1. Oct 16, 2006

VinnyCee

Find i1, i2 and i3:

My work so far:

KVL for left loop => $$5\,-\,v_1\,-\,v_2\,=\,0$$

KVL for right loop => $$v_2\,-\,v_3\,+\,3\,=\,0$$

KCL at center node => $$i_1\,=\,i_2\,+\,i_3$$

Now I plug the KVL's into the KCL equation and I get: $$\frac{5}{4}\,i_1\,=\,2\,i_2\,+\,\frac{11}{8}\,V$$

But how do I get the value of $i_2$?

I know this is a really stupid question, but can someone help me please?

Last edited: Oct 16, 2006
2. Oct 16, 2006

andrevdh

You can rewrite your first two equations using ohm's law. For the first one :

$$5 - 2i_1 - 8i_2 = 0$$

3. Oct 16, 2006

gunblaze

Express the v's in ur KVL and KVR in terms of its resistance and current. Use the relationship in KCR to help re express ur KVL for both loop and then hopefully u can solve for ur current.

4. Oct 16, 2006

VinnyCee

Yup that is how I got the last equation.

KVL for loop one => $$5\,-\,2\,i_1\,-\,8\,i_2$$

this turns into => $$i_2\,=\,\frac{5}{8}\,-\,\frac{1}{4}\,i_1$$

KVL for loop two => $$8\,i_2\,-\,4\,i_3\,+\,3\,=\,0$$

this turns into => $$i_3\,=\,2\,i_2\,=\,\frac{3}{4}$$

Now I substitute these into the KCL expression => $$i_1\,=\,i_2\,+\,i_3$$

Which produces => $$i_1\,=\,\frac{8}{5}i_2\,+\,\frac{11}{10}$$

But now I have no idea how to get $i_2$!!! Please help, this is supposed to be the easiest problem to do in the whole book.

Last edited: Oct 16, 2006
5. Oct 16, 2006

OlderDan

Substitute the last equation for i1 into the first KVL equation to get an equation with only i2. This is a typical linear equation problem with 3 equations for 3 unknowns

6. Oct 16, 2006

VinnyCee

$$i_1\,=\,5\,-\,2\,\left(\frac{8}{5}\,i_2\,+\,\frac{11}{10}\right)\,-\,8\,i_2$$

$$5\,-\,\frac{16}{5}\,i_2\,-\,\frac{22}{10}\,-\,8\,i_2\,=\,0$$

$$-112\,i_2\,=\,-28$$

$$i_2\,=\,\frac{28}{112}\,=\,0.25\,A$$

That is right! Thank you much, but now I want to know how you knew to do that substitution, otherwise I will just have the same problem on the next exapmle, etc.

7. Oct 16, 2006

OlderDan

There are various well known methods for solving simultaneous equations. Substitution is not bad when you have original equations with few unknowns. In this case, it was easy to recognize that you had two different equations with the same two unknowns. If all three equations had all three unknowns, you would want to use an elimination approach.

For three equations and three unknowns, the idea is to take one pair of equations and multiply or divide one or both to get the same coefficient in front of one of the unknowns. Then subtract one equation from the other to eliminate that unknown. (Or you could multiply or divide to get opposite sign coefficints for one unknown and then add.) Repeat this procedure with a different pair of equations to eliminate the same unknown as with the first pair. You will then have two equations with only two unknowns. You can solve those by a second elimination, or by substitution. Then work backwards using known values to solve for the others.
It takes some practice to master these techniques. A good Algebra II book should help you.

Last edited: Oct 17, 2006
8. Oct 17, 2006

VinnyCee

I think my problem is knowing which equations are independant and which are just another one stated in a different way.

The other problem I have is knowing where to apply what rule (KCL, KVL).

I guess it will just take practice, like everything else!