Series and Parallel Networks of Resistors

Series and Parallel Networks of Resistors

Now that you know how to handle simple series and parallel circuits, you are ready to tackle circuits which combine these two setups such as the following circuit:

An example of a series-parallel network. The dashed boxes indicate parallel sections of the circuit.

It is relatively easy to work out these kind of circuits because you use everything you have already learnt about series and parallel circuits. The only difference is that you do it in stages. In the figure above, the circuit consists of 2 parallel portions that are then in series with a cell. To work out the equivalent resistance for the circuit, you start by calculating the total resistance of each of the parallel portions and then add up these resistances in series. If all the resistors in the figure above had resistances of \(\text{10}\) \(\text{Ω}\), we can calculate the equivalent resistance of the entire circuit.

We start by calculating the total resistance of Parallel Circuit 1.

The value of \(R_{p1}\) is:\begin{align*}\frac{1}{R_{p1}} &= \frac{1}{R_1} + \frac{1}{R_2} \\R_{p1}&= \left(\frac{1}{10} + \frac{1}{10} \right)^{-1} \\&= \left(\frac{1+1}{10} \right)^{-1} \\&= \left(\frac{2}{10} \right)^{-1} \\&= \text{5} \, \Omega\end{align*}

We can similarly calculate the total resistance of Parallel Circuit 2:\begin{align*}\frac{1}{R_{p2}} &= \frac{1}{R_3} + \frac{1}{R_4} \\R_{p2}&= \left(\frac{1}{10} + \frac{1}{10} \right)^{-1} \\&= \left(\frac{1+1}{10} \right)^{-1} \\&= \left(\frac{2}{10} \right)^{-1} \\&= \text{5} \, \Omega\end{align*}

You can now treat the circuit like a simple series circuit as follows:

Therefore the equivalent resistance is:\begin{align*}R &= R_{p1} + R_{p2} \\&= 5 + 5 \\&= 10 \, \Omega\end{align*}

The equivalent resistance of the circuit in the series-parallel network above is \(\text{10}\) \(\text{Ω}\).