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http://www.webassign.net/hrw/hrw7_27-40.gif

I feel like I've tried everything to solve this problem; I just don't get it!

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- #1

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http://www.webassign.net/hrw/hrw7_27-40.gif

I feel like I've tried everything to solve this problem; I just don't get it!

- #2

collinsmark

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Use Kirchhoff's circuit laws to solve this. There are more than one way to do this problem, even when using Kirchhoff's laws. But no matter which way you go about it, you will end up with 2 equations and 2 unknowns. Solve the simultaneous equations and things will fall into place after that.

[Edit: Alternately, you might be able to use Thévenin/Norton equivalent transformations to grind your way to an answer. But I still suggest using Kirchhoff's laws if you have the choice.]

[Edit: Alternately, you might be able to use Thévenin/Norton equivalent transformations to grind your way to an answer. But I still suggest using Kirchhoff's laws if you have the choice.]

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vela

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collinsmark

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Of course you get to define your current loops yourself.

But let's suppose that you defined your current loops to be

Then by

Of course the particular resistor or resistors, who's current is a function of multiple unknowns completely depends on how you define your current loops at the beginning. And the details are a function of how you define the direction of the loops (clockwise or counterclockwise). The current loop definitions I used in the above paragraph is just one possible definition of many.

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LEFT LOOP:

E1-i1R1-(i2)R2-E2-i1ri=0

RIGHT LOOP:

-E3-i3R1+E2-(i2)R2-i3R1=0

I feel like a third variable is necessary. Because the R1's on either side aren't equal.

THere is one current going through the left, another through the right, and another through the middle.

I set a third equation to be i2+i1-i3=0.

But I wouldn't know where to go from there

- #7

collinsmark

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Well, that's not the way I would have done it. But let's use your method. I should still point out though that there is no need to defineOkay, so, defining the current to be traveling in a clockwise direction:

LEFT LOOP:

E1-i1R1-(i2)R2-E2-i1ri=0

But let's go ahead and use it anyway, even if it is unnecessary. But make sure you draw the direction of

So there is an error in your first equation! If you keep things in terms of

[tex] E_1 - i_1 R_1 + i_2 R_2 -E_2 -i_1 R_1 = 0 [/tex]

Using your method, solve forRIGHT LOOP:

-E3-i3R1+E2-(i2)R2-i3R1=0

I feel like a third variable is necessary. Because the R1's on either side aren't equal.

THere is one current going through the left, another through the right, and another through the middle.

I set a third equation to be i2+i1-i3=0.

But I wouldn't know where to go from there

But before I finish, I want to point out that using a third variable, such as

Based on your loop definitions of

Left Loop:

[tex] E_1 - i_1 R_1 - (i_1 - i_3)R_2 -E_2 -i_1 R_1 = 0 [/tex]

Right loop:

[tex] -E_3 - i_3 R_1 +E_2 -(i_3 - i_1) R_2 - i_3 R_1 = 0[/tex]

- #8

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Thanks, that actually helped alot!

And I looked over your method again, and it does make more sense.

And I looked over your method again, and it does make more sense.

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