Exercise:
The current vs. voltage and resistance vs. current characteristics of a small incandescent light bulb  V/. W is displayed in the two following graphs: center includegraphicswidthtextwidth#image_path:light-bulb-i-vs-v-# includegraphicswidthtextwidth#image_path:light-bulb-r-vs-i-# center According to the second graph the resistance is proportional to the current. abcliste abc Show that the proportionality between resistance and current is equivalent to the fact that the current is proportional to the square root of the voltage. Verify that the fit parameters are compatible. abc The light bulb and a RO resistor are connected in series to a voltage supply with VtotO. With the help of the current vs. voltage characteristic determine the current flowing through this circuit and the partial voltages across the resistor and the light bulb. abc Calculate the same quantities using the fit function for the current through the light bulb. abcliste

Solution:
abcliste abc From the expression for the resistance of the light bulb sscRL k I it follows for the voltage that sscDelta VL sscRL I k I^ Solving for the current I leads to I sqrtfracsscDelta VLk sqrtfracDelta V_k fracsscDelta VLDelta V_ sqrtfracDelta V_ksqrtfracsscDelta VLDelta V_ I_ sqrtfracsscDelta VLDelta V_ This is exactly the fit function in the current vs. voltage diagram. For the value of the fit parameter I_ we find I_ IfitCF sqrtfracVfitk IfitC approx resultIfitCP- This agrees with the fit parameter in the diagram. Remark: The insertion of Delta V_ leads to a more natural unit for I_ and removes the unit from the argument of the square root. abc The goal is to express the current through the resistor as a function of the voltage across the light bulb and add the corresponding graph to the current vs. voltage characteristic. Since the current through the resistor and the light bulb has to be the same the solution is given by the ersection of the two graphs. For a voltage sscDelta VL across the light bulb the voltage across the resistor is given by sscDelta VR sscDelta Vtot-sscDelta VL The current through the resistor can be expressed as I fracsscDelta VRRfracsscDelta Vtot-sscDelta VLR This is a straight line with axis ercept I IyF fracVtotR Iy approx IyP- and slope m slopeF fracR slope approx slopeP- We can add this graph to the current vs. voltage diagram of the light bulb: center includegraphicswidthtextwidth#image_path:light-bulb-i-vs-v-with-resistor# center The two graphs ersect at I resultIcalcP- sscDelta VL resultVLS The partial voltage across the resistor is sscDelta VR sscDelta Vtot-sscDelta VL Vtot-VLS resultVRS abc The total voltage in the series circuit is given by the of the partial voltages sscDelta VR across the resistor and sscDelta VL across the light bulb: sscDelta Vtot sscDelta VR+sscDelta VL R I+sscRL IR I+k I^ This is a quadratic for the current I: k I^+R I-sscDelta Vtot The solution is given by I IcalcF frac-R+sqrtR^+ k Vtot k Icalc approx resultIcalcP- The partial voltages can now be calculated as sscDelta VR VRF R Icalc VR approx resultVRS sscDelta VL VLF k Icalc^ VL approx resultVLS The results are in agreement with the graphical method above. abcliste