Using Y to Delta Transformation to Find Currents

rugerts

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
Homework 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.

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vela

Staff Emeritus
Homework Helper
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.

rugerts

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

Staff Emeritus
Homework Helper
In parallel with what?

rugerts

In parallel with what?
40 and 20

vela

Staff Emeritus
Homework Helper
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.

rugerts

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? "Using Y to Delta Transformation to Find Currents"

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