Kasey from Texas emails in with a puzzling issue from AP Physics 2. Paraphrasing:

In the circuits question from the 2015 AP Physics 2 exam [that's problem 2, available at this link], I get the solution to part ii using

*P=I*but I don't get the same answer using^{2}R,*P=V*. Does this equation not work? Shouldn't all versions of the power equation give the same answer? What am I missing?^{2}/R
Your instincts are correct that any form of the power equation should give the same answer. However, you have to be vigilant about using the values of voltage, current, and resistance that are relevant to the circuit element being considered.

Problem 2 on the 2015 AP Physics 2 exam shows a circuit diagram. Before the switch is closed, two identical resistors are in series with a battery. Closing the switch puts another identical resistor in parallel with one of the series resistors. The original AP question labels the important values with variables. For the purposes of this discussion, I'll use a 10 V battery with identical 10 Ω resistors.

The question asks a comparison of the power dissipated in that first resistor -- the one that remains in series with the battery -- before and after the switch is closed. I'll go through the calculation two ways: once with

*P=I*and once with^{2}R,*P=V*.^{2}/R**With**The first bulb takes the entire current in the circuit. Before the switch is closed, that's the voltage of the battery (10 V) divided by the equivalent resistance of the circuit (20 Ω), 0.50 A. So the power dissipated by the first resistor before the switch is closed is (0.50 A)

*P=I*:^{2}R^{2}* (10 Ω) =

**2.5 W**.

After the switch is closed, the circuit's equivalent resistance is now 10 Ω + the 5 Ω equivalent resistance of the parallel combination, for a total of 15 Ω. The total current is now (10 V) / (15 Ω) = 0.67 A; since the first resistor is still in series with the battery, it takes this current. So the power dissipated by the first resistor is (0.67 A)

^{2}* (10 Ω) =**4.5 W**. To answer the originally posed question, the power in the first resistor is multiplied by 1.8 when the switch is closed.**With**

*The voltage across the first bulb is not the entire voltage of the battery. Before the switch is closed, the two series resistors split the 10 V of the battery equally. (You could also see this by doing the complete Ohm's law analysis -- the total current in the circuit is 0.50 A as shown above; multiply*

**P=V**^{2}/R:*V = IR*for each resistor to get 5 V across each.) So the power dissipated by the first bulb initially is (5 V)

^{2}/ (10 Ω) =

**2.5 W**. Same as I calculated before.

After the switch is closed, the voltage across the first bulb changes. Now, the first bulb is in series with the parallel combination. Thus the voltage doesn't split evenly; more goes to the series resistor 'cause it has a larger resistance than the parallel combination. The equivalent resistance of the circuit drops, increasing the total current in the circuit to 0.67 A as shown above. To find the voltage across the first resistor (which takes all 0.67 A 'cause it's in series with the battery), use

*V*= (0.67 A) * (10 Ω) = 6.7 V. Now we can use the power equation with the voltage and resistance corresponding just to the first resistor: (6.7 V)^{2}/ (10 Ω) =**4.5 W**, again the same as before. The power in the first resistor is still multiplied by 1.8 when the switch is closed.**Why would my students get different answers with different forms of the power equation?**Most likely, someone used the voltage of the battery rather than just the voltage across the single resistor in

*P=V*Less likely but still possible* is that the student calculated the current in the first resistor using just the 10 Ω resistance rather than the equivalent resistance of the circuit.

^{2}/R.
*

*Possible? Having graded AP exams for 15 years, I'll tell you, every mistake in the universe is *possible*, just like any location of an electron in a hydrogen atom is *possible*. An individual student's answer merely resolves a wavefunction that maps probabilities of making each possible mistake.*
Greg, Do you think this particular question is pretty much the same as circuit questions from past exams? As in the past, these types of questions are easy if the students have done the related lab work and analysis. How do you do your circuit labs?

ReplyDelete1) Well, the situation is the same as in many previous exams. There are only so many circuits you can make that are simple enough to deal with on a 15-25 minute exam question. The response required, though, is more detailed than anything in AP B. That's what we should expect on P1 and P2.

ReplyDelete2) Circuit labs start with the exercise in this post: http://jacobsphysics.blogspot.com/2010/01/introduction-to-circuits-laboratory.html

Then continue with power in a bulb and switch problems like in this post. Eventually I introduce the P1/P2 vocabulary (i.e. "charge per unit time" simply means current, and "potential energy per charge" means voltage) and explicit reference to Kirchoff's laws. If I have time, I get out the TIPERS book and have students connect circuits to answer the questions experimentally.