Equivalent resistance

The diagram may be redrawn as below: The resistance of two resistors R wired in parallel is $\frac{1}{2}$ R .

So the resistance of the lower half of the circuit is R + $\frac{1}{2}$ R = $\frac{3}{2}$ R

The resistance between A and B is then given by:

$\frac{1}{\frac{1}{R} + \frac{2}{3R}}$ = $\frac{3}{5}$ R

Suppose each resistor was $1 \Omega$ and we applied a $1 V$ source from $A$ to $B$ . We immediately know almost all of the voltages in the circuit, as shown.

We don't know the voltage at $C$ . Use Kirchoff's Current Law to find this voltage.

$\frac{V_C-1}{1} + \frac{V_C-0}{1} + \frac{V_C-1}{1} = 0 \\ 3 V_C = 2 \\ V_C = \frac{2}{3}$

Find the current flowing into $A$ or out of $B$ . It's easy to calculate the current out of $B$ .

$I_{out} = \frac{\frac{2}{3} - 0}{1} + \frac{1 - 0}{1} = \frac{5}{3}$

The total resistance is therefore:

$R_{total} = \frac{V_{source}}{I_{out}} = \frac{1}{\frac{5}{3}} = \boxed{\frac{3}{5}}$

Of course, this would scale the same way regardless of the value of each resistance.

Going by first principles - Consider the circuit as given in the image. Total current in the circuit is (i1 + i2) getting distributed in the two arms. Further, i3 flows through the central arm. The potential drop in the loops 1 and 2 will be zero. The potential drop from A to B will be total current (i1 +i2)*effective resistance (3/5R). Hope this helps.

Process of elimination.

$\frac{R}{2}$ would require an equivalent circuit so that there are two equivalent resistances in parallel $R$ between A and B. Since $A$ is shunted across $R$ to $B$ there is no equivalent resistance network which will allow the Y network of $R$ resistors to also be $R$ , so it can be eliminated. It's worth pointing out that another resistor $R$ could be added to the circuit to get that result, but not with those 3 configured they way they are.

$\frac{3R}{5}$ we skip.

$R$ isn't a reasonable solution because there's for sure a divider between the $A \rightarrow R \rightarrow B$ path, so the entire resistance network it must be less than $R$ .

$4R$ isn't a reasonable solution for the same reason. We know there's a divider so it must be less than $R$ .

A quick spot check just says that the resistance can't be $\geq R$ and $\frac{R}{2}$ seems equally as unlikely.

The diagram is redrawn by giving letters. In new diagram, it is easier to see series-parallel resitances. The result is 3/5 R.

I see it as finding the number of shortest routes the current can take, adding up the resistances together for each path, take the inverse of it, then sum of all the inverse values, before inversing the sum of inverse values.

the solution is 3R/5 the point is the same as D , so the resistor AC and CD are parallel , Req1 = R //R , the resistor B is serial with Req1 so Req2 = Req1 + R = R//R + R , finally the resistor CB is parallel with Req2 and as a result : Req = ((R//R)+R)//R