Mesh Analysis: Simple to advanced tutorials for all students Part-1

Part-1

This article is the 1st part of our series Mesh Analysis: Simple to advanced tutorials for all students. For next parts to study, check the related topics with this post.

Mesh analysis is the best tool to solve complicated networks problems. In following paragraphs we shall understand how to use it in solving simple to complicated network problems.

Important note: To understand mesh analysis you must refresh your knowledge about KCL and KCL. Also I take it for granted that you are quite familiar with Ohm’s law – needless to say actually…!

Now to start with mesh analysis consider following simple network. Here we shall calculate the values of mesh currents I1 and I2.

Simple network to start with the study of mesh analysis

Here I1 is flowing through the loop (a-b-c-d-a) in clockwise direction while I2 is flowing through the loop (b-e-f-c-b) in anticlockwise direction.

So we shall write the equation for current I1 using Ohm’s law and KVL, as follows. Note that we shall start from point ‘a’ and we will follow the direction of current I1.

R1.I1 = 2.I1 → It is the voltage drop across R1 due to current I1 flowing through it.

Now this current I1 is also flowing through R2 along with the current I2 also. So there are two currents I1 and I2 flowing through R2 simultaneously.

R2.(I1 + I2) = 2.(I1 + I2) → We are taking the addition of the two currents since they are flowing in same direction through R2.

Now while going through the loop (a-b-c-d-a), we come at negative terminal of battery B1 first. So we shall take its value as negative, as follows.

(-4) → Taking the voltage value of battery B1 as negative.

Thus, the complete equation for the loop (a-b-c-d-a) using KVL will be –

R1.I1 + R2.(I1 + I2) + (-4) = 0

i.e. 2.I1 + 2.(I1 + I2) + (-4) = 0

Simplifying this equation we get –

2.I1 + 2.I1 + 2.I2 = 4

∴ 4.I1 + 2.I2 = 4 … (1)

Now in the same way, we shall obtain simplified equation for second loop i.e. for (b-e-f-c-b) loop, as follows –

Here also we shall follow the direction of I2, which is anticlockwise, starting from point ‘b’ –

R2.(I1 + I2) + (-8) + R3.I2 = 0

i.e. 2.(I1 + I2) + (-8) + 1.I2 = 0

Simplifying this equation we get –

2.I1 + 2.I2 + (-8) + I2 = 0

∴ 2.I1 + 3.I2 = 8 … (2)

Now solving equations (1) and (2) using simultaneous equation method –

4.I1 + 2.I2 = 4 … (1) 

2.I1 + 3.I2 = 8 … (2) … we shall multiply this equation by (-2)

(-4.I1) + (-6.I2) = -16 … (2) … after multiplying by (-2)

Our network shown for reference

So we have now two equations as follows –

4.I1 + 2.I2 = 4 … (1) 

-4.I1 -6.I2 = -16 … (2)

That gives us the following –

-4.I2 = -12 

∴ I2 = 3Amp

Now putting this value of I2 in equation (1) we can calculate the value of I1, which is given by –

2.I1 + 9 = 8 i.e. 2.I1 = -1

∴ I1 = -0.5Amp … This indicates that the direction of current we have taken is wrong, and it should be opposite i.e. anticlockwise. 

Important Note: In this example, I have deliberately reversed the direction of I1. So note here that, in Mesh Analysis, if the value of a current comes out to be negative, it shows that the direction which we have taken is WRONG and must be reversed. That’s all! No further action is necessary. You can try it for yourself by taking the direction of I1 opposite to get the value of I! as positive. That we shall discuss in next part.

We shall see about it in Part-2 of Mesh Analysis: Simple to advanced tutorials for all students.

Read Part-2 | Read Part-3

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