Series, Parallel, Kirchhoff

A network of resistors behaves as a single resistor: the current through the network is proportional to the potential difference applied. Calculating the equivalent resistance of a network starts for many at high school but may come back later with more complex cases also at university. For success in science and engineering it is crucial to be skilled in these calculations but also to develop an intuition on how to make quick approximate solutions. Being an expert in handling electrical resistance networks will come into great use for other types of networks that have similar behaviour: magnetic circuits, fluid dynamics, heat conduction, traffic flow, just to name a few.

To hear is to forget, to see is to remember, and to do is to understand, so let’s practice resistor networks hands-on!

This tutorial is great for the class-room, since it can be performed by students individually or in small groups: the equipment is cheap and common (a multimeter and some crocodile clips) and the consumables are very cheap (7 resistors per experiment - they cost less than a cent each when bought in 100-packs) - The students can keep the final resistor network as a souvenir.

There are two levels: for high-school students do all the series and parallel exercises, and skip the final assignment that requires Kirchhoff law’s. For bachelor students, the series and parallel networks provide a nice warm-up to the real work, which is to solve a multi-loop network with Kirchhoff’s laws.

• A multimeter - sold at hardware stores for as little as 5EUR
• It’s helpful to have crocodile clips at the end of the measuring leads - I did this by replacing the multimeter cables with banana cables and attaching crocodile clips. An alternative is to attach crocodile leads to the end of the multimeter leads.
• 2 more Crocodile leads - sold online for as little as 1 or 2 EUR for a pack of 10
• 7 identical resistors. 100Ohms or 1kOhm are fine. Common 0.25W 1% types will do. We’ll call the resistance value ‘R’ from now on, to get into the right habit of doing math with symbols.

Put the multimeter in the resistance setting, with the maximum just above the value of R. So set it to ‘200’ if R=100Ohm. Check that all 7 resistors are good!

How does a multimeter measure resistance? Internally, it has a current source, which will force a very small current through anything connected to the leads. The potential difference that results from it is proportional to the resistance: V=iR. Yes, that’s called Ohm’s law. The value displayed on the multimeter is simply the voltage, divided by the value of the current.

In electronic equipment, components are connected by soldering them to a printed circuit board (PCB). Here we use a simpler method: we wind the leads around each other. Cross the leads of two resistors and twist them around each other. Keep twisting until there are at least 10 turns and they are tightly connected. This is a quick way to make a connection that is both mechanically electrically strong.

This is our first network: A current that passes through both resistors will incur a voltage drop in both resistors. For the same current, the voltage will be the sum of the voltage on both resistors, and the equivalent resistance is the sum of all resistance values in series: Req=R1+R2. Since both resistors are equal we expect 2R as the equivalent resistance.

Go ahead and measure the resistance of the two resistors in series. Is the value that you find consistent with the calculation?

Repeat the previous step until you have a string of 6 resistors. What is the total resistance from one end to the other? Right, 6R. By clipping the multimeter leads at different positions, check that you can get all values of 1R, 2R, 3R, 4R, 5R and 6R.

Perform the final connection so that the 6 resistors form a ring. The multimeter leads can now be connected to 6 different positions, let’s call them a, b, c, d, e and f.

What resistance do you expect when measuring the resistance between a and d ? We now have two branches in parallel, each with a resistance of 3R. The equivalent resistance of resistors in parallel is calculated as the sum of inverses: 1/Req=1/R1+1/R2+1/R3+... . In case of two resistors, this simplifies to the product divided by the sum: Req=(R1*R2)/(R1+R2). And in case of n identical resistors, the parallel resistance is simply R/n. So we expect 3R/2=1,5R. Is that what you measure?

Of course, the multimeter leads can be connected at other points too: what about the resistance between point a and c? We then have one branch with a resistance of 4R and one branch with a resistance of 2R. So in parallel, this gives Req=(4R*2R)/(4R+2R)=(4/3)R or 1,33R. And between a and b? Req=(5R*R)/(5R+R)=(⅚)R=0,83R

Lets go back to the resistance between a and d. The resistance was 1,5R. We will make an additional connection with the crocodile lead between c and f. Before measuring or calculating, argue what you expect: will the resistance remain at 1,5R, or will it get higher or lower? Think of the equivalents: in a network of tubes, you add an additional zero-resistance tube. Or in a city with narrow streets, you open a new wide road, will the traffic increase or decrease? Correct, the extra 0-Ohm connection will make it easier for the current to flow, thus the resistance of the network can only go down.

It may not be immediately obvious, but the network can be redrawn such that it is more obvious that we have now two subcircuits in series: each subcircuit contains an R and a 2R in parallel. So we get Req=2*(R*2R)/(R+2R)=(4/3)R=1,33R. Indeed, the calculations confirm our intuition, the short-cut has reduced the resistance from 1,5R to 1,33R. If all is connected well, the measurement should confirm this

And now for the genius of the class: Let’s replace the shortcut between point c and f by a resistor between point c and f. We use one side of each crocodile clip to clamp the resistor in position, or, as in he picture, we wrap the leads of the resistor around point c and point f. It is no longer possible to calculate the equivalent resistance simply by finding combinations of series and parallel resistors! What then?

From the previous calculations and measurements we know the range in which the resistance must be: the present circuit cannot have a resistance more than 1,5R, because that’s what we get without the resistance between c and f. For the same reason it cannot have a resistance less than 1,33R, because that’s what we get when we short between c and f. So we’re somewhere in between 1,33R and 1,5R. Let’s measure: 1,4R, in range with expectation, but how would you calculate it?

Kirchhoff’s laws come to rescue: the first says that for every node the sum of currents must be zero and the second says that for every loop the sum of voltage differences must be zero. So let’s give names to the currents, make sure they add up to zero in all nodes and evaluate the voltages over some loops.

First of all, to calculate the behaviour of the circuit, we see what happens when we apply a potential difference V across it. There will be a current and we determine Req=V/i.

In point a, the current splits into two branches, we indicate these currents as i1 in the branch with resistance 2R, and i2 for the branch with resistance R. The network is perfectly symmetric, so also on the right side, the current through the 2R branch is i1 and the current through the R branch i2. In the central branch we’ll have i2-i1.

Now we apply Kirchhoff’s laws on some loops. Let’s start with the big loop which encompasses both the battery and one branch of resistors. We find V-i1*2R-i2*R=0.

Now we take a small loop that includes no battery but does include the central branch: -i2*R-(i2-i1)R+i1*2R=0. This equation gives us the ratio of i1 and i2: i2=(3/2)i1. We substitute it in the first equation and get V=3,5Ri1, or i1=(2/7)V/R. Then, i2=(3/2)i1=(3/7)V/R and i=i1+i2=(5/7)V/R. As the last step we find Req=V/i=(7/5)R=1.4R. Exactly like the measurement!

This exercise teaches the students a lot of skills in a single lesson, with hands-on practice and with a lot of fun: handling resistors, connecting cables, using a multimeter, making electrical connections, performing measurements. The abstract academic methods of calculations come to life in the student’s hands. The students are confronted with a difficult problem for which it turns out a solution does exist, although it requires non-trivial math: finding the solution to a linear system of two equations with two unknowns.

And, if you are ready for more, check out NerdSnipe's fun video about way more complex resistor networks: https://www.youtube.com/watch?v=nHUv80RsQns