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(eqV/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.
Shown is a schematic for the device. Positive voltage means that the electric potential of the left hand side of the device is higher than the right hand side. For this device, I0=25μA and V0=1.0V.
[Figure: Device schematic showing terminals labeled VL and VR]
Below is a circuit made up of these elements. The voltage supplied the circuit is sinusoidal, VAB=VA−VB=Vssinωt, and is also shown on answer sheets. The resistance is R=5.0Ω and Vs=5.0V.
[Figure: Circuit diagram with voltage source between points A and B, resistor R, and device between points C and D]
b. Sketch the potential difference VCD=VC−VD as a function of time on the answer sheet. For your convenience, VAB is shown in light gray. Assume that VAB has been running for a long time.
A capacitor is connected to the circuit as shown below. The capacitance is C=50mF.
[Figure: Modified circuit with capacitor C added in parallel with the device]
c. Sketch the new potential difference VCD=VC−VD as a function of time on the answer sheet. For your convenience, VAB is shown in light gray. Assume that VAB has been running for a long time.