W9_parallel_resonance.eps

Week 5: Resistance 183

Example 5.4.2: A Multiloop Resistance Problem

Although we will have many opportunities to use Kirchhoff’s Rules in thechapters to come, it
is worthwhile to apply it to an archetypical problem where it is necessary to use both rules to
determine the currents in a multiloop circuit with resistors and batteries. The problem doesn’t have
to be particularly difficult, but it does need to illustrate all of the steps required to solve problems
of this time, as well as some of the caveats – places where things onemight try don’t advance you
towards the solution.

I 1 I 3 I 2

Ω

Ω Ω

9 V 3 V

3 3

8

Figure 59: Use Kirchhoff’s Rules to find the three unknown currents:I 1 , I 2 , I 3.

In figure 59, we see a typical arrangement of batteries and resistors in a multiloop problem.
There arethree loopsandthree currentsvisible in the problem (can you see the three loops?). Our
job is tofind the three unknown currentsgiven the information on the figure. We have to do this by
writing Kirchhoff’s loop and current rulesalgebraically, using the unknown currents, and then solve
the resulting system of simultaneous equations to find the currents.

Thefirststep, however, is to identify the loops and choosetentative directionsfor the currents.
We don’t need to worry yet about whether or not they are correct– our eventual answers will tell
us the correct directions by means of theirsignsrelative to our initial assumptions. Let’s redraw
the figure, appropriately decorated with current directions and loops identified:

Ω

Ω Ω

9 V 3 V

3 3

8

I 1 I 3 I 2

3

1 2

Figure 60: Note the loops and current directions identified on the figure.

Now let us write (and identify) all four equations that we can obtain from Kirchhoff’s Rules in
this problem:

9 − 3 I 1 − 8 I 3 = 0 (loop 1) (354)

3 − 3 I 2 − 8 I 3 = 0 (loop 2) (355)

9 − 3 I 1 + 3I 2 −3 = 0 (loop 3) (356)

I 1 +I 2 −I 3 = 0 (junction) (357)

Recall that the potentialdecreaseswhen we go across a resistorin the direction of the current. We do
not write the equation for the bottom junction because it is just -1times the top junction equation
and hence not independent.

We immediately notice that there is a wee problem – we havefour equationsand onlythree
unknowns!This means that our equations cannot all be independent. If you examine the first three
equations, a moment of reflection should convince you that the third equation (for loop 3) is the