# Delta to Wye conversion

I have a circuit like this. And, I want to calculate resistance from here. I was following these steps.

I was calculating resistance for left side circuit.

$$R_1=\frac{2 × 2}{2+2+4}=0.5\Omega$$ $$R_2=\frac{2 × 4}{2+2+4}=1\Omega$$ $$R_3=\frac{2 × 4}{2+2+4}=1\Omega$$

Then, I took series and parallel circuit. Then, calculated equivalent of resistance.

$$R_s1=0.5+3=3.5$$ $$R_s2=1+3=4$$ $$R_p=(\frac{1}{3.5}+\frac{1}{4})^-1=1.8677$$ $$Requivalent=1.8677+1=2.8667$$

I am not sure if it is correct. If it is wrong than how can I proceed?

## 1 answer

No, the answer can be seen by inspection in a few seconds, and it's not 2.8667. We don't just give answers to homework problems here, so I'll only make a few comments on the problem and your attempted solution:

- While this is on topic here, you would probably get better response on the Electrical Engineering site.
- This question should be in the
**Problems**category, not here in the**Q&A**category. There is no excuse, since you were clearly told that when you explicitly asked about it here. - You never actually stated what the problem is. It seems you want to find the equivalent resistance between the two connections to the circuit? This should be stated explicitly.
- Always show component designators. This may be less of an offense here, but over on EE this is not tolerated. Copying the circuit from someplace else is no excuse.
*You*are responsible for what*you*post here. Add component designators or redraw the circuit altogether. It doesn't matter how you achieve this, only that you do. - It should be immediately obvious that 2.8667 can't possibly be right when the problem is asking for a resistance, since resistance isn't dimensionless. Units matter. If I was marking this, you'd lose points for sloppiness with units.
- Not quite as bad as above, but in most cases there should be a space between the value and its unit. For example "0.5 Ω" is correct, "0.5Ω" is not. NIST has a good publication on all this stuff.
- Instead of mechanically following some recipe, stop and actually think about this circuit. In particular, note the symmetry between the top and bottom legs. That allows for some simplification.
- If you applied a known voltage across this circuit and the 4 Ω resistor wasn't there, what would be the voltage between the two resistors of each leg? What current flows thru the 4 Ω resistor?

In my other post I had added main question. I forgot to add it here.

So fix it. Trying to excuse it instead of fixing it is a waste of everyone's time. We don't care why, only what is.

In MathJax it is not possible to give spaces between any units and numbers

I find that hard to believe. I don't know much about MathJax, but surely there is a way to insert a hard space.

What current flows thru the 4 Ω resistor?I don't have the value.

The point was to make you think about the circuit, not the actual numbers.

Try this exercise: Consider what happens when the left end of this circuit is held at 0 V and the right end at 5 V. If the 4 Ω resistor weren't there, what voltage results at each of the two junctions between the resistors? What voltage is therefore across where the 4 Ω resistor would be? You can convert this to a Thevenin voltage source. What current would flow thru the 4 Ω resistor if it were then connected as the circuit is actually shown?

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