Strategies for More Complicated Circuits

Up to this point, we've worked with very simple two-junction circuits, for ease of calculation and presentation. Solving larger circuits can become mathematically complicated, often forcing us to solve multiple simultaneous equations at once. Sometimes you just have to bite the bullet and do the algebra (or break out a matrix, if you've had some linear algebra). But there are some strategies to make the process simpler. To be specific, we will consider this circuit.

Labelling all the currents

The first thing we should do when solving a circuit is to label all the different currents. It can be easy to overlook one or more currents unless one uses a systematic approach. Here's mine:

No two currents into or out of any particular junction can have the same label.

This animation shows the process.

Start with one junction, and assign a different variable (and direction) to each of the currents going into or out of it. Proceed to the next junction and do the same thing, but don't label a current that has already been defined. When you have done this with every junction, you should have labelled all the currents. This method is particularly useful for catching currents like \(I_3\) and \(I_6\) in the movie, currents which don't pass through any battery or resistor. Here's the final result: there are nine variables to solve.

If you look carefully, you may notice that \(I_3=I_6=I_9\). If you see it, feel free to eliminate \(I_6\) and \(I_9\) right away. If you don't you'll still be able to solve the circuit systematically.

Next we'll try to eliminate some of these variables as easily as possible.

Using resistance reduction

One way we can simplify circuits is by using resistance reduction (Section 11.5), replacing resistors that are in series or parallel with their equivalent resistance. In the circuit above, for instance, the two resistors at the bottom (each with resistance 2Ω and conductance ½℧) are clearly in parallel, and can be replaced with a single resistor with conductance ½+½=1℧ (or 1Ω). This eliminates three variables: \(I_7\), \(I_8\), and \(I_9\), which makes it easier to solve for the rest of the currents. Once you find \(I_6\), you can solve for \(I_7\) and \(I_8\) by using a loop rule around the two parallel paths: $$-2I_8+2I_7=0 \implies I_7=I_8$$ and a junction rule where \(I_6\) enters that loop: $$I_6=I_7+I_8 \implies I_7=I_8=\frc2I_6$$

Simple loops

When you use the loop rule, a variable is added to the equation every time you pass through a resistor, because passing through a resistor adds \(\pm IR\) to the tally. If a loop only passes through one resistor, then the equation will only have one variable, which you can solve for right away. I call these simple loops, and it's good strategy to look for them first.
Find the simple loops in this circuit.
There are three of them:
  • The blue loop on the left passes through the 4V battery and a single 2Ω resistor, so the loop equation is \(4-2I_2=0\), giving us \(I_2=2\u{A}\).
  • The green loop on the right is similar. If we go counterclockwise around the loop, we get the equation \(4+2I_4=0\). Notice the plus sign, because our loop goes up the resistor according to the arrow we drew. Thus \(I_4=-2\u{A}\). The minus sign simply means that the current flows up and to the left through that 2Ω resistor, and that's perfectly fine.
  • The last loop, in orange, might have been harder to see, but it only includes the 1Ω resistor at the bottom. (A simple loop can have multiple batteries because batteries just introduce constants into the equation, not variables.) This loop gives us the equation $$4+4-1I_6 \implies I_6=8\u{A}$$

Apply junction rules

Once the simple loops have been found, it can be useful to redraw the circuit, with all the known currents written in, and then look to see if there are any other currents that can be determined via junction rule. You can write out the junction rules by hand, but a visual inspection is often simpler and faster.
For instance, look at the lower-left junction, labelled (a). We see that an 8A current and a 2A current flow into it, while current \(I_1\) flows out of it. Clearly then, \(I_1=10\u{A}\). Similarly, at the junction labelled (b), current \(I_5\) flows in while current \(8+2=10\u{A}\) flows out, so \(I_5=10\u{A}\) as well. Once we know \(I_5\), we can look at the junction (c), where current \(2+I_3\) flows in and \(5\u{A}\) flows out. Clearly \(I_3=3\u{A}\), and the circuit is solved.

Sometimes simple loops and junction rules will not suffice, and you will need to introduce some new loop rules, and maybe solve simultaneous equations. But use these techniques to eliminate as many variables as possible first, and the algebra will be that much easier.

If you found any of the above confusing, this video (which solves the same circuit in the same way) may help.