2015 USA Physics Olympiad Exam Part A 6 Out of symmetry,the current through each vertical resistance of 2R must be the same,as well as the currents through each vertical resistance of 4R.This will give us a few equations: Is 12+I4 I2=I6+I4 144R 122R+166R Solve, Is 16+214 414=2(I6+I4)+616 or I4=416 and Is=916 or 6= c.It turns out that it is possible to replace the above circuit with a new circuit as follows: B From the point of view of any passive resistance that is connected between A and B the circuits are identical.You don't need to prove this statement,but you do need to find It and Rt in terms of any or all of R and Is. Solution Use the previous results.For a short,all of the current will flow through AB,so It=I6=Is For an open circuit,the potential across AB will be IsR,so Rt=9R. Copyright C2015 American Association of Physics Teachers2015 USA Physics Olympiad Exam Part A 6 Out of symmetry, the current through each vertical resistance of 2R must be the same, as well as the currents through each vertical resistance of 4R. This will give us a few equations: Is = I2 + I4 I2 = I6 + I4 I44R = I22R + I66R Solve, Is = I6 + 2I4 4I4 = 2 (I6 + I4) + 6I6 or I4 = 4I6 and Is = 9I6 or I6 = 1 9 Is c. It turns out that it is possible to replace the above circuit with a new circuit as follows: It Rt A B From the point of view of any passive resistance that is connected between A and B the circuits are identical. You don’t need to prove this statement, but you do need to find It and Rt in terms of any or all of R and Is. Solution Use the previous results. For a short, all of the current will flow through AB, so It = I6 = 1 9 Is. For an open circuit, the potential across AB will be IsR, so Rt = 9R. Copyright c 2015 American Association of Physics Teachers