Using Kirchoff's Rules

1. Feb 27, 2008

ttiger2k7

[SOLVED] Using Kirchoff's Rules

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Junction Rule, Loop Rule, V=IR

3. 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?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
2. Feb 27, 2008

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

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

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

3. Feb 27, 2008

ttiger2k7

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

http://img297.imageshack.us/img297/7287/kirchoffeditck6.gif [Broken]
(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?

Last edited by a moderator: May 3, 2017
4. Feb 27, 2008

No I don't think so. The resistances are neither in series nor parallel. You'll have to use KVL.
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 $$i_4$$ .

5. Feb 27, 2008

ttiger2k7

Thanks. Got it!

12 V - 4.2*2 - 3I_4 = 0

I_4 = 1.20 A