# Using Y to Delta Transformation to Find Currents

## Homework Statement:

Use a Y-to-Δ transformation to find the unknown quantities for the circuit.
Find:
1) io
2) i1
3) i2
4) power delivered by ideal voltage source

## Relevant Equations:

Shown below are the equations for delta to y transformations and vice verse. Also the current divider equation and Ohm's law.  Can someone explain why I can't simply use a current divider once I've found the equivalent resistance and source current for the entire circuit? This would look like i0 = 0.044*(113.53/210). Req = 113.53.

If it helps, the correct answers appear to be: i0 = 8.28 mA, i1 = 23.6 mA, i2 = 35.8 mA, P = 0.220 W.

## Answers and Replies

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vela
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The logic is: if you have a current divider circuit, then I = I_total (R/Req).

You don't have a current divider, so the formula doesn't apply.

You might try finding the Norton equivalent of the circuit connected to the 210-ohm resistor. You would then have a current divider. You'll find I_nort is not equal to the source current you calculated, and Rth is nowhere near Req.

The logic is: if you have a current divider circuit, then I = I_total (R/Req).

You don't have a current divider, so the formula doesn't apply.

You might try finding the Norton equivalent of the circuit connected to the 210-ohm resistor. You would then have a current divider. You'll find I_nort is not equal to the source current you calculated, and Rth is nowhere near Req.
Why don't I have a current divider for 210 ohm resistor? It is, after all, in parallel. And parallel circuits divide current up?

vela
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In parallel with what?

In parallel with what?
40 and 20

vela
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It doesn't really make sense to say the 210-ohm resistor is in parallel with the 40-ohm and 20-ohm combination as those two resistors can't be combined into one element because of the 50-ohm resistor connected to the node in between them.

Anyway, the current divider rule applies to a specific configuration of resistors, and you simply don't have that here.

It doesn't really make sense to say the 210-ohm resistor is in parallel with the 40-ohm and 20-ohm combination as those two resistors can't be combined into one element because of the 50-ohm resistor connected to the node in between them.

Anyway, the current divider rule applies to a specific configuration of resistors, and you simply don't have that here.
Ok. I thought current divider works for parallel configurations because parallel configurations divide up current. Would a current divider then work for something more along these lines? 