Somewhat embarrassingly, I got this wrong at first:

(Dammit, Jim, I’m a mathematician, not a physicist.)

Have a go at it yourself, if you’re so inclined; after the line come spoilers.

### Kirchhoff and Ohm

In the bad old days, I used to help out at a Physics homework centre – the work they were doing was largely mechanics, and it gave me a chance to develop some Ninja skills. For electronics problems like this, I ended up with a model that generally worked for me. I’ll share it with you, but I think this is the kind of problem where coming up with your own analogy is more powerful that borrowing someone else’s.

• Potential ($V$) is a value that belongs to the junctions between sections of wire. It’s high next to one end of the battery and low at the other. (I tend to think of a circuit a bit like a waterfall, with potential as the height above the ground. It doesn’t make the sums work, but ‘height’ has the same sort of effect as potential does.)
• Current ($I$) is a value that belongs to the bits of wire between junctions, and is a flow: what goes into a junction must equal what comes out. This is Kirchhoff’s law.
• Resistance ($R$) links the difference in potential between two junctions and the current flowing in the wire joining them, and satisfies $V=IR$. This is Ohm’s law.

These three things are enough to figure out the problem – and most of the resistance/current/potential problems that cross my path.

### This problem

We’re not given a value for the potential difference, so let’s suppose the left-hand junction of this system has a potential of $V$ and the right-hand junction $0$. Also let’s suppose that there’s a current of $I_0$ flowing in from the left end and out at the right end (what goes in must come out. Kirchhoff said that.)

There are two other junctions: let’s call the potential at the junction between the first two resistors $V_{12}$ and the potential at the junction between the second pair $V_{23}$.

However, the ends of two resistor-less wires must each be equal, so $V_{23}=V_1$ and $V_{12} = 0$. This makes everything much simpler!

The potential difference across each resistor is $V$, so the current across each resistor is $\frac{V}{R_i}$.

The total current arriving at the right-hand end is therefore $I_0 = V\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right)$.

Now all we need to do is find the equivalent resistor! We know $R_e = \frac{V}{I_0}$, and we just worked out $I_0$.

$\frac{V}{I_0} = \frac{1}{\frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3}} = \frac{R_1 R_2 R_3}{R_1 + R_2 + R_3}$.

### Where I went wrong

I made the mistake of assuming the current would only flow left to right – I think the nugget to this problem is realising that the current flows right to left through the middle resistor.

Now I’ve seen the error, it’s obvious, but I couldn’t see it for looking at first.

Did you do it the same way? I’d love to hear about your approaches.