

E

_{L}=200V

Show that the equivalent delta-connected load is 7.846 - 2.77j

## Homework Equations

Z

_{Y}=(1/3)Z

_{Δ}

## The Attempt at a Solution

Here's my thought process:

For part (i), attempt to convert Y to Δ,

So using the relevant equation above

Z

_{Δ}of transformed Z

_{Y}is 12 + 9i

However, that's not done because I need to combine the two delta loads.

Hence, I use the product/sum rule to obtain the combined delta:

((12+9i)(6+8j))/((12+9i)+(6+8j))

which obtained 4.1598 + 4.405j

__which is wrong__

I also used https://www.physicsforums.com/threads/delta-and-star-transformation-ac-circuits.791979/ as guidance for this problem but can't quite point out where my mistake is.