31 October 2012

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

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