This article is the 2nd 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.
In previous Part-1 we have calculated the values of two mesh currents as follows –
I1 = -0.5Amp and I2 = 3Amp
The value of I1 came out negative, because we had taken its direction as clockwise. Now we shall again solve the problem with the same network, given below, but now we shall take the direction of I1 as anticlockwise.
Now we shall see that the value of I1 also comes out to be positive with I2 also, as given below.
Here I1 is flowing through the loop (a-d-c-b-a) in anticlockwise direction while I2 is flowing through the loop (b-e-f-c-b) in anticlockwise direction.
So the equation for current I1, in this loop, is given by –
4 + R2.(I1 – I2) + 2.I1 = 0 … Here I1 & I2 are opposite to each other and we are calculating for I1. So we have taken the term (I1 – I2).
i.e. 4 + 2.(I1 – I2) + 2.I1 = 0
4.I1 – 2.I2 = (-4) … (1)
Now in the same way, we shall obtain equation for second loop i.e. for (b-e-f-c-b) loop, as follows –
2.(I2 – I1) + (-8) + I2 = 0 … Here we are calculating for I2. So we have taken the term (I2 – I1).
∴ -2.I1 + 3.I2 = 8 … (2)
Now solving equations (1) and (2) using simultaneous equation method. For that we shall multiply equation (1) by ‘3’ and equation (2) by ‘2’, as follows –
12.I1 – 6.I2 = -12 … (1)
-4.I1 + 6.I2 = 16 … (2)
That gives us the following –
8.I1 = 4
∴ I1 = 0.5Amp … Now this value is positive and so the direction of I1 is correct.
In the same way if we calculate the current I2, then it will come out as I2 = 3Amp.
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