# Homework Help: Node voltage analysis help

1. May 6, 2010

### James889

Hi,

I have this circuit which i am trying to write KVL equations for, but i just can't seem to get it right...

[PLAIN]http://img146.imageshack.us/img146/3374/upg255.png [Broken]

The equations i have written so far are:
For v1: $$\frac{v_1-v_2}{5} +\frac{v1}{10} +1 = 0$$
For v2: $$\frac{v_2-v_1}{5} - \frac{v_2}{20} -0.5i_x = 0$$

Im not 100% sure this one is correct.
$$i_x=\frac{v_1-v_2}{5}$$

Some pointers on whether these are correct or not, and in the case, what's wrong, would be appreciated.

Best regards
/James

Last edited by a moderator: May 4, 2017
2. May 6, 2010

### Staff: Mentor

First, those are KCL equations, summing the currents leaving (or going into) each node.

Second, you have a sign error at least in the first equation (I didn't check further). Remember to be consitent -- sum the currents leaving each node (or entering, but I prefer leaving). What should the sign be on the current source term in the first equation for the node V1?

Last edited by a moderator: May 4, 2017
3. May 7, 2010

### James889

Hi,
Hm, as the 1A current is entering node 1 the sign should be a +, should it not?

4. May 7, 2010

### Redbelly98

Staff Emeritus
Okay, but then you need a "-" sign for the currents leaving the node through the 10Ω and 5Ω resistors.

5. May 8, 2010

### Staff: Mentor

So as RB says, if you insist on summing the currents entering the nodes, then you will need to adjust your equations to change the voltage subtractions to give you the currents entering the nodes.

Again, I prefer to sum the currents leaving each node, since that makes the voltage subtractions (divided by the resistances between the node voltages) more intuitive. It's easier to throw a "-" sign on a current source value in the sum, versus inverting voltage differences, IMO. But in the end, it's up to personal preference. Whatever is intuitive and accurate for you.

6. May 8, 2010

### James889

This equation conforms with v1 being 6v, v2 being 4v and ix = 0.4A.
$$-\frac{v_2-v_1}{5}-\frac{v_2}{20}=0.5i_x$$

This is how i would write the second one.
$$\frac{v_1}{10} - \frac{v_2-v_1}{5} +1 = 0$$

Putting v1 = 6 and v2 = 4 yields 1.2 instead of 0.

However if you put a minus sign in front of the first term it works out.

This does not make any sense. How can the current go from v2 to v1, and from v1 to v2 at the same time

7. May 8, 2010

### The Electrician

You're still not getting your signs right.

When you're working on (summing currents at) a given node, say V1, and you have a resistor R connected from that node to ground, if you write a term V1/R, you are assuming positive currents are leaving the node. If you write a term -V1/R, you are assuming positive currents are entering the node.

With respect to currents from one node to another, if you're working on the V1 node and you write a term (V1-V2)/R you are assuming positive currents are leaving the node. If you write a term (v2-V1)/R, or equivalently -(V1-V2)/R, you are assuming positive currents are entering the node. If you are working on node V2, then the term (V2-V1)/R assumes positive current leaves the V2 node.

Either choice is ok, but you need to be consistent once you've made your choice.

This equation:$$\frac{v_1}{10} - \frac{v_2-v_1}{5} +1 = 0$$

is inconsistent.

The two terms $$\frac{v_1}{10}$$ and $$-\frac{v_2-v_1}{5}$$

are consistent with the assumption that positive currents are leaving the node, but the term

$$+1$$ is consistent with positive currents entering the node.

Last edited: May 8, 2010