deletions | additions       diff --git a/Once_we_had_taken_values__.tex b/Once_we_had_taken_values__.tex  index 46c5d6a..efe82e9 100644  --- a/Once_we_had_taken_values__.tex  +++ b/Once_we_had_taken_values__.tex 
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Once we had taken values across the photo-diode from $0$ to $120$ mV in steps of $10$ mV, had our error in all the values, we were able to start calculating the current from both multi-meters. In order to find the current from the voltage across the photo-diode, we only had to use Ohm's Law (V = IR) because we knew the value of R, 10K ohms. Finding the current for Vsq was harder because we had to account for the change through the amplifiers and multiplier. The equation we used to find the filtered current was: $10 \textrm{ Vsq} = 2 \textrm{ e} \textrm{ idc } \Delta f (\textrm{G1} \textrm{ G2} \textrm{ Resistance})^2$, where e is the charge of the electron that we are looking for, R is the resistance of $10K$ ohms, and G2 varied as stated above. We solved for $2 \textrm { idc}$ and plotted $2\textrm{ idc}$ versus the current found from the photo-diode. We found the slope of the data points because we wanted to find e. Our value was $1.64 10^{-19}$ coulombs. This happens to be only $2.37$ Our measuremnt is less than $2.4 \%$  percent different from the actual value of the charge of an electron.