# Homework Help: Kirchoff's Laws

1. May 16, 2007

### fffff

1. The problem statement, all variables and given/known data
http://img353.imageshack.us/img353/4445/kirchoffkq4.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

for the first loop (2v battery)

the effective resistance is 33/7
the voltage is 2v

so current=v/i
=14/33 A

Now loop 2 is the problem.
i dont know how to add up the all the resistance.

or am i using the wrong approach here.

Last edited by a moderator: May 2, 2017
2. May 16, 2007

### fffff

http://img515.imageshack.us/img515/9714/kirchoffhf5.jpg [Broken]

ok for the first lloop

2=3(I1) + 4(I2) +3(I3)

the second loop

1=4(I4) +3(14)

but its still wrong

Last edited by a moderator: May 2, 2017
3. May 16, 2007

### fffff

the problem im having iswith the I3 part of the second loop
does current even flow there? or does i4 behave like i2 ?

4. May 16, 2007

### antonantal

I2 and I4 are the same because there is no node between them. Try writing Kirchhoff's loop equations for the two loops indicated and the node equation for the blue node.
[URL=http://img183.imageshack.us/my.php?image=kirchoffhf5zk0.jpg][PLAIN]http://img183.imageshack.us/img183/7638/kirchoffhf5zk0.th.jpg[/URL][/PLAIN]

Last edited: May 16, 2007
5. May 16, 2007

### fffff

ok for the first loop

2=3(I1) + 3(I3)

second
1=4(14) + 3(14)

6. May 16, 2007

### fffff

I1=I2+I3

but dosend I4 go through the same direction as I1

Last edited: May 16, 2007
7. May 16, 2007

### antonantal

I don't know where that 14 came from. Just forget about what you wrote in your attempt and solve this by using only Kirchhoff's equations.

The directions you choose are arbitrary. Only after you solved the equations you will know the real direction of each current.

But for the voltage sources you must consider the direction of the voltage. Look again at the direction of the voltage from the 1V source and see what sign it should have in the equation you wrote.

8. May 16, 2007

### fffff

Yes, I2=I4
Thats was what I meant
But are my equations correct

let me edit my 2nd equation

1=4I2 +3(12-I3)

Last edited: May 16, 2007
9. May 16, 2007

### Mindscrape

Hint: what is the constraint on I3?

Have you ever heard of mesh current analysis? You might be able to find some helpful examples in there.

10. May 16, 2007

### antonantal

You defined I3 as the current that flows through the 3ohm resistor from the small loop. Why do you add I4 to it? I4=I2 is the current through the 4ohm resistor.

11. May 16, 2007

### fffff

I didnt add I4 to it

1=4 I2 + 3(I2-I3)

(12-13) would be the net current flowing into the 3 ohm resistor in the small loop

12. May 16, 2007

### fffff

or do you consider them separately.

say

1 V = 4 I2 + 3 I3

13. May 16, 2007

### fffff

Then for the first loop

2= 3 I1 + 3 i3

14. May 16, 2007

### fffff

oh sorry for the small loop i3 is negative

15. May 16, 2007

### antonantal

Exactly! Now,you have 2 equations and 3 unknown currents. You need one more equation. That's where the node equation comes in.

16. May 17, 2007

### Mindscrape

I3 doesn't matter because it is a combination of I1 and I2.