# Delta-to-Y, Y-to-Delta

1. Oct 16, 2007

### FilthyOtis

I'm really having troubles figuring this stuff out. Here's a question and what I've done..

a) Calculate current through R5

b) Determine Voltage across R5

I haven't gotten a) yet so haven't even tried b). I made a crude drawing of the original circuit in mspaint.

So... I did a delta-to-y on resisters 1, 2 and 5. Ra = R5, Rb = R2, Rc = R1

running the formulas

R1 = [(18)(12)] / [18 + 12 + 24] = 4 ohms

R2 = [(24)(12)] / 54 = 5.33 ohms

R3 = [(24)(18)] / 54 = 8 ohms

I redrew the picture with the 4ohms at the top splitting into the 5.33 on the left, 8 on the right and continuing down into the other 2 original resistors.

at this point I combined them all into one resistor by going
|| = parallel

4 ohms + [(5.33 ohms + 6 ohms) || (8 ohms + 6 ohms)] = 10.32 ohms

total current(I) = 15 V / 10.32 ohms = 1.46 A

and this is where my problem occurs... I don't know how to work backwards through this delta-to-Y transformation and properly find out how the current is splitting up. Can anyone help make sense of this? Thank you!

- Otis

#### Attached Files:

• ###### Circuit.GIF
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2. Oct 27, 2007

### lightgreen

Hi, the circuit drawn is known as brigde circuit. I'm not sure whether your working is wrong or correct. But I try to solve this question by the method below. It will be simpler. The V5 is 1.25v with I= 52.08mA.

First, i knew that V5= R3(15)/(R3+R1)-R4(15)/(18+6)
= 5-3.75
= 1.25v

Here a note attached here, and I hope it is useful for you.
Please correct me if I'm wrong.

#### Attached Files:

• ###### Bridge Circuit.pdf
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162.8 KB
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3. Oct 27, 2007

### meopemuk

I think the standard way to solve this problem is to use the Kirkhoff current law, i.e., the sum of all currents flowing in and out of the node is zero. Denote x and y the unknown voltages on the left and right node, respectively. Assume that currents flow down trough resistors R1, R2, R3, and R4, and that the current flows from left to right through resistor R5. (It doesn't matter if this guess is wrong: your calculated currents will be negative then.) Then zero current condition in the node x leads to the following equation

$$\frac{15-x}{12} -\frac{x}{6} - \frac{x-y}{24} = 0$$

The similar condition for the node y yields

$$\frac{15-y}{18} -\frac{y}{6} + \frac{x-y}{24} = 0$$

Now you have 2 equations with two unknowns. The rest is just math.

Eugene.