# Homework Help: Nodal and Mesh Analysis

1. Feb 29, 2012

### beanus

1. The problem statement, all variables and given/known data
Find the node voltage method equations. Do not solve
Find the mesh current method equations. Do not solve

2. Relevant equations
None

3. The attempt at a solution
Using nodal analysis:
Node v1 : -Is4+(v1-v2)/2+v1/20-10=0
Node v2 : Is4+(v2-v1)/2+3+(v2-30)/5=0
Looking at this I feel like v1 is just -10 and v2 is just -30. But that can't be right can it?

Using mesh analysis:
I set up my meshes from left to right, and then up. So the left most mesh is i1, the one to the right of it is i2, the far right is i3, the top is i4.

mesh 1 : -30+V_(is2)+5i1=0
mesh 2 : 20(i2-i4)+2(i2-i3)-V_is2=0
mesh 3 : 10i3+5_vr1+2(i3-i2)=0
mesh 4 : 10+20i4=0

I understand the principles of this, but this circuit is really confusing me. Thanks in advance for the help!

EDIT: The 5Vr1 is actually 5*V_r1

2. Feb 29, 2012

### Staff: Mentor

You're mixing voltages and currents in the Node v1 equation. "10" is a voltage.

When writing node equations you should first identify a reference (ground) node. That node will be the zero reference for all the other node voltages that you determine. In this case the node that you labeled V2 looks to be a good candidate. Next you should pick out any nodes whose potential differences are constrained by fixed voltage sources. These can be combined into "supernodes". Currents into and out of supernodes reference only a single supoernode potential, with appropriate fixed offsets (due to the fixed voltage supplies). In your circuit your node labeled V1 has a fixed 10V supply connected to the node to its left, so that comprises a supernode.

Here's a revised diagram:

In mesh 3 the 5_vr1 is a current source, so you'll need to insert a new variable for its unknown voltage. This current supply places a constraint on the mesh current (in fact it fixes the mesh current to be 5*vr1).

You'll also have to write the constraint equation that goes with Is2 --- it fixes a relationship between your i1 and i2. You should also write the constraint equation that relates vr1 to i1.

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3. Feb 29, 2012

### beanus

Ok I revised my node equations
v2-v1=10
(v2-30)/5-3=0
v1/2-Is4=0
I'm not sure how to go about taking care of is4 in this case

For the meshes,
Everything ok was except mesh 3?
This is what I got for the constraint: Is4=I2-I3.

Thanks for the help !

4. Feb 29, 2012

### Staff: Mentor

Rewrite your node equations; you've really only got one node, the supernode I indicated, so V2 won't appear in it (because V2 is just V1+10). Be sure to replace "5*Vr1" with an appropriate expression based upon the voltage that appears across R1 due to the supernode voltage across the branch containing R1.

Is4 is not I2 - I3. Is4 is I3. (well, -I3 if I3 is taken to be clockwise). Is4 will be constrained by I1, right? since I1 flows through R1. You still need to write the constraint that Is2 imposes.

5. Feb 29, 2012

### beanus

Still having trouble understanding the whole constraint thing, but this is what I've done.

For the nodes
v2=v1+10
(v1+10-30)/5-30=0
(v1)/2-5(5*i1)=0

For the meshes
Is4=5(5*i1)

6. Feb 29, 2012

### Staff: Mentor

Okay, you've still got a voltage (30) stuck in a current equation. What's it there for?
Also, there should be one node equation that represents the entire supernode (KCL sum of currents entering/leaving that supernode). Then the constraints are applied, such as the Is4 current source being equal to 5 x the voltage across R1. You can write an expression for that voltage if you determine the current through R1 in terms of the supernode voltage at the top of its branch.

What about Is2? What does it constrain?

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7. Feb 29, 2012

### beanus

v2=v1+10
(v1+10-30)/5+v1/2-5(5*i1)=0

Is the current through the left branch equal to 3?

So the second equation becomes
(v1+10-30)/5+v1/2-75=0?

Or is the current in the left branch equal to (v1+10)/5? or (v1+10)/5+3?

8. Feb 29, 2012

### Staff: Mentor

You're getting there

Since there's no known "i1", you'll have to write an expression that will will give you Vr1 based upon the given variables. In this case you can write a KVL expression for the branch starting at V1, passing through Vs1 and R1. The sum must be zero.

The current in the left branch (which would be your "i1") is given by:

(V1 + 10 - 30)/R1

which was the first term in your first equation above.

Be sure to include Is2 which also flows into the supernode

9. Feb 29, 2012

### beanus

so v1+10-30+5*i1+3(?)=0 Do I multiply by one since there's no resistor on that branch?
v1+5i+3(?)=20

10. Feb 29, 2012

### Staff: Mentor

I think we're diverging here. Probably my fault.

what are you trying to calculate? The Is2 should go into your node equation, not the calculation for Vr1.

11. Feb 29, 2012

### beanus

I was trying to make the KVL equation like you said. But the Is2 is current so I have to multiply by a resistance to make it a voltage.

I'm so confused right now.

So my meshes are OK, it's just my constraint equation that I'm having trouble with? I am really trying to understand what you're saying.

12. Feb 29, 2012

### Staff: Mentor

Let's concentrate on the node equation first.

The KVL equation that you want is for the branch containing R1. The reason is you want to find an expression, based upon the working values and variables, to replace Vr1 (for Is4). As I mentioned, the path is from V1 down through the 10V battery and the 30V battery and then R1 and then ground. That's it. That's all. Vr1 can be extracted from that.

13. Feb 29, 2012

### beanus

So 10-30+iR1=0
i1(5)=20
i1=4 ?

14. Feb 29, 2012

### Staff: Mentor

You've left out V1 (currently an unknown). And you really don't need i1 at this point, just Vr1.

V1 + 10 - 30 - Vr1 = 0

Vr1 = V1 + 10 - 30

Vr1 = V1 - 20

Use that in the Is4 expression so you can do away with Is4 as an independent variable. How does your node equation look now?

15. Feb 29, 2012

### beanus

2v1+5v1-250v1+960=0

16. Feb 29, 2012

### Staff: Mentor

Okay, that's a little in-between start and finish. I was expecting to see a bit more of the derivation. Let's avoid confusion and start with the initial steps:

Node Equations:

Supernode: $\frac{V1}{2} - Is4 - Is2 + \frac{V1 + 10 - 30}{5} = 0$

Where:
$~~~~~Is4 = 5 Vr1$

$~~~~~Vr1 = V1 + 10 - 30$

$~~~~~Is2 = 3$

Replacing Is2 and Is4:

$\frac{V1}{2} - 5(V1 + 10 - 30) - 3 + \frac{V1 + 10 - 30}{5} = 0$

which can be simplified and solved for V1 (if you were going to solve the circuit)

Within the supernode:
$V2 = V1 + 10$

$I_{R2} = 10/20$ from V2 to V1

17. Feb 29, 2012

### Staff: Mentor

Quick question. I've been ignoring the line with the big arrow at the top of the circuit. Is it significant? It just occurred to me that it might represent an infrequently used version of a ground symbol... in which case your ground point is being forced upon you and your choice of independent nodes would change...

18. Feb 29, 2012

### beanus

Oh duh. Vr1 is just found by doing the regular node stuff the branch with r1.

So that means the mesh equations are
-30+V_is2+5i=0
20(i2-i1)+2(i2-i3-Vis2) = 0 (add the first two equations to cancel Vis2)
10i3+is4+2(i3-i2) = 0
10+20i4 = 0
is4 = 5(v1+10-30)

Is that right?

19. Feb 29, 2012

### beanus

That is the ground point that I set for it. I made this circuit in paint based on something I found in the book. I just wanted to do this problem for practice because it seemed difficult.

Our book uses the black delta as ground.

20. Feb 29, 2012

### Staff: Mentor

Ah. Alright. I hope my change of reference node didn't confuse you.