2015 USA Physics Olympiad Exam Part A 5 Question A2 Consider the circuit shown below.Is is a constant current source,meaning that no matter what device is connected between points A and B,the current provided by the constant current source is the same. 4R 2R 2R ● ● 2R3 A B 4R a.Connect an ideal voltmeter between A and B.Determine the voltage reading in terms of any or all of R and Is. Solution An ideal voltmeter has infinite resistance,so no current flows between A and B. Out of symmetry,the same current must flow down each leg,so the current in a leg isIs/2. Assume the potential at the bottom is zero. The potential at A is the same as the junction to the left of A,or,by simple application of Ohm's law VA-52R-LR. The potential at B is found the same way,so R=2I,R. VB =2 The difference is VA-VB=-Is R. The negative is not important for scoring purposes. b.Connect instead an ideal ammeter between A and B.Determine the current in terms of any or all of R and Is. Solution An ideal ammeter has zero resistance.So the problem is simply finding the current through the effective 6R resistor that connects the two vertical branches.This current will flow to the left. Copyright C2015 American Association of Physics Teachers2015 USA Physics Olympiad Exam Part A 5 Question A2 Consider the circuit shown below. Is is a constant current source, meaning that no matter what device is connected between points A and B, the current provided by the constant current source is the same. 2R 4R 4R 2R Is 4R 2R A B a. Connect an ideal voltmeter between A and B. Determine the voltage reading in terms of any or all of R and Is. Solution An ideal voltmeter has infinite resistance, so no current flows between A and B. Out of symmetry, the same current must flow down each leg, so the current in a leg is Is/2. Assume the potential at the bottom is zero. The potential at A is the same as the junction to the left of A, or, by simple application of Ohm’s law VA = Is 2 2R = IsR. The potential at B is found the same way, so VB = Is 2 4R = 2IsR. The difference is VA − VB = −IsR. The negative is not important for scoring purposes. b. Connect instead an ideal ammeter between A and B. Determine the current in terms of any or all of R and Is. Solution An ideal ammeter has zero resistance. So the problem is simply finding the current through the effective 6R resistor that connects the two vertical branches. This current will flow to the left. Copyright c 2015 American Association of Physics Teachers