Circuit misconceptions, and an advance copy of a quiz

On a problem set last week, I gave students the simple circuit shown to the right.  I asked what would happen to various parts of the circuit when I decreased R₂.  One of the questions in particular said, "What will happen to the current flowing from the battery when the value of R₂  is decreased?"

The most common answer:  

"The current will not change, because it's the same battery, so it will always provide the same current." 

Silly students, a battery provides a constant VOLTAGE, not a constant current -- but that's a common misconception in the first week of circuits.

The second most common answer:

"The current will not change, because R₂ is the farthest resistor from the battery, and so the current hasn't reached R₂ yet.

Silly student with a common misconception again.  The "distance from the battery" should never be used to justify anything associated with circuitry, because "distance" from a battery is irrelevant.

I decided to use a quiz to bust these misconceptions.  I've often announced the topic of a quiz the night before, in the hopes that students will target some studying.  This time, I actually sent out the quiz below via email, along with a quick note that discussing the questions in advance was encouraged.  

Did it work?  Yes, in that I've pretty much eliminated the misconceptions I've listed (for now -- I'll have to try again in a couple of months during our review time).  Sure, a few students did poorly, because either they (a) didn't prepare at all, or (b) convinced themselves or their friends that the current of a battery is always constant.  Either way, this exercise was useful!  For the students in category (b), they will never make this mistake again.  Someone in category (a) hangs his head in shame when his classmates tells him, "Jeez, Will, Mr. Jacobs gave us this exact copy ahead of time, it was easy points!"

Common misconceptions -- parallel resistors

Parallel resistors each take the same voltage, which is equal to the total.

Now ask a student: "Two 100 ohm resistors are connected in parallel to a 12 V battery.  Determine the voltage across one of the two resistors."  What does the student say?

Generally, that student reasons, "Parallel resistors take the same voltage.  The battery provides 12 V to two resistors equally, so that's 6 V across each."  D'oh.

How do I attempt to remedy this misconception?  Give everyone the chance to predict and then MEASURE the voltage across several resistors, as in this laboratory exercise.  When a student comes to my desk for me to sign off on his correct measurement, I throw the common misconception in his face.  I say, "Hey, Will, that doesn't make sense.  Seems to me, you've got two parallel resistors here, you should only get half the battery's voltage across each."  Will generally has two points to his rebuttal:  (1) "That's not the correct rule, Mr. Jacobs, the voltage across parallel resistors is equal to the total." And, most importantly, (2) "That's not what I measured.  I get the same voltage across everything."

Two Masses and a Pulley, and a New Misconception

The badly sketched picture to the right shows a classic mechanics problem.  Two equal masses are connected by a string over a pulley.  In this case, the table is frictionless.

Typically, a student is asked to determine the tension in the rope and the acceleration of the masses.  Great -- that's (mg)/2 and g/2.*  This problem is richer, though, than a mere calculation might suggest.  Take a look at a quiz I gave the other day:

* The quick way to get this is to consider both objects as a single system.  The net force on that system is the weight of the hanging mass, mg; the mass of the system is 2m.  By Newton's second law, a = g/2.

1.       A block of mass m is attached over a pulley to another hanging mass m, as shown above.   The surface is frictionless.  The system is released from rest.

     

(a)    What is the direction of the hanging mass’s acceleration?  Explain.

(b)   Is the acceleration of the hanging mass greater than, less than, or equal to g?  Explain.

(c)    Is the tension in the rope greater than, less than, or equal to mg?  Explain.

2.       A block of mass m is attached over a pulley to another hanging mass m, as shown above.   The surface is frictionless.  This time, the top block is given an initial velocity to the left and released. 

(a)    What is the direction of the hanging mass’s acceleration?  Explain.

(b)   Is the acceleration of the hanging mass greater than, less than, or equal to g?  Explain.

(c)    Is the tension in the rope greater than, less than, or equal to mg?  Explain.

Ideally, 1(a) is answered with a kinematic approach -- the hanging mass is speeding up and moving down, so acceleration is also down.  For 1(b), I've defined "free fall" as the situation in which no forces besides weight are acting.  Since a tension acts upward on the hanging mass, the mass is not in free fall and the acceleration is less than g.*  And in 1(c), acceleration is downward, so net force must also be down.  That means down forces greater than up forces, so the tension is less than the weight.

* Okay, sure, if the upward tension is twice the block's weight, the acceleration could be g, upward.  That's highly unlikely in hanging-block-and-pulley problems.  

Of course, question 2 is identical to question 1!  The hanging mass is moving up but slowing down, so acceleration must still be downward.  (Or, one could argue that the block on the table still experiences only one horizontal force, that of tension, so its acceleration must be to the right; the blocks must move as a unit, so the hanging block has downward acceleration.)  Once it's established that acceleration is still down, questions 2(b) and 2(c) follow as in 1(b) and 1(c).

By far the most common misconception here is that the net force must be in the direction of movement.  A student will commonly get question one reasonably correct, but then say "the block is moving upward, so up forces must be bigger than down forces."  This question is just one more salvo in my arsenal aimed at that piece of nonsense.  

Another typical misconception is that in question 1, since the hanging block is falling near earth, its acceleration must be g.  That's taken care of with a request to state the definition of free fall and a sheepish look from the student.

And, a common mistake is to justify (a) and (c) with circular reasoning:  The acceleration is downward because the weight is greater than the tension; the tension is less than the weight because the acceleration is downward.  This student doesn't earn full credit, but I'm not worried so much about his comprehension. 

I discovered a new misconception today, though.  One of my brighter students said acceleration was equal to g, and he stated the definition of free fall accurately.  He asked, "since the surface is frictionless, the block on the table doesn't require any force to move.  So why won't the rope will be slack, the tension zero, and gravity the only force on the hanging block?"

At first I was flummoxed.  I set up two carts on my track, and showed him that the string was in fact not slack.  But why on earth would he think that no friction leads to a slack rope?

In further conversation, I discovered that he was referring back to our class's multiple conversations about how no net force is necessary for motion at constant speed in a straight line.  A mass on a frictionless track, once moving, keeps moving, even without any tension to pull it.  My student wasn't processing that this block on this surface was accelerating, not moving at constant speed.  Once I pointed out how the blocks must move together, and therefore accelerate together, he got it.

Bert, Ernie, Oscar, and nailing first-month misconceptions

I don’t care how good your lectures are, I don’t care how well your textbook presents the material, your students are going to internalize misconceptions about physics concepts.

Sure, some of these can be avoided by careful planning – for example, I don’t introduce Coulomb’s law for the force between electrical point charges at all. If I do, every problem that says anything about electricity automatically brings forth F = kQQ/r². Instead, I start with the concept of an electric field; several days later, I explain that a point charge can CREATE an electric field by the equation E = kQ/r²; and then the equation for the force on a nearby charge follows from F = qE.

Nevertheless, even with my oh-so-careful presentation, I have to hammer the idea that the equation for the electric field created by a point charge only applies in that particular circumstance. They see questions on fundamentals quizzes asking when the equation is valid; they lose mongo points if they ever even write the equation in an illegal circumstance. My specially planned lectures, which evolved from 13 years of experience, still do not preempt misconceptions.

My attack strategy for misconceptions must necessarily take shape early in the course. In the very first month, we usually present one of the more conceptually difficult topics of the year: Newton’s second law. No matter how careful my presentations, students still try to set any old force (rather than the net force) equal to ma; they still assume that a force is required for motion to occur, assuming in their brains that F_net = mv rather than ma.

I don’t believe in a magic technique to eliminate misconceptions. Rather, we must fight a war of attrition, converting one student at a time through a multi-faceted approach. Docking points on homework, asking the same question numerous different ways on fundamentals quizzes, multiple choice questions, in-class reminders, et cetera are all legitimate weapons in our attack on misconceptions. Today’s post demonstrates another part of the arsenal: make the student recognize the misconception when someone else writes it. Then, make that student write out an explanation for why this someone else has said something silly.

Below is a problem I use as a quiz in AP physics, or on a concepts test in general physics.

(The saddest part is that I often have to explain the use of the term “muppet.” RIP, Jim Henson.)

Bert, Ernie, and Oscar are discussing the gas mileage of cars. Specifically, they are wondering whether a car gets better mileage on a city street like Route 15 in Orange, or on a freeway like I 64. All agree (correctly) that the gas mileage of a car depends on the force that is produced by the car’s engine – the car gets fewer miles per gallon if the engine must produce more force.

Below is the statement made by each muppet – one statement is completely correct, and two contain errors. Identify the correct statement. For the others, explain thoroughly the error in physics (Free body diagrams may be useful).

Bert says: Gas mileage is better on I 64. On route 15 in town the car is always speeding up and slowing down because of the traffic lights, so since Fnet=ma and acceleration is large, the engine must produce a lot of force. However, on I 64, the car moves with constant velocity, and acceleration is zero. So the engine produces no force, allowing for better gas mileage.

Ernie says: Gas mileage is better on route 15. On route 15, the speed of the car is slower than the speed on the freeway. Acceleration is velocity divided by time, so the acceleration on route 15 is smaller. Since Fnet=ma, then, the force of the engine is smaller on route 15 giving better gas mileage.

Oscar says: Gas mileage is better on I 64. The force of the engine only has to be enough to equal the force of friction and air resistance – the engine doesn’t have to accelerate the car since the car maintains a constant speed. Whereas on route 15, the force of the engine must often be greater than the force of friction and air resistance in order to let the car speed up.