Here is a Mesh Analysis Practice Problem to practice basic mesh analysis and circuit analysis skills. This is a fairly basic problems without super-meshes. These are the types of problems you could find in the beginning of a Circuit Analysis 1 type of class. Typically around chapter 2 or 3 of your textbook.
There are several questions that could be asked for this problem.
Solve these questions using mesh analysis:
- What is the voltage across R2?
Continue reading to see a step by step solution on how to solve this question!
One big concept about mesh analysis is that it is based off of Kirchhoff’s voltage law. If you are unsure of what that is, I would strongly taking some time and understanding it.
Solutions using Mesh Analysis
This question could also easily be answered using nodal analysis but being that this is a mesh analysis practice problem we’ll use mesh analysis! To find the voltage drop across R2 we need to find the current through the resistor.
Step 1: Draw arrows to get a better visualization of what is going on. You’re always going to want the arrows to go in the same direction and the typical teaching convention is to draw them clockwise.
Step 2:Now lets create an equation for each of the loops. Lets start with the left/red loop. The arrow is pointing into the negative terminal of the voltage source so that would be a negative 3.3. Next we find the change in voltage across R1. Using ohms law we know that v=i*r. I1 is still unknown though so we have 1000(I1). Next is R2. Let first just worry about the effect of I1 on R2. Using the same technique that we just did we will get 100(I1). At the same time though we have I2 opposing I1 through R2. So we have to subtract the change in potential. We get -100(I2). When we combine those we get the following equation:
Step 2: Once the equation for the left loop is done we can move onto the right loop. We are going to use the same technique to get the second equation. Simply by following the direction of the arrow. Entering the negative terminal so -5. Voltage change across R3, 2700(I2). Voltage change across R2, 100(I2). There is I1 following in the opposite direction across R2, so we have -100(I1). When we combine those we get the following equation:
Step 3: We can take the two ‘cleaned up’ equations from step one and from step two and solve them simultaneously. Solving these two equations could be done by hand using algebra, using a matrix, or using a simultaneous equation solver on your calculator.I used an Ti Nspire CX CAS Calculator to solve both equations. I1 was found to be 3.173mA and I2 was found to be 1.899mA.
Step 4: Now that we have the value for both I1 and I2. We are going to take I1-I2. This is will give us the current through the resistor. 3.173-1.899 = 1.274mA. 1.274mA will be the net current through the resistor.
Step 5: Using ohms law we are able to finish up this mesh analysis practice problem. Using ohms law we know that v=I*R. We have the I and R so we can solve for v. 0.001274*100=0.1274v.
The Answer to this Mesh Analysis Practice Problem
So the final answer is 0.1274v
I hope this mesh analysis practice problem helped you understand the content better! If you have any questions feel free to leave a comment below. This mesh analysis practice problem was check for correctness using circuitlab.com
What else can help out with problems like these?
A textbook with more problems similar to these. I have done quite a few problems like these and I really feel it helped me understand this content. Several of my college classes used Fundamentals of Electric Circuits and it was a fairly good book for this sort of stuff!
A calculator that is able to solve simultaneous equations. A Texas Instruments TI-89 or Texas Instruments Nspire CX CAS could be very beneifical to quickly solving similar mesh analysis pratice problem and exam problems.