- We start by looking for groups of resistors that are in series or in parallel. We see that the 1Ω and 2Ω resistors are in parallel, so we can replace them with a single resistor having the sum of their conductances: \(\def\ohm{\u{\Omega}} 1\mho+\frc2\mho={3\over2}\mho\). (In the figure I've written the conductance of both resistors underneath their resistances for reference.) Similarly, the \(\frc2\) and \(\frc4\) resistors are in parallel, so can be replaced with a resistor with conductance \(2\mho+4\mho=6\mho\).
- For complicated systems, it can be handy to redraw the system everytime you replace a set of resistors, as I've done in the next step. Here I see that the \({2\over3}\u{\Omega}\) and the \({1\over3}\u{\Omega}\) resistors are in series, so can be replaced with a single resistor with resistance \({2\over3}\ohm+{1\over3}\ohm=1\ohm\).
- Finally, the 1℧ resistor and the 6℧ resistor here are in parallel, so we can combine them into a \(1\mho+6\mho=7\mho\) resistor...
- ...and a 7℧ conductance corresponds to an equivalent resistance of \(\boxed{\frc7\ohm}\).

Two resistors are in series when the same current passes through both. If you can trace a path from one resistor to the other without passing through any junctions, then those resistors are in series. (Passing through batteries or other resistors is allowed.)

Two resistors are in parallel if they have the same potential drop across them. You can show that resistors A and B are in parallel if you can draw a path from one end of A to one end of B without passing through a battery or another resistor (junctions are okay), and you can draw a path between the opposite ends of A and B as well.

We solve this using Kirchhoff's rules. First we label all the possible currents, as shown. Solving for six unknowns is difficult, especially if you haven't had linear algebra, and there are many ways to do it. I'll show you a quick way to solve the equations, and then an even quicker way using symmetry.

We start with two junction rules: $$I=I_1+I_2 \quad\mathrm{ and }\quad I=I_3+I_4$$ which leads to the equation $$I_1+I_2=I_3+I_4 \quad (1)$$ Next we apply the loop rule around the outer square, giving us $$-2I_1-2I_3+2I_4+2I_2=0 \implies I_1+I_3=I_2+I_4 \quad (2)$$ Subtract equation 1 from equation 2 and we get $$I_3-I_2=I_2-I_3 \implies 2I_3=2I_2 \implies I_3=I_2$$ If \(I_2=I_3\), then we can subtract them from both sides of equation (1) to get $$I_1=I_4$$ Now let's find the loop rule for the lower-right triangle: $$-2I_5-2I_4+2I_3=0 \implies I_5=I_3-I_4 \quad (3)$$ But the junction in the upper-right corner gives us the equation $$I_1=I_3+I_5 \implies I_5=I_1-I_3 \quad (4)$$ Add (3) and (4) together and we get $$2I_5=I_1-I_4$$ But \(I_1=I_4\)! Thus \(I_5=0\): no current flows through that diagonal resistor. Knowing that, it's easy to see that \(I_1=I_3\) and \(I_2=I_4\). From the junction in the upper-left corner we have $$I=I_1+I_2=I_1+I_4=2I_1 \implies I_1=\frc2I$$ and so does \(I_4\), \(I_2\), and \(I_3\). According to the tally rule, the potential difference between the upper-left corner and the lower-right corner is $$-2I_2-2I_4=-2\left(\frc2I\right)-2\left(\frc2I\right)=-2I$$ and this is equal to \(-\Dl V\). Thus the equivalent resistance of this set of resistors is $$R_{eq}={\Dl V\over I}={2I\over I}=\boxed{2\ohm}$$

That was a lot of work! Instead, we could have noticed that this circuit has symmetry: if we grabbed the two terminals and twirled them around by 180°, all the resistors stay the same. If the current \(I_5\) points down and to the left, then rotating it will make it point up and to the right, which is a contradiction: you can't have all the same resistors, but have the current through the middle resistor point up and down at the same time. Thus \(I_5=0\). That leads us to say that \(I_2=I_4\), \(I_1=I_3\), and the current \(I\) must split evenly into \(I_1\) and \(I_2\) due to symmetry.