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If I want to calculate the resistance between \$a\$ and \$b\$, that is, \$R_{ab}\$. How do I calculate the equivalent resistance of this kind of circuit? All resistors are \$22Ω\$

enter image description here

I know we we ignore the resistors between two point, take a and b for example, if we want to calculate the equivalent resistance between \$a\$ point and \$b\$ point, but in here, I am not sure about which resistors should I ignore first, R1~R5 resistor?

The answer is \$R_{ab}=10Ω\$, but I don't know how to calculate it! Can anyone teach me how to calculate the resistor between \$a\$ and \$b\$, that is, \$R_{ab}\$?

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As (nearly) always, try and simplify: -

Move from top left to top right by using two star to delta transformations to produce: -

  • Ra, Rb, Rc

and

  • Rd, Re, Rf

enter image description here

Bottom picture is the simplified combination of the the above right.

Can you take it from here by doing another star to delta transform on 3 more resistors?

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  • \$\begingroup\$ @shineele - are we done with this one now? \$\endgroup\$
    – Andy aka
    May 14 '20 at 9:15
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I think the solution would be to change the deltas to stars. So use the Star-Delta conversion. and after you should simplify it.

something like this

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There is an alternative much easier way, compared to star-delta, to solve for the resistors. (By the way, your answer is incorrect, the equivalent resistance is \$16\Omega.\$)
Since all the resistors are same symmetry arguments can be applied. By symmetry, the current going in branch \$ae\$ is expected to be same as in \$eb\$. Similarly, current in \$ac\$ is same as current going in \$db\$. This is shown in the figure below. This also implies the current in \$ao\$ is same as in \$ob\$. Moreover, the current through \$co\$ has to equal that through \$od\$.

enter image description here

Clearly, no current flows through \$eo\$ and from \$ao\$ to \$oc\$ or \$od\$. You can thus remove the resistor R3 and the connection at node o as shown below, where the 'x' means the connection is removed: enter image description here

Assuming all resistors are equal to R, there are now three parallel branches with resistor values: \$2R, 2R\$, \$2R+\frac{2R}{3} = \frac{8R}{3}\$ $$R_{eq} = \frac{R.\frac{8R}{3}}{R+\frac{8R}{3}} = \frac{8R}{11} = 16\Omega$$

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  • \$\begingroup\$ Please don't hand out solutions to homework problems. We expect the OP to do a significant amount of work themselves, and try to just give them hints so they learn how to solve problems on their own. \$\endgroup\$ May 12 '20 at 20:10
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    \$\begingroup\$ @ElliotAlderson He already knew the answer. Clearly, he was stuck. Anyways, it is a question and answer site and I don't care if OP is doing work himself or not. Also the question is not tagged as homework so I don't know how you concluded it is a homework. I just wanted to show him another approach which is less computationally intensive. \$\endgroup\$
    – sarthak
    May 12 '20 at 21:16

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