# How do I calculate the equivalent resistance of this kind of circuit?

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\$$

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}\$$?

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

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?

• @shineele - are we done with this one now? May 14 '20 at 9:15

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

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\$$.

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:

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}\$$ $$R_{eq} = \frac{R.\frac{8R}{3}}{R+\frac{8R}{3}} = \frac{8R}{11} = 16\Omega$$

• 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. May 12 '20 at 20:10
• @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. May 12 '20 at 21:16