# Homework Help: Better understanding of KVL

1. Mar 19, 2017

### Just_enough

« thread moved from technical forum; so template is absent »

So I'm trying to get a better understanding of KVL by looking up practice problem and I'm having some difficulties. I'm attempting to do this problem
my first step is to get an equation for Ia (32v-10ia+8ib = 0) an Ib (12ib+20v-8ia = 0), adn then solve for either b or a first and plug into the other equation, but this is where the problem occurs. on the site http://fourier.eng.hmc.edu/e84/lectures/ch2/node2.html it got a current for each mesh to be something completely different. I was wondering how they got 4 and 1 for the currents?

Last edited by a moderator: Mar 19, 2017
2. Mar 19, 2017

### Staff: Mentor

3. Mar 19, 2017

### Just_enough

What do you mean? That is a pratice example so I'm assuming their answers and equations are correct and if mine doesnt match up then I'm doing something wrong

4. Mar 19, 2017

### Staff: Mentor

Do yours "match up" or not?

5. Mar 19, 2017

### Just_enough

at first, no and the only thing that didnt match up was that i got my 20v to be the opposite sign of what they had

6. Mar 19, 2017

### Staff: Mentor

Your equations are identical with theirs. You can multiply everything on both sides by –1 and the equation still holds.

7. Mar 19, 2017

### Just_enough

no not what I mean, on the B side, i have all the other sign matching up except the 20v so multiply by -1 wont make it correct, but that's not my question I just want to know how they got their currents for each of the meshes

8. Mar 19, 2017

### Staff: Mentor

32v-10ia+8ib = 0 and
12ib+20v-8ia = 0
then they are indeed identical with those in the worked example. All signs correspond.

You are asking how to solve 2 simultaneous equations?

If we use yours, which I'll write out again neatly:
32 - 10ia + 8ib = 0 ...... (i)

12ib + 20 - 8ia = 0 ...... (ii)

Multiply eqn (i) by 4
Multiply eqn (ii) by -5

Then add the corresponding sides of both new equations, and the resultant term for ia has a coefficient of 0, i.e., ia disappears from the equation, and with only one unknown in the equation you can solve for it to find the value for ib.