Find I: Solving Electrical Circuits

[cdotter]

Homework Statement

[PLAIN]http://img688.imageshack.us/img688/1452/45481808.png

Find I at all branches.

Homework Equations

$\sum I=0$ for a junction.
$\sum V =0$ for a closed loop.

The Attempt at a Solution

I found the solution but I don't understand why they chose certain variables like why the arrows are in a certain direction. From my understanding it's completely arbitrary but when I do it my own way I get a completely different answer.

Could someone tell me if this would give the correct answer?

The arrows and currents $I_1$, $I_2$, and $I_3$ are all drawn in by me.

$I_3$ = $I_1$ + $I_2$ per the junction law. I could then find I for the top inner loop and I for the outer loop, and solve the two simultaneous equations. The other two currents could then be found by back-substitution.

 

[tiny-tim]

hi cdotter!

(btw, two h's in Kirchhoff! )

I found the solution but I don't understand why they chose certain variables like why the arrows are in a certain direction. From my understanding it's completely arbitrary but when I do it my own way I get a completely different answer.

you should get the same answer whichever way you do the arrows

the only difference is that if you put say the I₁ arrow the wrong way round, you get a result for I₁ multiplied by -1

but why are you using round arrows next to the wires? that's really confusing …

put arrows on the wires!

if you're still getting the wrong answer, show us your full calculations, and then we'll see how to help ​

 

[cdotter]

I don't know, that's just the way my physics professor does it.

For the top inner loop:

$$10V-I_1 \cdot 3 \Omega -((I_1+I_2) \cdot 4 \Omega) + 5V - ((I_1+I_2) \cdot 1 \Omega) - I_1 \cdot 2 \Omega=0$$
$$15V-I_1\cdot 10 \Omega - I_2 \cdot 5 \Omega=0$$
$$\Rightarrow I_2=3A-I_1 \cdot 2$$

For the outside loop:

$$10V-I_1 \cdot 3 \Omega - I_2 \cdot 10 \Omega - I_1 \cdot 2 \Omega=0$$
$$10V-I_1 \cdot 5 \Omega - I_2 \cdot 10 \Omega=0$$

Substitute in the top inner loop term:

$$10V-I_1 \cdot 5 \Omega - (3A-I_1 \cdot 2) \cdot 10 \Omega=0$$
$$10V-I_1 \cdot 5 \Omega - (30V-I_1 \cdot 20 \Omega)=0$$
[tex]\Rightarrow I_1= \frac{20V}{15 \Omega} = 1.33 A[/itex]<br /> <br /> According to my textbook, this isn't right. It should be 1.60 A. Where am I going wrong?[/tex]

 

[tiny-tim]

hi cdotter!

I don't know, that's just the way my physics professor does it.

yup, well this is what happens if you don't use straight arrows on every available line…

it's too easy to make a mistake, in this case you haven't noticed that the arrows are going the opposite way at the bottom of the outer loop!

try again, and in future i suggest you ignore your professor and go for the straight arrows!

(alternatively, only use the loops with the round arrows on them, in this case the top loop and the bottom loop)

 

[cdotter]

[PLAIN]http://img602.imageshack.us/img602/1452/45481808.png

[PLAIN]http://img703.imageshack.us/img703/8694/giflatex.gif

The answer is still wrong. It should be 1.60A. Where am I making a mistake?

edit: My algebra is a bit wrong. It should be 2.7A. Still wrong.

 

[tiny-tim]

hi cdotter!

your 7ΩI₁ should be 10ΩI₁

(btw, you don't have to do everything in cyclic order …

it's much easier to add the 2Ω to the 3Ω, and the 1Ω to the 4Ω, right at the start, and you're less likely to make a mistake if you do that! )

 

[cdotter]

hi cdotter!

your 7ΩI₁ should be 10ΩI₁

(btw, you don't have to do everything in cyclic order …

it's much easier to add the 2Ω to the 3Ω, and the 1Ω to the 4Ω, right at the start, and you're less likely to make a mistake if you do that! )

You can do that even though they're on different sides of the voltage source? That would make it much, much easier.

 

[tiny-tim]

Yup, so long as they have the same current through them!

 

[cdotter]

I'm finally getting 1.6A! Thank you very much for your help and patience, tiny-tim.