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% Document Info
\author{cadmiumium}
\pdfinfo{
  /Title (generation-and-transmission-of-electricity.pdf)
  /Creator (Cheatography)
  /Author (cadmiumium)
  /Subject (Generation and Transmission of electricity Cheat Sheet)
}

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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Generation and Transmission of electricity Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{cadmiumium} via \textcolor{DarkBackground}{\uline{cheatography.com/135751/cs/28194/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}cadmiumium \\
  \uline{cheatography.com/cadmiumium} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 2nd June, 2021.\\
   Updated 2nd June, 2021.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
\raggedright
\raggedcolumns

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% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{magnetic flux}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Φ=B⊥A} \tn 
% Row Count 1 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{·B=magnetic field strength (T) \newline ·A=area perpendicular to the magnetic field (m$^{\textrm{2}}$) \newline ·Φ=magnetic flux (Wb)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Faraday's Law}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{ε=-N(∆Φ/∆t)} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{ε=EMF(V), N=number of turns in coil} \tn 
% Row Count 2 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{·(∆Φ/∆t) is the derivative of Φ with respect to time. \newline ·ε graph should be negative when Φ-t graph has +ve gradient \newline ·ε graph should be positive when Φ-t graph has -ve gradient}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Lenz's Law}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{An induced current will flow in a direction such that the magnetic field created by the current will oppose the change in flux that induced the current.} \tn 
% Row Count 4 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Right Hand coil rule:   \newline ·thumb: direction of induced magnetic field  \newline ·fingers: direction of induced current}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{generators and alternators}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f=1/T} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{f=frequency (Hz), T=period of revolution(s) - time taken to complete a full cycle} \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{*max ε when coil is parallel to magnetic field, ie. greatest rate of change} \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{*DC current can only be produced in presence of a split ring commutator} \tn 
% Row Count 7 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Alternator (AC): sinusoidal} \tn 
% Row Count 8 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{DC generator: modulus of AC} \tn 
% Row Count 9 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{·Φ(t)=acos(2πft) \newline ·ε=Φ'(t)=-2aπfsin(2πft)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{electricity recap}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{V=I·R - V(V), I(A), R(Ω)} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{P`supply`=V·I - also: power rating} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{P`dissipated`=I$^{\textrm{2}}$·R=V$^{\textrm{2}}$/R} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Power (W) is the rate of change of energy with respect to time} \tn 
% Row Count 5 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{P=∆E/∆t - gradient of E-t graph} \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{series circuit: \newline ·current (I) is the same through the whole circuit \newline ·flow: from positive to negative terminal \newline ·total resistance (RT) is the sum of individual resistances: R`1`+R`2`+… \newline ·total voltage supplied to a circuit must be equal to the total voltage used around the circuit (sum of voltage drops): V`supply`=V`1`+V`2`+…}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Transformers, comparing AC and DC}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{V`RMS`=V`peak`/√2 , I`RMS`=I`peak`/√2} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{P`avg`= V`RMS`⋅I`RMS` = I`RMS`$^{\textrm{2}}$⋅R = V`RMS`$^{\textrm{2}}$/R - avg power delivered by sinusoidal signal} \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Transformers:} \tn 
% Row Count 4 (+ 1)
% Row 3
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\mymulticolumn{1}{x{5.377cm}}{P`in`=P`out`} \tn 
% Row Count 5 (+ 1)
% Row 4
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\mymulticolumn{1}{x{5.377cm}}{\seqsplit{VI`primary`=VI`secondary`}} \tn 
% Row Count 6 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{V`1`/V`2`=N`1`/N`2`} \tn 
% Row Count 7 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{I`1`/I`2`=N`2`/N`1`} \tn 
% Row Count 8 (+ 1)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{when V`1`\textgreater{}V`2`: Step-down transformer} \tn 
% Row Count 9 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{when V`1`\textless{}V`2`: Step-up transformer} \tn 
% Row Count 10 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{*transformers will not work for converting a DC voltage} \tn 
% Row Count 12 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Transmission of power (Power systems)}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Power loss:} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{P`loss`=I`line`$^{\textrm{2}}$·R`line`} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Voltage drop:} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{V`drop`=I`line`·R`line`} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\seqsplit{I`load`/I`line`=N`1`/N`2`}} \tn 
% Row Count 5 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{·almost all wires have some resistance  \newline ·as electricity passes through the wires, it causes them to heat up, resulting in power loss, and a decrease in the voltage that is available at the load.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{High voltage transmission}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{} \tn 
% Row Count 0 (+ 0)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{·we can reduce power loss by lowering the current in the line \newline ·we can keep the same supply power by increasing the supply voltage \newline ·this is done using transformers.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}