In these tutorials, we are going to be looking at **mesh analysis**. This is a circuit analysis technique which can easily be applied for analysing circuits comprised of a number of ‘loops’ or *meshes*. We will start with simple circuits, and then move on to more difficult ones in other parts.

First, lets start with an example problem:

Considering figure 1 above, if we wanted to find the value of the current flowing through the 20Ω one option would be to simplify the circuit and use Ohm’s law as follows :

- We can redraw the circuit as (combining the 5Ω and 10Ω resistor and adding as they are in series):
- We could then combine the 10Ω and 15Ω resistors in parallel to be equivalent to a single 6Ω resistor (as (10 x 15)/25 is 6Ω)
- We can then of course add the 20Ω to the 6Ω to calculate that the total resistance is 26Ω
- Using Ohms law we know that the current is the voltage divided by the resistance, so the current through the 20Ω resistor is 13/26 =
**0.5A**

This is fine, but if we then wanted to know the current through the five Ohm resistor instead of the twenty Ohm, things get even more complicated! That’s where mesh analysis comes in.

#### Using Mesh Analysis to solve the problem

Mesh analysis relies on Kirchoff’s voltage law – that the voltages within a closed loop sum to zero.

We can therefore draw onto our circuit two clockwise current loops – I have labelled them i1 and i2 below:

Using Ohm’s law we can now form equations for the sums of the voltages in the two meshes.

###### In the first mesh (the enclosed square on the left)

We have -13V + (i1 x 20Ω) + (i1 x 10Ω) from the ten volt battery and from Ohm’s law as V = i x R. We also have another voltage, this time in the opposite direction from the other two which is caused by i2 (i2 x 10Ω).

As all of the voltages sum to zero, we can bring this together and say that: 0 = -13V + (i1 x 20Ω) + (i1 x 10Ω) – (i2 x 10Ω), or rearranging and simplifying: **13**** = (i1 x 30) – (i2 x 10)**

###### In the second mesh (the enclosed square on the right)

We can apply the same technique here, but this time there is no voltage source so we can say: 0 = (i2 x 10Ω) + (i2 x 10Ω) + (i2 x 5Ω) – (i1 x 10Ω) [By the same reasoning as before]

We can rearrange and simplify this to be **i2 x 25 = i1 x 10**

We now have two equations which we must solve simultaneosly – **i2 x 25 = i1 x 10 **and **13 = (i1 x 30) – (i2 x 10). **You can solve these however you like, but for this example, we can use simple reasoning:

- Multiplying the first equation by 3, we get
**i2 x 75 = i1 x 30** - Notice the
**i1 x 30**appears in both now, so we can swap one for another - So
**13 = (i2 x 75) – (i2 x 10)** - So
**13 = 65 x i2** - So i2 = 13 /65 =
**0.2A [This is the current through the five Ohm resistor]**

And, as **i2 x 75 = i1 x 30**, **i1 = (0.2 x 75) / 30 = 0.5A Which is the same as we obtained using Ohm’s Law above**

If you have any questions, please do comment below