# Resistors-1

Once again, I've been thinking about the xkcd nerd sniping question. In preparation for relearning the maths required to solve the problem - here is the simplest possible instance I can think of that is even remotely related to the problem.

Solution

The circuit can be viewed as two sets of resistors in series, connected together in parallel. Each branch has two resistors in.

$$R_{series} = R_1 + R_2 + R_3 + \dots$$ $$\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$ the upper branch between A and B has two resistors in series, so has an effective resistance of $2\Omega$. Similarly, the lower branch has an effective resistance of $2\Omega$ Their combined resistance, as they are in parallel is given by $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{2} = 1$$ $R$ is therefore $1\Omega$.