Series-Parallel Circuits

As you might have guessed, most real circuits are not just series or just parallel. They are large, complicated messes that involve some components in series, some in parallel, and others in even stranger connections. When solving for a voltage, current, resistance or power in a real circuit we almost always have to use the techniques developed so far to simplify these circuits by replacing components which are connected in series or parallel with equivalent components until the system is simple enough to analyse. No new information will be provided in this topic, however this topic will be a test of your ability to understand and apply everything we've covered up until now. Series parallel circuits are an exercise in creative problem solving and can be intimidating at first since there are no formulas to follow to get your answer. However, some practice and the ability to mentally step back and consider your options before diving in, and you'll master it just like students before you.

Series-parallel circuits are circuits which have resistors in both series and parallel configurations, often mixed together in different ways.

Series-parallel circuits do not need any new techniques, you simply apply simple series or parallel simplifications over several steps in order to get the final answer.

To tackle a series-parallel problem you simply look for two resistors which are either in series or parallel and you replace them with an equivalent resistor, repeat this process until the circuit looks like something you know how to solve.

Find the equivalent resistance of the circuit below.

So in this example we're just simplifying resistors until we get to one, pretty straight forward. We'll look through the circuit and try to find two resistors in series or parallel, combine them using the appropriate formula, then repeat until we're finished. So looking at the circuit we can see that no resistors are in series but \(R_2\) and \(R_3\) are in parallel so let's replace them: \(R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{500 \times 230}{500 + 230} \approx 157.53\Omega\) Now \(R_1\) and \(R_{23}\) are in series so we can combine them to get: \(R_{eq} = R_1 + R_{23} = 100 + 157.53 = 257.53\Omega\) So the equivalent resistance is: \(R_{eq} = 257.53\Omega\)

Find the voltage across resistor \(R_3\) in the circuit below.

As with most problems in series parallel circuits there are a few different ways we can tackle this. We know from our work on parallel circuits that the voltage across \(R_2\) is the same as the voltage across \(R_3\) and that this is the same as the voltage across the equivalent resistor \(R_{23}\) (found by combining \(R_2\) and \(R_3\) in parallel). So we're going to combine \(R_2\) and \(R_3\) into an equivalent resistor \(R_{23}\) and then use our voltage divider knowledge to find \(V_{23}\) which will be the same as \(V_2\). Got it? Good. \(R_{23} = 157.53\Omega\) so using our trusty voltage divider formula we know that \(V_{23} = 15 \frac{157.53}{257.53} \approx 9.18\)V And so \(V_2 = V_{23} \approx 9.18\)V