deletions | additions       diff --git a/Once_we_had_taken_values__.tex b/Once_we_had_taken_values__.tex  index 3f7ca4a..8c18c64 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, $10 K \Omega$. Finding the current for $V_{sq}$ 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:   \begin{equation}  10 V_{sq} = \left( 2 e i_{dc} \Delta f \right) \left( G_1 G_2 R \right)^{2}   \end{equation}, where $e$ is the charge of the electron that we are looking for, R is the resistance of $10 K k  \Omega$, 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. \textbf{can you say this more clearly and concisely? }Your readers probably don't need to be told this much detail, particularly if you rewrite the shot equation as $<\delta i^2> / \Delta f $ vs $2 i_{dc}$ with a slope of $e$.