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\author{MostAncientDream}
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  /Title (alvl-p2-capacitance-ch6.pdf)
  /Creator (Cheatography)
  /Author (MostAncientDream)
  /Subject (Alvl P2: capacitance (ch6) Cheat Sheet)
}

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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Alvl P2: capacitance (ch6) Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{MostAncientDream} via \textcolor{DarkBackground}{\uline{cheatography.com/168994/cs/42330/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}MostAncientDream \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 8th February, 2024.\\
   Updated 8th February, 2024.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Definitions}}  \tn
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\seqsplit{Capacitance} & the charge stored per unit pd \tn 
% Row Count 2 (+ 2)
% Row 1
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capacitor & component that stores charge. composed of two parallel conducting plates with a dielectric between them \tn 
% Row Count 6 (+ 4)
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\seqsplit{dielectric} & an insulating material placed between the two plates of a capacitor in order to increase the amount of charge it can store \tn 
% Row Count 11 (+ 5)
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time constant & time taken for a capacitor to dishcarge 37\% of its inital charge \tn 
% Row Count 14 (+ 3)
% Row 4
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 & it is equal to the product of capacitance and the resistance of a fixed resistor (that the capcitor is being discharged through) \tn 
% Row Count 19 (+ 5)
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\mymulticolumn{1}{x{5.377cm}}{- the positive side of the battery attracts electrons from one side of the initally uncharged plate  \newline % Row Count 3 (+ 3)
-{}-\textgreater{} this causes the left plate to become positively charged \newline % Row Count 5 (+ 2)
- the negative side of the battery repels these new electrons to the right of the parallel plates  \newline % Row Count 7 (+ 2)
-{}-\textgreater{} causing the right to become positively charged \newline % Row Count 9 (+ 2)
it can also be used to recity ac current by bring the current down gradually instead or sharply (? refer to graph) \newline % Row Count 12 (+ 3)
these plates store these electrons as charge so if the circuit breaks connection with the battery a separate circuit connected to the capacitor can continue to operate with the capacitor acting as a battery until it runs out of charge \newline % Row Count 17 (+ 5)
maximum stored charge = pd across the battery% Row Count 18 (+ 1)
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capacitance & Q/V (farad) \tn 
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V & EQ \tn 
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energy stored in a capacitor & 1/2 QV \tn 
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 & 1/2 CV\textasciicircum{}2\textasciicircum{} \tn 
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V/Vo (V over V inital) & e \textasciicircum{}-t/RC\textasciicircum{} \tn 
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C & Er Eo A / d \tn 
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 & Er -relative permitivity of a dielectric/insulator between plates \tn 
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 & Eo -relative permitivity of free space \tn 
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 & A area of plates \tn 
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 & d distance between plates \tn 
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- & - \tn 
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\mymulticolumn{2}{x{5.377cm}}{in a capcitor half the energy is always lost to heat either in resistor or wires etc \newline (refer to energy = 1/2 QV - this is where the missing energy is going)}  \tn 
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capacitor & series & parallel \tn 
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charge & same & 2x \tn 
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pd & split & same \tn 
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C total & 1/Ct = 1/C1+1/C2 & Ct = C1+C2 \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Decay and Time constant}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{compared to the V-t graph of a charging capacitor (similar to x\textasciicircum{}1/2\textasciicircum{}) which changes to a curved exponential decreasing graph of I-t \newline % Row Count 3 (+ 3)
the V and I-t graphs for a discharging capacitor are the same (same shape as charging I-t) \newline % Row Count 5 (+ 2)
V/Vo = e\textasciicircum{}-t/RC\textasciicircum{} \newline % Row Count 6 (+ 1)
r- resistance \newline % Row Count 7 (+ 1)
c- capacitance \newline % Row Count 8 (+ 1)
when t = RC we get e\textasciicircum{}-1\textasciicircum{} which is 0.37 (37\%) \newline % Row Count 9 (+ 1)
therefore when the time equals the resistance, we get 37\% of the original voltage \newline % Row Count 11 (+ 2)
this is why RC is our time constant - tc% Row Count 12 (+ 1)
} \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Capacitance and dielectrics}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{if you were to increase the distance between the two plates the capacitance would decrese \newline % Row Count 2 (+ 2)
two things can follow from that: \newline % Row Count 3 (+ 1)
1. if the battery was connected - you have a constant v \newline % Row Count 5 (+ 2)
as E = 1/2CV\textasciicircum{}2\textasciicircum{}  \newline % Row Count 6 (+ 1)
the energy would decrease \newline % Row Count 7 (+ 1)
2. if the battery was disconnected - you have a constant Q \newline % Row Count 9 (+ 2)
as E = 1/2Q\textasciicircum{}2\textasciicircum{}/C  \newline % Row Count 10 (+ 1)
energy increases \newline % Row Count 11 (+ 1)
(2 makes sense as if you try to separate two charged plates that are attracting each other then you are putting energy into the system to do this)% Row Count 14 (+ 3)
} \tn 
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% That's all folks
\end{multicols*}

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