Solving Kirchoff's Law for Loop 2: A Homework Challenge!

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Homework Statement

http://img353.imageshack.us/img353/4445/kirchoffkq4.jpg

Homework Equations

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 don't know how to add up the all the resistance.

or am i using the wrong approach here.

 

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http://img515.imageshack.us/img515/9714/kirchoffhf5.jpg

ok for the first lloop

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

the second loop

1=4(I4) +3(14)

but its still wrong

 

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

 

[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]

 

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ok for the first loop

2=3(I1) + 3(I3)

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

 

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I1=I2+I3

but dosend I4 go through the same direction as I1

 

[antonantal]

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

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.

but dosend I4 go through the same direction as I1

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.

 

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

let me edit my 2nd equation

1=4I2 +3(12-I3)

 

[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.

 

[antonantal]

1=4I2 +3(12-I3)

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.

 

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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

 

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or do you consider them separately.

say

1 V = 4 I2 + 3 I3

 

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Then for the first loop

2= 3 I1 + 3 i3

 

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oh sorry for the small loop i3 is negative

 

[antonantal]

or do you consider them separately.

say

1 V = 4 I2 - 3 I3

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

 

[Mindscrape]

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