Simple circuit analysis using KVL

[PainterGuy]

Hi,

Please have a see on this circuit diagram:
http://img192.imageshack.us/img192/7726/meshkvl.jpg

I used these equations:
1: I₁ = I₂ + I₃ (Using KCL at junction e)
2: 6 - 4I₂ + 4 - 2I₂ = 0 (Using KVL around afeba)
3: 3 - 6I₃ - 4 - 4I₂ = 0 (Using KVL around dcbed)

I₂ = 2.8 A

I₁ = 4.5 A

I₃ = 1.7 A

I don't understand where I'm wrong. Perhaps the error lies in approach. Help me out please.

Cheers

 

[BackEMF]

Hi,

Please have a see on this circuit diagram:
http://img192.imageshack.us/img192/7726/meshkvl.jpg

I used these equations:
1: I₁ = I₂ + I₃ (Using KCL at junction e)
2: 6 - 4I₂ + 4 - 2I₂ = 0 (Using KVL around afeba)
3: 3 - 6I₃ - 4 - 4I₂ = 0 (Using KVL around dcbed)

I₂ = 2.8 A

I₁ = 4.5 A

I₃ = 1.7 A

I don't understand where I'm wrong. Perhaps the error lies in approach. Help me out please.

Cheers

Hey PainterGuy. One thing, that's probably just a typo, you've used I₂ when you should have used I₁ at the end of the second formula:

It should be

2: 6 - 4I₂ + 4 - 2I₁ = 0 (Using KVL around afeba)

You also have a sign wrong in the last formula. Try writing it again, but first mark the polarity of the voltage drops across all the resistors in your circuit diagram - put a "+" sign where the current flows into the resistance, and a "-" sign where it flows out.

 

[PainterGuy]

Hey PainterGuy. One thing, that's probably just a typo, you've used I₂ when you should have used I₁ at the end of the second formula:

It should be

2: 6 - 4I₂ + 4 - 2I₁ = 0 (Using KVL around afeba)

You also have a sign wrong in the last formula. Try writing it again, but first mark the polarity of the voltage drops across all the resistors in your circuit diagram - put a "+" sign where the current flows into the resistance, and a "-" sign where it flows out.

Hi BackEMF,

Many thanks for the help.

1: I₁ = I₂ + I₃ (Using KCL at junction e)
2: 6 - 4I₂ + 4 - 2I₁ = 0 (Using KVL around afeba)
3: 3 - 6I₃ - 4 + 4I₂ = 0 (Using KVL around dcbed)

Is it correct now? Someone used mesh analysis to solve it. I'm confused which method to use where. Could you please provide some guide when to use which method out of these three: 1: Simple combination of KVL and KCL (as I did above), 2: Nodal analysis (branch current analysis), 3: Mesh analysis?

Much grateful for all this help.

Cheers

 

[BackEMF]

...Is it correct now? Someone used mesh analysis to solve it. I'm confused which method to use where. Could you please provide some guide when to use which method out of these three: 1: Simple combination of KVL and KCL (as I did above), 2: Nodal analysis (branch current analysis), 3: Mesh analysis?

Much grateful for all this help.

Cheers

Yeh it looks right to me.

As for which method to pick, it often comes down to preference. But obviously sometimes one method will be easier than another. For example, if the circuit has mostly voltage sources and few current sources, I'd use Nodal Analysis (each voltage source reduces the number of equations by one), similarly for Mesh Analysis when there are many current sources.

The combination of KVL and KCL that you can do by inspection (i.e. not following any algorithm in particular) can be quite easy for small circuits, often you can see an optimal way of writting the equations, but for large circuits this is generally never the best way to go.

Case in point: you wrote 3 equations, whereas Mesh Analysis would have left you with only two to solve (admittedly one of your three was easy!).

Also Mesh analysis will only work for planar networks. Nodal analysis will work for every network, hence Circuit Simulators base their equations on a variant of this - Modified Nodal Analysis.

 

[PainterGuy]

Hi BackEMF, :)

It is nice of you to reply me and give advice on what methods to you. Many thanks.

In the case of the circuit if my first post, yes I would have used Mesh analysis but what good it would have done for me. Instead of I3, (I1 - I2) would be used because I3=I1-I2. Sorry it is not possible for me to use subscripts because I'm on mobile. Please correct if wrong.

I almost all the times get confused between nodal analysis (i think it's also called branch current method), mesh analysis. There is another problem the book I have has two different methods by the name: branch current method, and nodal analysis. This means according to the book by Boylested these two methods are different not same.

Cheers

 

[NoPoke]

Mesh analysis = use KVL around a loop in the circuit. keep adding loops until all paths are included at least once.

Nodal analysis = use KCL for each node other than the "ground" or reference node.

Mesh analysis yields a matrix [V] = [R]
Nodal analysis yields a matrix = [Y][V]


Until you are familiar with them I'd recommend sticking with mesh analysis. Drawing current loops just seems to be more natural .

So from your drawing I'd use two loops I1 = current in the loop afeb and I3 = current in cbed. The current you have labeled as I2 is thus = I1-I3

KVL for I1 loop : E1+E2 = (R1+R2) x I1 - R2 x I3
KVL for I3 loop : E3 - E2 = (R2+R3) x I3 - R2 x I1

substitute actual values
10 = 6xI1 - 4xI3
-1 = 10xI3 - 4xI1

so eliminating I1..
20 = 12xI1 - 8xI3
-3 = 30xI3 -12 x I1

17 = 22x I3
I3 = 17/22 Amp

substitue back
-1 = 170/22 - 4 x I1
I1 = (170+22)/88 = 192/88 = 96/44 = 48/22 = 24/11 = 2 + 2/11 Amp

And I2 = I1 - I3 = 48/22 - 17/22 = 31/22 = 1 + 9/22 Amp

====

check: confirm node voltage is the same

I1 x 2 - 6 = 48/11 -66/11 = -18/11 V
4 - I2 x 4 = 44/11 - 62/11 = -18/11V
3 - 6 x I3 = 33/11 - 51/11 = -18/11 V

 

[PainterGuy]

Many, many thanks NoPoke.

Cheers