Zen and the art of predicting voltage across series resistors

I have a circuit in which a 14 V battery is connected to a 15 ohm and a 25 ohm resistor in series.  What's the current through and the voltage across each resistor?

In my honors-level classes, I teach a mathematical solution using the VIR chart.  They calculate the equivalent resistance of 40 ohms; use ohm's law on the total circuit to get a current of 0.35 A; recognize that series resistors each take that same 0.35 A current; then multiply across the rows of the chart with ohm's law to get 5.3 V and 8.7 V across the resistors.

In conceptual, though, we don't use a calculator, and I want to minimize (not eliminate) calculation, anyway.  So we approach this problem slightly differently.  

Take a look at the worksheet we've used in laboratory to learn how to deal with series resistors.  Students must justify their answers to each question thoroughly.

1. Which resistor carries a larger current through it?

We start with a not-so-subtle reminder of the rule that each resistor carries the same current. 

2. Which resistor takes a larger voltage across it?

Now, I insist on an equation justification with ohm's law.  V = IR; I is the same for each because series resistors carry the same current.  By the equation, then, the larger 25 ohm resistor takes the larger voltage.

3. What is the equivalent resistance of the circuit?

Fact: the equivalent resistance of series resistors is the sum of the individual resistances.  Just add 'em up to get 40 ohms.

4. Calculate the current through the circuit.

Only now do we do a calculation.  We've learned that ohm's law applies for the total voltage and resistance in a circuit.  So, we simply say that I = V/R, with V = 14 V and R = 40 ohms.  This makes the current 14/40 amps. Yes, I allow that as the answer -- we aren't using calculators, and I don't want to mess with issues of significant figures and decimals.  Look, go ahead and haul me before the Klingon Death Tribunal for my sins.  I'm looking forward several years.  When they see circuits in a senior honors class, they'll be able to call this 400 milliamps.  For now, I'm happy that they know which values to plug in to ohm's law.

5. Estimate the voltage across and current through each resistor in the chart below.  You may use fractions for current, but not for voltage.  Answers without units earn no credit.  

Here's the Zen.  Not calculating current through each resistor -- that's the same 14/40 amps through each.  The Zen is the estimate of voltage.  I'm NOT teaching them to multiply the resistance of each resistor by the current.  Nor am I teaching them to proportionalize the voltage according to the resistance.  Nope.  I'm just saying "estimate."  

All I am looking for is an answer that fits the facts they've already stated:  the voltage across each must add to 14 V, and the 25 ohm resistor must take greater voltage.  Some students will guess 2 V and 12 V; some will guess 6 V and 8 V.  I don't care.

6.  Now, set up the circuit, and MEASURE the voltage across each resistor, and across the battery.  Record your results here.

And now the Zen must be reconciled with reality.  They make the voltage measurements, and see how the voltage is distributed:  in this case, about 5 V and 9 V.  Some students got the estimate right -- they get candy.  I praise everyone else for a "good guess;" we look to see whether their guess was high or low for the 25 ohm resistor.  

I don't teach anything, still; instead, I hand out a new sheet, with different resistors.  They fill it in again, all the way from the beginning, steps 1-6.

As you may or may not suspect, by the second or third time most students are getting pretty dang close to the right voltages.  Some folks discover the proportionality rule for themselves.  Others just recognize that "close" resistors demand "close" voltages, and "far apart" resistors demand disparate voltages.  

To me, this process is teaching good physics.  I've taught the calculations for series resistors for ages, and I've been repeatedly frustrated by students who can make calculations well but can't answer simple conceptual questions like "which takes the bigger voltage."  And as well by students who frustrate themselves because they predicted 8.25 V but only measured 8.22 V.

By the second day of filling out these sheets and measuring voltages, these freshmen are getting almost bored with the process.  That's the sign I'm looking for.  When I start to see faces saying "Gawd, not again with the voltage question..." that's when I know it's time to move on.  I give the faster guys a sheet with three, not two, resistors; then, after a multiple choice quiz, we move on to resistors in parallel.  

I'll teach parallel resistors the exact same way.

GCJ

Using a lookup table for a conceptual physics lab

In conceptual physics, I want to do an experiment with a 60 Hz frequency generator and waves on a string.  The setup is shown in the picture to the right: the hanging mass is varied, varying the tension in the string and thus the wave speed and the wavelength.  We move the generator left and right until the standing waves are clear; then we measure the wavelength with a ruler.

I want to plot wave speed vs. wavelength, so that the slope of the straight-line graph will be the 60 Hz frequency.  

Problem is, I don't have an instrument to measure wave speed on the string.  In AP physics, I'd just show the students the equation 

and let them figure out the wave speed for themselves.

Well, this is 9th grade conceptual physics. Most of my students either have not completed algebra 1; most wouldn't know a square root if it bit them on the arse.*  I can not expect my class to be able to plug into this formula.  But I still need them to be able to graph a wave speed, knowing only the mass of the hanging mass.

*That happened to me once.

One thought I had was to create a quick app to make the calculation:  On an iphone or ipad, it could ask "What's the hanging mass?"  Then, using the linear mass density value I measured for the string before class, I could program the app* to spit out "the wave speed is 3000 cm/s."  Yes, I know I could do something like this in excel or on wolfram alpha, perhaps, but anything beyond a mass input in grams followed by a speed output in cm/s is too complicated for me.

*That is, if I knew how to program ios apps.  Hey, now, if I had access to a 1985 version of applesoft basic, I'd pwn all of ya in a programming contest.  And I'd have that "app" ready in five minutes.

Without the ability to make the program I want, I realized that I could go all 1940s and just create a lookup table.  Excel will do the calculation... in fact, I learned how to get excel to round the speeds to two significant figures.  So I put mass values from 5 g to 300 g in one column.  I made excel use the equation above to calculate the wave speed in units of cm/s.  

Then I just printed the two columns.  I'll hand this out to each lab group.  I think it's totally reasonable to expect freshmen to use this table to relate the hanging mass to the wave speed... then to graph wave speed on the vertical, and the measured wavelength on the horizontal.

GCJ