How Do You Calculate I_2 and I_4 Using Kirchoff's Rules?

[ttiger2k7]

[SOLVED] Using Kirchoff's Rules

Homework Statement

http://img266.imageshack.us/img266/4120/kirchoffho1.gif
In the circuit, I_1 = 1.6 A. Find the value of: (a) the current I_2 and (b) the current I_4.

Homework Equations

Junction Rule, Loop Rule, V=IR

The Attempt at a Solution

Small loop, top left:

2V + (2ohms)(I_2) - (4ohms)(1.6 A) = 0
(2 ohms)(I_2) - (6.4) = -2
2 ohms * I_2 = 4.4
I_2 = 2.2 A

That also means that the current in the entire circuit is 4.4 Amps.

...Right?

That is as far as I got. Assuming my answer for I_2 is correct, all I have to do is find I_4, which I am stumped on how to do. Should I combine the two resistors on the top left, since they're in series?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Google_Spider]

Small loop, top left:(moving anti-clockwise)

2+ 4*1.6-2i_2=0
i_2 =4.2 A

Should I combine the two resistors on the top left, since they're in series?

I don't think any two resistors in this circuit are in series

 

[ttiger2k7]

So that means, I got something that *roughly* looks like the following:

http://img297.imageshack.us/img297/7287/kirchoffeditck6.gif
(horrible MS paint editting by me)

So that means total I will be 5.8 (1.6 + 4.2).

Should I now do a loop rule for the whole circuit, calling the smaller loop Ir (where r is the total resistence of the right loop)? Then just solve for R using r and I?

 

[Google_Spider]

So that means, I got something that *roughly* looks like the following

No I don't think so. The resistances are neither in series nor parallel. You'll have to use KVL.

Should I now do a loop rule for the whole circuit, calling the smaller loop Ir (where r is the total resistence of the right loop)? Then just solve for R using r and I?

In the circuit, I_1 = 1.6 A. Find the value of: (a) the current I_2 and (b) the current I_4.

The question doesn't ask you to find R. Only i_2 and i_4
Just use loop rule in lower big loop to find i_4 .

 

[ttiger2k7]

Thanks. Got it!

12 V - 4.2*2 - 3I_4 = 0

I_4 = 1.20 A