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Linear Algebra

Linear Equations

Application: Kirchhoff’s Laws and Circuits

                       

Throughout this chapter, we’ve seen how linear algebra can be used to tackle systems of linear equations. This is important because of the various contexts in which systems of linear equations appear: even besides areas of math, several areas of physics rely on our ability to solve these systems.

One important area is circuits, which measures how electricity flows throughout various devices. Mathematically analyzing these circuits is extremely important for evaluating their efficacy: too little power into a device and the device won’t work; too much, and the device will overload. As we’ll see throughout this quiz, linear algebra provides the toolbox for analyzing circuits.

Let’s first define some of the basic modules present in circuits. In order to get to the actual math part, we won’t go through all of them, but we’ll touch on the important ones. Let’s start our exploration by looking at voltage (rather than current just yet).

First, the basic idea is that current travels through wires, in such a way that the voltage of each contiguous segment of wire (measured in volts, or V), known as a node, is the same. For example, the following circuit has two nodes, each with their own voltage:

Here, the boxes represent non-wire modules that we’ll look at shortly

Here, the boxes represent non-wire modules that we’ll look at shortly

How many nodes are in the following circuit?

The main purpose of analyzing circuits is to determine which nodes have which voltages, which in turn tells us how the voltage changes across modules, and thus tells us how much voltage the module uses.

The first of these modules is a simple battery, or voltage source. A battery with voltage \(v\) enforces the difference between the positive node (the node connected to the positive part of the battery) and the negative node (the node connected to the negative part of the battery) is \(v\). For example,

Here, the difference between the voltages at the green node and the red node is 10 volts.

Here, the difference between the voltages at the green node and the red node is 10 volts.

The circuit below contains three batteries, with voltages \(a, b\), and \(c\). Which of the following must be true?

Before we look at our next module, we have to introduce the idea of current (measured in amperes, or A). Current is the flow of electricity through wires, which can in turn power modules such as lights. As such, current travels over modules as well as through wires.

An important property of current is Kirchoff’s law: the amount of current going into a node is the same as the amount of current going out of a node. For instance, one possible circuit could be the following:

We are now ready to introduce the second major module we’ll be working with: a resistor. Resistors relate voltage and current via Ohm’s law, which states the following: if \(v\) is the difference in voltage across the nodes connecting to the resistor, \(i\) is the current flowing across the resistor, and \(R\) is the strength of the resistor \((\)measured in ohms, or \(\Omega),\) we have \[v = iR.\] The following circuit contains a battery and two resistors of strengths \(3 \Omega\) and \(2 \Omega,\) respectively. What is the value of the current travelling across the \(2 \Omega\) resistor?

The last few problems illustrate a remarkable fact: using just batteries and resistors, we can induce a current across a module, which allows the module to use that current for power. This is why circuits are so important!

Unfortunately, circuits are usually much more complicated than the simple examples we’ve seen here. In the next couple problems, we’ll work through a much more complex circuit--and see how linear algebra ties into the analysis.

Over the next few problems, we’ll analyze the following complex circuit:

Here, a lightbulb is modeled as a \(50 \Omega\) resistor. Our goal will be to find the current travelling across the bulb (remember, if the current is too little, the bulb won’t turn on; if the current is too much, the bulb will burn out).

First of all, how many nodes are in this circuit?

From the previous problem, we see there are 4 nodes. We can assume for simplicity that the pink node has \(0\) V voltage. This means there are three nodes we don’t know the voltage of, and thus there are 3 variables we will need to solve for. As we’ve seen previously, this means we need to find 3 equations linking them.

Suppose the red node has voltage \(v_1\) V, the green node has voltage \(v_2\) V, and the blue node has voltage \(v_3\) V. Analyzing the red node first, which of the following is true?

Now let’s look at a more complicated equation, by analyzing the green node. Recall that Ohm’s law allows us to write the currents in terms of \(v_1, v_2\), and \(v_3\). For instance, the current across the \(50 \Omega\) resistor (from the green node to the red node) is \(\frac{v_2 - v_1}{50}\) (make sure you see why!).

Finding the three currents and applying Kirchoff’s law to the green node, which of the following equations must be true?

In the last problem, we saw that applying Ohm’s and Kirchoff’s laws to the green node gives \[\frac{v_2 - v_1}{50} + \frac{v_2 - 0}{300} + \frac{v_2 - v_3}{100} = 0,\] which can be rewritten as \[-6v_1 + 10v_2 - 3v_3 = 0.\] Applying the same process to the blue node, which of the following equations is true?

We’ve now reduced to a system of three equations: \[\begin{pmatrix}1&0&0\\-6&10&-3\\0&-1&3\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix} = \begin{pmatrix}5\\0\\0\end{pmatrix}.\] Now we can finally find the current traveling across the bulb. What is this current?

In this chapter, we explored the mathematical analysis of circuits, which is integral to correctly setting up electrical systems. As we saw, using just the simple modules of batteries and resistors, we are able to induce a current that can power a lightbulb (or many other things, like the computer you are reading this on!).

As we saw, simple circuits can be analyzed “by hand,” but more complex circuits require a chunk of linear algebra to fully understand. Fortunately, the tools we’ve developed throughout this chapter reduce the problem to a systematic process, from which we can understand precisely what setup is necessary to--for instance--light a bulb.

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