2018 USA Physics Olympiad Exam Part A 5 Question A2 For this problem,graphical answers should be drawn on the answer sheets graphs provided.Supporting work is to be written on blank answer sheets.Incorrect graphs without supporting work will receive no partial credit. The current I as a function of voltage V for a certain electrical device is I=l0e-qVo/kBT (eaV/KBT-1) where g is the magnitude of the charge on an electron,kB is Boltzmann's constant,and T is the absolute temperature.To and Vo are non-zero positive constants.Throughout this problem assume low temperature values kBT<qVo. a.On the answer sheets,sketch a graph of the current versus voltage for low temperature values kBT<gVo,clearly indicating any asymptotic behavior. Solution The current is simply proportional to eT-1,which is a shifted exponential.Then I always has the same sign as V,and vanishes when V vanishes.The current grows quickly for high V and approaches a constant for low V. This answer is acceptable,but we can use the condition kBT<qVo to simplify the graph. For negative V,we have I/Io≈e-g6/kBT which is extremely small.For positive V,we have I/L0≈e4V-o)/ksT which is extremely small when V<Vo and extremely large when V>Vo.Then 0V<, as shown below.Accounting for finite temperature,which is not necessary for full credit, simply rounds the corners in all of the graphs. Copyright C2018 American Association of Physics Teachers2018 USA Physics Olympiad Exam Part A 5 Question A2 For this problem, graphical answers should be drawn on the answer sheets graphs provided. Supporting work is to be written on blank answer sheets. Incorrect graphs without supporting work will receive no partial credit. The current I as a function of voltage V for a certain electrical device is I = I0e −qV0/kBT  e qV /kBT − 1  where q is the magnitude of the charge on an electron, kB is Boltzmann’s constant, and T is the absolute temperature. I0 and V0 are non-zero positive constants. Throughout this problem assume low temperature values kBT  qV0. a. On the answer sheets, sketch a graph of the current versus voltage for low temperature values kBT  qV0, clearly indicating any asymptotic behavior. Solution The current is simply proportional to e qV /kBT − 1, which is a shifted exponential. Then I always has the same sign as V , and vanishes when V vanishes. The current grows quickly for high V and approaches a constant for low V . This answer is acceptable, but we can use the condition kBT  qV0 to simplify the graph. For negative V , we have I/I0 ≈ e −qV0/kBT which is extremely small. For positive V , we have I/I0 ≈ e q(V −V0)/kBT which is extremely small when V < V0 and extremely large when V > V0. Then I I0 ≈ ( 0 V < V0, ∞ V > V0 as shown below. Accounting for finite temperature, which is not necessary for full credit, simply rounds the corners in all of the graphs. Copyright c 2018 American Association of Physics Teachers