## Shooting circuits

I’ve posted before about how I struggle teaching complex circuits (really just circuits that contain batteries and resistors in ways that can’t be analyzed with parallel and series tricks). There you’ll read about how I find that if I just give my students one of the unknowns for free it allows them to show me how well they understand the basic principles of circuits without getting bogged down in the math of, for example, five equations and five unknowns.

I’ve shared the ideas from that post a bunch and occasionally I get feedback that it robs students the ability to actually solve the circuits from scratch, since I’m giving them one of the unknowns for free. This post is about thoughts I’ve had about that, including some more substance to my ideas at the end of that post about guessing and checking.

## Bridge circuit

The gateway drug that demonstrates the need of tools beyond series and parallel tricks is the bridge circuit:

The problem with this circuit is that you can’t model the resistors as a combination of series and parallel elements. Go ahead, try, I’ll wait!

… nope R1 and R2 are not a parallel pair

… nope R1 and R4 are not a series pair

… etc

Ok, now that you’re on board with that, the question is how to analyze such a circuit using the basic principles that went into developing the series and parallel tricks, namely that current flowing into a node flows back out again (conservation of charge or “no piling up!”), batteries raise the voltage from one side to the other by the EMF of the battery, and resistors reduce the voltage from one side to the other in a way that’s proportional to the current flowing through them (and the proportionality constant is conveniently named “resistance”).

Other answers to that question include:

• Kirchhoff’s laws (do a bunch of loops and a bunch of nodes and hope you have the right mix that enables a successful linear algebra solution)
• Mesh approaches that are really the same thing, with just a little different focus
• Go in the lab and measure everything

My answer, as noted in that last post (it was 5 years ago!), is to make a guess for one of the currents and then follow through the ramifications of that guess until you reach a discrepancy. For the circuit above, for example, I would (note that when I say “voltage” I actually mean the voltage difference between that point and the bottom of the battery):

• Start by making a guess for the current through R1
• That enables me to calculate the voltage at the left node
• That enables me to calculate the current through R4
• Those two currents enable me to calculate the current through R3.
• That enables me to calculate the voltage at the right node
• That enables me to calculate the current through R2 (because I know the voltage drop across it
• HERE COMES THE COOL PART
• That enables me two ways to calculate the current through R5:
• One way is to consider the voltage drop across it (which we know) and then determine the current
• The other is to use the current flowing into the right node and make sure nothing piles up

Unless you make a lucky guess, those two calculations will not be the same. I’m calling their difference a “discrepancy”.

So what now? Well, as I stated in the last post, do all that again with a different guess and find out how the discrepancy changes. Since it’s a linear circuit, you then “just” need to extrapolate from those two data points to find out what guess would yield a zero discrepancy.

When I wrote about this 5 years ago, I gave a nod to the fact that it’s a lot of work to do all that. But now that I’ve actually tried it a few times, it really isn’t! The first pass is when you establish the relationships, and the second is easy if you use a tool like a spreadsheet. It also turns out that if your first guess is zero and your second guess is one the extrapolation is really easy as well.

What I mean by that last point is that if d0 and d1 represent the two discrepancies for a guess of zero and one respectively, the correct current is simply d0/(d0-d1).

Here’s an example. Let’s say that R1=1 ohm, R2 = 2 ohm etc and that V=10. Here’s the first pass assuming the current through R1 (labeled I1) = 0:

• I1=0
• Vleft=10
• I4=10/4=2.5 down
• I3= 2.5 left
• Vright= 10+2.5*3 = 17.5
• I2=(17.5-10)/2=3.75 up
• I5a=17.5/5=3.5 down
• I5b=6.25 up

So d1=6.25 – (-3.5)=9.75 (also note that Vright gives you a clue this is a bad guess since you wouldn’t expect any part of the circuit to have a voltage higher than the battery)

Here’s the second pass with I1=1:

• I1=1
• Vleft=10-1*1=9
• I4=9/4=2.25 down
• I3=2.25-1=1.25 left
• Vright=9+1.25*3=12.75
• I2=(12.75-10)/2=1.375 up
• I5a=12.75/5=2.55 down
• 15b=1.25+1.375=2.625 up

So d2=2.625-(-2.55)=5.175. Getting better.

That means that the correct current through R1 is 9.75/(9.75-5.175)=2.13 amps.

Yes, that seems to have gotten ugly, I admit. But repeating identical calculations is what spreadsheets are built for. Here’s one I built for this problem (note that I decided down or right would be considered positive):

Note the zero discrepancy in cell D9! Hmm, I wonder if certain people in my life will read that last sentence and let me have it.

So now I’m starting to think this method has some merit. We’re always talking about the value of spreadsheets in physics teaching (usually lab, but still) and now with this approach you’ve really only got to see what students do for the formulas in column B to see if they get the physics!

Not only would you look at their formulas, but the order they go through the circuit is important and makes me feel that this approach is closer to “problem solving” than “exercise” that Ken Heller is always pestering me about. What I mean is that in the usual Kirchhoff procedure students are given an excellent algorithm that has simple choices involved: What loops and nodes should I do? What direction should I go around the loops? Contrast that with carefully seeing what new piece of information you can discern from the previous step as is needed in this method. I think it involves more decision making. It also has some great teachable moments like above when I pointed out a voltage that was higher than physically possible.

## Why I call it shooting circuits

I was sharing this approach with a colleague yesterday and she said it reminded her of the shooting method for solving second order differential equations. Here’s an example of how I use that to solve for the quantum states of a hydrogen atom. In that method you start with a guess of the wavefunction at one side of a quantum well and then look to see how it screws up on the other side. Then you make an adjustment to your guess and try to extrapolate the results so that it doesn’t screw up on the other side.

So this is a lot like that. What the heck, we’ll call it shooting circuits!

## Series/parallel comparison

Consider this incredibly common circuit:

First let’s consider the work necessary to calculate the current through all the resistors:

• Combine R2 and R3 into Req1
• Combine R1 and Req1 into Req2
• Determine the current through Req2
• Recognize that the current through R1 and Req1 is that same current
• Determine the voltage at the node by finding the voltage drop across R1
• Determine the current through R2 and R3 similarly using the now known voltage drops across them.

It’s interesting that many students think they’re done at step 3 (or possibly 2). They groan when you tell them that they still have to reconstruct the circuit to find all the currents.

Now let’s do it the new way, again using R1=1, R2=2, R3=3 and V = 10:

So it’s 4 different statements of physics (A2:A5) and we’re done! All four of those statements demonstrate the student’s mastery of either Ohm’s law for a resistor or the node law. But remember that the order is interesting too! Can you do it in a different order? Does it work if you choose to make your guess for one of the other currents? Give it a try!

## Pitfalls

I’ve been playing with this quite a bit and haven’t really found many pitfalls. One minor one involves the most basic parallel circuit (one battery, 2 parallel resistors). If you guess the current through one of the resistors you immediately get a discrepancy regarding the voltage drop across that resistor. That’s cool, as then you can apply the method, but you learn nothing about the other resistor! So then you’d have to repeat for that one, I guess. I think that means that a complex circuit that basically has two parallel parts might suffer from that problem.

Here are some starters for you:

• I like this method, but would it work for . . . ?
• I think this method sucks and here’s why . . .
• What’s wrong with ending a sentence with a number and then an exclamation point?
• What circuit drawing software do you use? They really look great!
• Mathematicians would call this method . . .
• Can you tell me more about how you can assume the discrepancy is a linear function of the original guess?
• Of course you can describe a bridge circuit as series and parallel! Here’s how . . .
• Seriously, you want students to do their homework with a spreadsheet!? You’re an idiot
• I’m not sure I understand your pitfall situation. Can you describe it better?
• Here are 7 more pitfalls I thought of within 10 seconds of reading this:
• I clicked through to the old post (which you oddly called your last post – what the heck?) and gave up on you when I saw you hate Kirchhoff’s loop law!
• Would this work for the “you have a cube made of resistors . . .” problem? I hate that problem.

Professor of physics at Hamline University in St. Paul, MN
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### 8 Responses to Shooting circuits

1. ProfeJMarie (Janet Rundquist) says:

No, certain people in your life will not let you have that. What goes around comes around! I mean, if you’re going to ask us to look at cell D9!, show us the actual cell, D362880.

Sheesh.

2. bretbenesh says:

d0/(d0-d1) is a pretty result.

• Andy Rundquist says:

It’s funny that it gave me pause at first because it’s unitless and it’s supposed to represent a current. But that’s because built into it is the “1 Amp” second guess.

3. Andy Rundquist says:

I tried the method with the “resistors on the edge of a cube” problem and found that I had to guess 2 currents to be able to figure everything else out. After some thinking I’m convinced that guessing (0,0), (1,0), and (0,1) for those two while tracking *2* discrepancies allows you to figure everything out. So my guess is that each circuit has a number of guesses, I’ll call it a rank for now, so all the ones in this post are rank 1 and the cube is rank 2. I guess a single battery, single resistor circuit is rank 0. From the perspective of testing students knowledge, that “rank” is also the number you should give them for free if you’re trying to avoid much math/linear algebra.

4. James Gell says:

On the description for the spreadsheet for the current calculation in yellow for the bridge circuit I think the formula should be B9/(B9-C9). Also, I think A4 should be I3 = I1 – I4.

Interesting ideas here. I like the freebie idea in the previous article. Much better to have the students thinking about the physics than simply applying rules and beating their way through a calculation.

• Andy Rundquist says:

Yep to both, thanks! I fixed the C9 in the text but the other is just a typo in the label column of the spreadsheet (I did the right formulas in columns B, C, and D).

5. Dave Wiggins says:

It’s far easier to use what engineers call the loop current method (or mesh analysis). I stumbled across it years ago and taught it in my physics classes (I don’t teach anymore). I was amazed we don’t teach it in physics. A quick Google search should gather the relevant details.

• Andy Rundquist says:

Yeah I once tried to teach a computer how to solve arbitrary circuits and stumbled on that approach. I love it, but I’m intrigued by this approach because it so clearly lets you see how well students can both display their mastery of the simple relationships (ohm’s law and the node law) *and* see how well they can use clues to move through the circuit as they look for relationships.