Electrical Circuit Analysis

Introduction

In this post, we will use Kirchhoff’s Voltage and Current Laws (KVL and KCL) to solve a circuit. If you are not familiar with KVL and KCL, please read the previous posts, because we are just going to jump right into it!.

Kirchhoff’s Voltage Law
Kirchhoff’s Current Law

Also, we’ll be using a TI-84 Plus calculator in our lesson. If you don’t own one, I have a link to a free online version. It is a little touchy on the buttons, but it works. The screenshots of the calculator functions were taken from this site.

TI-84 Plus Online Calculator

Example

This example shows a circuit with a 24V battery source, four resistors, and three different loops. We need to find both the voltage drop across all resistors and the current through all resistors, using KVL and KCL. I will work the problem below with the final answers at the end. I encourage you to solve the problem on your own and then check your work.

Setup

In this problem, we have 8 unknown values. We need to find four different currents and four voltage drops in the circuit. Let’s start with finding our equations.

KVL Equations:

In the example there are three different loops, so that results in three different equations. We have four variables to find and in order to find four variables, we need four different equations. You always need the same number of equations as variables to solve the system of equations.

Since we need one more equation to fill out our system, let’s look at KCL. With KCL, we can create an equation at any of the four nodes shown on the diagram. For simplicity, we will use Node 1 (N₁).

KCL Equation:

Systems of Equations

A system of equations is a group of equations that all have the same variables to solve. These equations are solved simultaneously to find solutions for the variables so that all equations are true. For simplicity sake, we are going to use a calculator to solve this system of equations, but we need to make sure that our input values are correct.

Matrix Setup

In order to solve the system of equations, we need to set everything up in a matrix. We will have all the variables with a current (I) associated with them, to be on the left side, and all variables with a voltage (V) to be on the far right side. With four variables to solve, our matrix will be a 4×5.

It is also important to keep the variables in the same order. Since we have variables I₁ through I₄, we will list them in numerical order from left to right. Some of the equations don’t have all the variables, but we still need to account for them in the matrix, so we will put a zero in their place.

Below are the four equations rearranged so that everything is accounted for and in order.

Once we have the equations in order, we can put the values into a matrix. When putting the values into the matrix, we only put the coefficients (the numbers in front).

One you have your values in the matrix, you can use your calculator’s RREF (Reduced Row Echelon Form) function under matrix and math.

TI-84 Matrix RREF

Below are the steps to doing the matrix and rref functions on a TI-84 Plus calculator.

Step 1: Enter the Matrix

Step 2: Select RREF

Once you select rref(), you’ll have to go back into the matrix menu and select the matrix you want to use. In our case it was matrix [A].

Press ENTER to solve.

Results

The answers are in column 5 (far right). When value they correspond to is shown by the number 1 in the matrix to the left. A 1 in the first column indicates that it is the value for I₁.

We will round the values to three decimal places.

I₁ = 5.992 A
I₂ = 1.247 A
I₃ = 1.558 A
I₄ = 3.117 A

Now let’s solve for the voltage drop across the resistors.

Self Check

Do our answers make sense? We need to remember that we have some rounding error, since the numbers didn’t come out even (more like real life). We can check if each of the loops come out to a zero voltage (minus the rounding errors).

Check that the voltage drop across the resistors that are in parallel (R₂-R₄) are the same. Another check you can do is for the total current in the circuit. You’ll have to simplify the circuit to get the equivalent resistance and then solve for the current. That current should be equal to I₁.

Thank You

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