DC Networks and Kirchoff's Laws

[gneill]

I have follewed up this point quite clearly, however I don't understand the reason why we have to multiply equation 1 by 3 ? If somebody could explain this too me or even send me a link where I can understand why.

It's a technique for solving the two simultaneous equations. The idea is to make the coefficients of one of the variables the same in both equations so that when the difference of the two equations is taken that variable drops away leaving a single equation in one unknown (the other variable).

In this case the poster chose to eliminate the I1 variable. To that end he looked at the coefficients of I1 in each equation: 4 and 3. A number that has both as factors is 12, so he multiplied the two equations by suitable values to yield 12I1 in each.

 

[biggdogg9]

Cheers gneil

 

[waltarno]

Hi All,

I too have this question and am a little confused. For the value of L2 I get -0.4545 but if I use this figure to get L1 I get 0.590875 which is nowhere near what you guys are saying.

If I use 0.4545 instead I get somehwere near the answers suggested on here.

Could someone please explain what happens to the minus (is it disregarded because it is indicative of current direction?) or have I done something school boy style?

Thanks

 

[waltarno]

maths used to get to this point :-

(12L₁+9L₂)-(12L₁+20L₂)=3-8

-11L₂=-5

L₂=-5/11

L₂=-0.4545

4L₁+3L₂=1

4L₁+(3*-0.4545)=1

4L₁-1.3635=1

4L₁=1+1.3635

L₁=2.3635/4

L₁=0.590875

 

[gneill]

maths used to get to this point :-

(12L₁+9L₂)-(12L₁+20L₂)=3-8

-11L₂=-5 ← negative on both sides[/color]

L₂=-5/11 ← Incorrect sign[/color]

L₂=-0.4545

4L₁+3L₂=1

4L₁+(3*-0.4545)=1

4L₁-1.3635=1

4L₁=1+1.3635

L₁=2.3635/4

L₁=0.590875

 

[waltarno]

Thanks for the quick reply, obviously didn't spot that!