Is My Delta to Wye Conversion Calculation Correct?

AI Thread Summary
The discussion focuses on verifying a Delta to Wye conversion calculation for resistances in a circuit. The original calculations for resistances yielded values of 0.5Ω, 1Ω, and 1Ω, but there were errors in applying these values to the circuit components. It was clarified that the resistance connected to point A should be 0.5Ω, not 1Ω as previously stated. Additionally, a simpler method was suggested, highlighting that the 4Ω resistor could be disregarded due to symmetry, leading to a final equivalent resistance of 2.5Ω. Proper labeling and understanding of the circuit components are crucial for accurate calculations.
Istiak
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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.
 
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Istiakshovon said:
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.)
 
haruspex said:
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.)
1620572803653.png

1620572817775.png


I don't know if I am wrong.
 
Istiakshovon said:
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## .
 
haruspex said:
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##
 
Istiakshovon said:
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.
 
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