# 3. Circuit Analysis

#### by George Sephton on October 3, 2014 at 1:40 pm | 0 comments.

## Removing Components

Sometimes the most effective way to analyse a circuit is to remove some of the components. Let’s take the example below: we have 2 branches from a current source which, with the resistors, creates a current divider circuit. However unlike our previous example, we have another current source on one of the branches.

Let’s apply Kirchoff’s Current Law at the highlighted node:

Since current does not change with voltage, we can say that and we know that therefore making the resistance equal to infinity (since ).

If the resistance is equal to infinity then we can remove the source without there being any change in the characteristics of the circuit – an empty space in the circuit, known an open circuit and will have a resistance of infinity. To give you an example of why this is, consider taking a wire of a circuit and cutting it in 2 and then holding them a centimetre apart. The resistance between those 2 wires is infinite because it will never be able to conduct voltage – unless of course you are dealing with extremely high currents, which is another story.

So let’s first turn off source A:

And then source B:

With these 2 circuits that have a source missing, it is easy to see where the current would flow. Using this current flow we can see that will always have current flowing though it and therefore we can say that:

When we have turned off the sources we are left with a single current source in each example and so *i _{a}* and

*i*will equal

_{b}*i*and

_{A}*i*respectively.

_{B}## Example

Let’s take the circuit above and use it as an example to illustrate this concept.

Let’s now turn off source A and source B and we can calculate the value of *i _{2}*

Since we have 10mA in the opposite direction we can write:

*i*pointing downwards, this was an arbitrary direction and had we decided to point it upwards, the equation still would have worked correctly.

_{2}## Circuit Analysis

So we’ve briefly looked at how we can remove parts of a circuit and create an open circuit in order to analyse it and also how we can remove components and create a closed circuit in order to analyse it. Using these 2 principles and Kirchoff’s Voltage and Current Laws we will be looking at 2 ways of analysing circuits. The first way is called Mesh analysis and although is not used much these days, it will help to understand the next method which is called Nodal Analysis. Afterwards we will be looking and how we can convert one type of circuit into another for easier analysis and we’ll be doing this using Norton and Thevenin’s Equivalent Circuits.

### Mesh Analysis

Sometimes analysing a circuit can be easy, for example:

*i*:

Although a lot of people argue that no one uses Mesh Analysis, there is no harm in learning it and it will definitely help you understand Nodal analysis.

Consider this circuit:

All the resistors have a resistance of 1kΩ

This circuit is quite difficult simplify and then analyse and so instead we use *Mesh Analysis*. Imagine we want to find the current through R_{3}. We first need to consider the 2 different loops – always draw these loops flowing from the positive end of the source to the negative end. Remember that if we drew the loop going the wrong way then we will simply end up calculating a negative current. We don’t have to draw it going from positive to negative, we can draw it any way we want, but in these tutorials we will always go from positive into negative.

So let’s add our current loops:

We label this as *i _{1}* and

*i*and now make 2 equations for each mesh (a mesh is the loop that we have just drawn). So in this example, our resistors are have the value of 1kΩ and each voltage source is 5V – usually the resistors aren’t always the same and neither are the sources but we’ll keep it simple for our example.

_{2}We should note that if the loop of the mesh goes into the positive side of the source then it needs to be written as negative in the equation. Since in our examples we always make the loops go into the negative side of the source then we don’t need to worry but should you choose to do it differently make sure you correct this or you will get some serious errors!

We now need to solve our equations and we will do this by writing ohm’s law in Matrix form. Although we don’t have a set of maths tutorials, we are going to assume that you know how to deal with matrices but if you don’t understand it then there’s a lot of information online and also a lot of matrix calculators that you can use – these aren’t complicated matrices but simple 2×2 matrix multiplication. For larger circuits with more meshes, there will be a large matrix to solve but we won’t be looking at these in this tutorial.

First we need to fill in the values of our sources and resistances:

We then need to write the matrix form of ohm’s law

(This is to be calculated)

We then must use the following syntax to write our resistances:

*i*= 0.125A

_{1}*i _{2}* = 0.125A

Now that we have found the total current in each mesh we can find the currents on each branch.

We have added our known currents and our only unknown is the current through R_{2}, which in this case is very simple to calculate:

### Resistance Matrices

As we saw above, the syntax for a resistance matrix when using two meshes is:

### Nodal Analysis

Nodal Analysis is very similar to Mesh Analysis but instead of calculating the current for each mesh in a circuit, we can use it to calculate the current and certain points. Nodal analysis focuses a lot more on removing components to see how current flows and then uses Kirchoff’s Laws to calculate the current at each node.

Consider the circuit below:

We have 3 nodes that we will be looking at: A, B, C (note that the nodes on the opposite side of the resistors R* _{A}*, R

*, R*

_{B}*are all at Ground, 0V). Let’s consider node A first and apply Kirchoff’s Current Law. There are 4 branches off it that go to 2 current sources and 2 resistors. By using ohm’s law for the resistors we can say that:*

_{C}- 1mA and 2mA are the 2 current sources, both are flowing into the node so we have them as positive; remember we could just have easily made them negative as long as any currents flowing out of the node are positive, but in this example the currents flowing in are positive.
- We then have ; this is ohm’s law to find the current through the 1kOhm resistor: . We know that the current is negative because the node below the resistor is ground and a current can’t flow out of ground so it must be flowing out of the node (this is illustrated above).
- Finally we have ; again this is from ohm’s law where the voltage is . Since below R
and R_{A}is 0V, we can say that the nodes A and B are at voltages V_{B }and V_{A}respectively, therefore the voltage drop across the 2kΩ resistor will be from V_{B}to V_{B}(again this is illustrated above)._{A}

So now that have calculated Kirchoff’s Current Law for node A, we will continue to find it for node B and C and we can then form a simultaneous equation to calculate the voltage at each node. We won’t go into so much detail for these 2 nodes since node A was so detailed – it’s basically the same stuff.

For node B:

So now let’s use all the knowledge we’ve learned in this tutorial to learn about Norton and Thevenin’s Equivalent Circuits in our next tutorial.