184 Week 5: Resistance
equation for loop 2 minus the equation for loop 1. This ischaracteristicof multiloop problems –
the sum or difference of interior loops always adds up to exterior loops as the inner/shared voltages
cancel.
This is very important to remember when we solve the simultaneous equations –adding loop
equations to eliminate variables does not make progress towards the solution! It just gives you
another loop equation. In order to make progress, youmustuse the junction equation(s) and a
subsetof the loop equations. Let’s dump the equation for loop 3 and keep only the three we need
to solve the problem. With a bit of rearrangement, we get:
3 I 1 + 8I 3 = 9 (358)
3 I 2 + 8I 3 = 3 (359)
I 1 +I 2 = I 3 (360)There are many ways to proceed to find a solution to this linear system. One can line up the
I’s, form a matrix equation, and invert the matrix using more or less standard determinants and
linear algebra. One can line up theI’s and do Gauss elimination (being careful to use the junction
rule before the loop rules) followed by back substitution. Or in the case of systems as simple as this
one, one can just use substitution to eliminate one of the currentsusing the junction equation, then
eliminate one of the two remaining currents (followed by back substitution), a sort of sloppy Gauss
elimination. Being a sloppy kind of guy (and not wanting to teach a course in linear algebra on top
of everything else) I’m going to illustrate the solution of this problem with this latter approach, but
if you are down with using Cramer’s Rule (the fancy name for the firstapproach) so am I.
So we substituteI 3 =I 1 +I 2 into the two voltage equations:3 I 1 + 8I 1 + 8I 2 = 9 (361)
3 I 2 + 8I 1 + 8I 2 = 3 (362)or
11 I 1 + 8I 2 = 9 (363)
8 I 1 + 11I 2 = 3 (364)If we multiply the top equation by 11 and the bottom equation by 8, weget:
121 I 1 + 88I 2 = 99 (365)
64 I 1 + 88I 2 = 24 (366)If we subtract the second equation from the first, we get:
57 I 1 = 75 (367)or
I 1 =
75
57
= 1. 316 (368)
(in Amperes). We substitute this back into:
11
75
57
+ 8I 2 = 9 (369)
so
I 2 = (9− 1175
57
)/8 =− 0. 785 (370)
Finally
I 3 =I 1 +I 2 = 1. 316 − 0 .785 = 0. 531 (371)