A problem on equivalent resistance

We can get the equivalent resistance of a circuit by connecting an ideal source of tension and measuring the current that pass through the source.

By the simmetry of the problem we would get the distribution of currents shown in the (sorry for the quality)

By using Kirchoff Law of Tensions following the path shown below

and then going though the source and doing this path

we obtain the following equations:

$3i_i - 2(-i_1 + \frac{I}{2}) - (I - i_1) = 0$

$V_x = 3i_1 + 2\frac{I}{2} + I - i_1$

By solving the first one we get $i_1 = \frac{I}{3}$

Using this in the second we get $V_x = \frac{8}{3}I => \frac{V_x}{I} = \frac{8}{3}$

So the equivalent resistance is $R = \frac{8}{3}\Omega$

The problem can be readily solved using the wye-delta transformation. Let the top two unlabeled nodes be $C$ and $D$ from left to right and the bottom two, $E$ and $F$ . The left 1-2-3 $\Omega$ delta is transformed to $R_a = \frac 12 \ \Omega$ , $R_c = 1 \ \Omega$ , and $R_e = \frac 13 \ \Omega$ . Similarly, for the right delta, $R_b = \frac 12 \ \Omega$ , $R_d = 1 \ \Omega$ , and $R_f = \frac 13 \ \Omega$ .

The resultant resistance between $A$ and $B$ :

$\begin{aligned} R_{ab} & = R_a + (R_c + 2 + R_d) || (R_e + 2 + R_f) + R_b \\ & = \frac 12 + \left(\frac {10}3 || \frac {10}3 \right) + \frac 12 \\ & = \frac 12 + \frac 53 + \frac 12 \\ & = \frac 83 \approx \boxed{2.7} \ \Omega \end{aligned}$

The 1, 2 and 3 ohm resistors on both sides are delta connected. Convert it to wye connection. R13 = (1)(3)/(1+2+3) R12 = (1)(2)/(1+2+3) R23 = (2)(3)/(1+2+3)

Then. R12 from side A, R23 from side B and the 2 ohm resistor in the middle is series. And same as the resistances below. Since they are connected in parallel and their resistances are the same, divide it by 2 to get its resistance equivalent.

Finally, those remained are series, 0.5 + 1.66666666667 + 0.5 = 2.66666667 or 2.7 ohms.

i found other solutions so tedious so i'll upload an easy one, connect a 100 volt battery(for ease of calculations)

assume potientals to be x and y and then using symmetry for potential drop solve for x and y use ohm's law and the calculate

notice that the end meshes look like DELTA configurations...hence they can be converted to the corresponding STAR configurations and then simplify and get the answer as 2.6667 ohm.

$$ The electrical circuit diagram in this problem can be transformed by using $\text{Y-}\Delta$ transformation technique . After transformation, we can calculate the equivalent resistance between $\text{A}$ and $\text{B}$ by using basic series and parallel resistors configuration.

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$\text{\# }\mathbb{Q.E.D.}\text{ \#}$