Is My Delta to Wye Conversion Calculation Correct?

[Istiak]

Homework Statement

https://i.stack.imgur.com/dbNPL.png
Find resistance of the image

Relevant Equations

$$R_A=\frac{R_{AB}R_{AC}}{R_{AB}+R_{AC}+R_{BC}}$$

I was following these [steps](https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/delta-y-and-y-conversions/).

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.

 

[haruspex]

Homework Statement:: https://i.stack.imgur.com/dbNPL.png
Find resistance of the image
Relevant Equations:: $$R_A=\frac{R_{AB}R_{AC}}{R_{AB}+R_{AC}+R_{BC}}$$

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

Looks like you applied the three resistances you calculated for the Y to the wrong components of it.
Draw the circuit after the conversion and post it to the thread.

(Btw, there is a much easier way to answer this question.)

 

[Istiak]

Looks like you applied the three resistances you calculated for the Y to the wrong components of it.
Draw the circuit after the conversion and post it to the thread.

(Btw, there is a much easier way to answer this question.)

I don't know if I am wrong.

 

[haruspex]

View attachment 282800
View attachment 282801

I don't know if I am wrong.

You are wrong.
Suppose we label the points in the left half of the given diagram:
A is on the left
B is above the 4 Ohm
C is below the 4 Ohm.
Then in the equation
$$R_A=\frac{R_{AB}R_{AC}}{R_{AB}+R_{AC}+R_{BC}}$$
$R_{AB}$ refers to the 2 Ohm at top left and $R_{BC}$ is the 4 Ohm.
This gives $R_A=0.5\Omega$, where $R_A$ is the resistance connected to point A.
In post #3 you have $R_A=1\Omega$ .

 

[Istiak]

You are wrong.
Suppose we label the points in the left half of the given diagram:
A is on the left
B is above the 4 Ohm
C is below the 4 Ohm.
Then in the equation
$$R_A=\frac{R_{AB}R_{AC}}{R_{AB}+R_{AC}+R_{BC}}$$
$R_{AB}$ refers to the 2 Ohm at top left and $R_{BC}$ is the 4 Ohm.
This gives $R_A=0.5\Omega$, where $R_A$ is the resistance connected to point A.
In post #3 you have $R_A=1\Omega$ .

So, the answer is $2.5 \Omega$

 

[haruspex]

So, the answer is $2.5 \Omega$

Yes.
The simple way is to observe the symmetry. There cannot be any current in the $4\Omega$ resistor, so you can throw it away.