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  /Subject (CircuitsFinal2014 Cheat Sheet)
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    {\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{CircuitsFinal2014 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Lini} via \textcolor{DarkBackground}{\uline{cheatography.com/21323/cs/4073/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Lini \\
  \uline{cheatography.com/lini} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 8th May, 2015.\\
   Updated 12th May, 2016.\\
   Page {\thepage} of \pageref{LastPage}.
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  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{4}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 1 - Basics}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{- {\bf{Electric current = (i):}} time rate of change of charge, measured in amperes (A). \newline % Row Count 2 (+ 2)
- {\bf{Charge = (q):}} integral of i \newline % Row Count 3 (+ 1)
- {\bf{Voltage (or potential difference) = (V):}} energy required to move a unit charge through an element \newline % Row Count 6 (+ 3)
- {\bf{Power}} = (W): vi = (i\textasciicircum{}2)R \newline % Row Count 7 (+ 1)
- {\bf{Passive sign convention:}} when the current enters through the positive terminal of an element (p = +vi) \newline % Row Count 10 (+ 3)
{\bf{Remember:}}  \newline % Row Count 11 (+ 1)
+Power absorbed = -Power supplied -{}-\textgreater{} sum of power in a circuit = 0 \newline % Row Count 13 (+ 2)
- {\bf{Energy (J)}} = integral of P% Row Count 14 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 2}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Ohms Law:}} v=iR \newline % Row Count 1 (+ 1)
{\bf{Conductance (G)}} = 1/R = i/v \newline % Row Count 2 (+ 1)
{\bf{Branch:}} single element such as a voltage source or a resistor. \newline % Row Count 4 (+ 2)
{\bf{Node:}} point of connection between two or more branches \newline % Row Count 6 (+ 2)
{\bf{Loop:}} any closed path in a circuit. \newline % Row Count 7 (+ 1)
{\bf{Kirchhoff's current law (KCL):}} algebraic sum of currents entering a node (or a closed boundary) is zero. \newline % Row Count 10 (+ 3)
{\bf{Kirchhoff's voltage law (KVL):}} algebraic sum of all voltages around a closed path (or loop) is zero. \newline % Row Count 13 (+ 3)
{\bf{Voltage D:}} v1 = ((R1) / (R1 + R2)) * v \newline % Row Count 14 (+ 1)
{\bf{Voltage D:}} v2 = ((R2 / (R1 + R2)) * v \newline % Row Count 15 (+ 1)
{\bf{Current D:}} i1 = (R2 * i) / (R1 + R2) \newline % Row Count 16 (+ 1)
{\bf{Current D:}} i2 = (R1 * i) / (R1 + R2)% Row Count 17 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 3 - Methods of Analysis}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Nodal Analysis:}} want to fine the node voltages \newline % Row Count 2 (+ 2)
Step 1:  \newline % Row Count 3 (+ 1)
select reference node \newline % Row Count 4 (+ 1)
    - assign voltages v1 -{}-\textgreater{} vn  to  \newline % Row Count 5 (+ 1)
      remaining nodes \newline % Row Count 6 (+ 1)
Step 2:  \newline % Row Count 7 (+ 1)
apply KCL to each node  \newline % Row Count 8 (+ 1)
    - want to express branch currents in  \newline % Row Count 9 (+ 1)
      terms of voltage \newline % Row Count 10 (+ 1)
Step 3:  \newline % Row Count 11 (+ 1)
solve for unknowns \newline % Row Count 12 (+ 1)
{\bf{Important:}} \newline % Row Count 13 (+ 1)
{\bf{current flows from high to low (+ ==\textgreater{} -)}} \newline % Row Count 14 (+ 1)
{\bf{SuperNode Properties}} \newline % Row Count 15 (+ 1)
1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages \newline % Row Count 18 (+ 3)
2. Supernode had no voltage of its own \newline % Row Count 19 (+ 1)
3. Supernode requires the application of both KCL and KVL \newline % Row Count 21 (+ 2)
{\bf{Mesh Analysis}} \newline % Row Count 22 (+ 1)
Step 1: \newline % Row Count 23 (+ 1)
Assign mesh currents or loops \newline % Row Count 24 (+ 1)
Step 2: \newline % Row Count 25 (+ 1)
Apply KVL \newline % Row Count 26 (+ 1)
    - use OHMS LAW to express voltages in terms of the mesh current \newline % Row Count 28 (+ 2)
Step 3: \newline % Row Count 29 (+ 1)
Solve for the unknown \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 3 - Methods of Analysis (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Supermesh}} \newline % Row Count 1 (+ 1)
    - when two meshes have an independent or dependent CURRENT source  \newline % Row Count 3 (+ 2)
      between them% Row Count 4 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 4 - Circuit Theorems}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Superposition}} \newline % Row Count 1 (+ 1)
principal states that the VOLTAGE ACROSS or CURRENT THROUGH an element in a linear circuit is the SUM of the VOLTAGES OR CURRENTS that are caused after solving for each INDEPENDENT source separately \newline % Row Count 5 (+ 4)
{\emph{How to solve a superposition circuit}} \newline % Row Count 6 (+ 1)
Step 1: Turn OFF ALL independent sources except for ONE ==\textgreater{} find voltage or current \newline % Row Count 8 (+ 2)
Step 2: Repeat above for all other independent sources \newline % Row Count 10 (+ 2)
Step 3: Add all voltages/currents together to find final value \newline % Row Count 12 (+ 2)
{\bf{Thevenin's Theorem}} \newline % Row Count 13 (+ 1)
V(th) = V(oc) \newline % Row Count 14 (+ 1)
circuit with Load: I(L) = V(th) / (R(th) + R(L)) ==\textgreater{} V(L) = R(L) {\emph{ I (L) ==\textgreater{} (R(L) / ((R(th) + R(L)) }} V(th)) \newline % Row Count 17 (+ 3)
{\bf{Norton's Theorem}} \newline % Row Count 18 (+ 1)
R(n) = R(th) \newline % Row Count 19 (+ 1)
I(n) = i(sc) ==\textgreater{} (sc) = short circuit \newline % Row Count 20 (+ 1)
I(n) = V(th) / R(th) \newline % Row Count 21 (+ 1)
{\bf{Maximum Power Transfer}} \newline % Row Count 22 (+ 1)
max power is transferred to the LOAD RESISTOR when the LOAD RESISTOR is EQUAL to the THEVENIN RESISTANCE:  \newline % Row Count 25 (+ 3)
R(L) = R(th) \newline % Row Count 26 (+ 1)
p(max) = V(th)\textasciicircum{}2\textasciicircum{} / 4R(th)% Row Count 27 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 6 - Capacitors and Inductors}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Capacitors}} \newline % Row Count 1 (+ 1)
q = C * v \newline % Row Count 2 (+ 1)
{\bf{capacitance:}} ratio of the charge on one plate to the voltage difference between the two plates \newline % Row Count 4 (+ 2)
i(t) = C(dv/dt) \newline % Row Count 5 (+ 1)
v(t) = 1/C {[}Integral: i(T)dT + v(t0)){]} \newline % Row Count 6 (+ 1)
T = time constant \newline % Row Count 7 (+ 1)
energy (w) = .5Cv\textasciicircum{}2\textasciicircum{} \newline % Row Count 8 (+ 1)
{\bf{Important:}} \newline % Row Count 9 (+ 1)
{\bf{VOLTAGE of a capacitor cannot change instantaneously}} \newline % Row Count 11 (+ 2)
{\bf{Capacitors in Series:}} 1 / Ceq = 1/C1 + 1/C2 + 1/Cn \newline % Row Count 13 (+ 2)
{\bf{Capacitors in Parallel:}} Ceq = C1 + C2 + Cn \newline % Row Count 14 (+ 1)
{\bf{Inductors}} \newline % Row Count 15 (+ 1)
v = L(di / dt) \newline % Row Count 16 (+ 1)
i = (1/L) {[}Integral: (v(T)dT + i(t0){]} \newline % Row Count 17 (+ 1)
energy (w) = .5Li\textasciicircum{}2\textasciicircum{} \newline % Row Count 18 (+ 1)
{\bf{Important:}} \newline % Row Count 19 (+ 1)
{\bf{CURRENT through an inductor cannot change instantaneously}} \newline % Row Count 21 (+ 2)
{\bf{Inductors in Series:}} \newline % Row Count 22 (+ 1)
Leq = L1 + L2 + Ln \newline % Row Count 23 (+ 1)
{\bf{Inductors in Parallel:}}  \newline % Row Count 24 (+ 1)
1/Leq = 1/L1 + 1/L2 + 1/Ln% Row Count 25 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 7 - First Order Circuits}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Source Free RC Circuits}} \newline % Row Count 1 (+ 1)
v(t) = V0 * e\textasciicircum{}-t/T\textasciicircum{}  ==\textgreater{} T = RC \newline % Row Count 2 (+ 1)
{\emph{How to Solve SOURCE FREE RC CIRCUITS}} \newline % Row Count 3 (+ 1)
Step 1: Find v0 = V0 across the capacitor \newline % Row Count 4 (+ 1)
Step 2: Find T (time constant) \newline % Row Count 5 (+ 1)
{\bf{Source Free RL Circuits}} \newline % Row Count 6 (+ 1)
i(t) = I0 * e\textasciicircum{}-t/T\textasciicircum{} ==\textgreater{} T = L / R \newline % Row Count 7 (+ 1)
vr(t) = iR = I0 * Re\textasciicircum{}-t/T\textasciicircum{} \newline % Row Count 8 (+ 1)
{\emph{How to Solve SOURCE FREE RL CIRCUITS}} \newline % Row Count 9 (+ 1)
Step 1: Find i(0) = I0 through the inductor \newline % Row Count 10 (+ 1)
Step 2: Find T (time constant) \newline % Row Count 11 (+ 1)
{\bf{Step response of an RC circuit}} \newline % Row Count 12 (+ 1)
v(t) = V0 when t \textless{} 0 \newline % Row Count 13 (+ 1)
v(t) = Vs + (V0 - Vs)e\textasciicircum{}-t/T\textasciicircum{} when t \textgreater{} 0 \newline % Row Count 14 (+ 1)
v = vn + vf ==\textgreater{} vn = V0e\textasciicircum{}-t/T\textasciicircum{}, vf = Vs(1-e\textasciicircum{}-t/T\textasciicircum{}) \newline % Row Count 16 (+ 2)
OR \newline % Row Count 17 (+ 1)
v(t) = v(infinity) + {[}( v(0) - v(infinity){]}e\textasciicircum{}-t/T\textasciicircum{} \newline % Row Count 19 (+ 2)
{\emph{How to solve a STEP RESPONSE OF AN RC CIRCUIT}} \newline % Row Count 20 (+ 1)
Step 1: Find initial capacitor voltage v0 (t \textless{} 0) \newline % Row Count 21 (+ 1)
Step 2: Find final capacitor voltage v(in) (t \textgreater{} 0) \newline % Row Count 23 (+ 2)
Step 3: Find T (time constant) (t \textgreater{} 0) \newline % Row Count 24 (+ 1)
{\bf{Step response of an RL circuit}} \newline % Row Count 25 (+ 1)
i(t) = i(infiniti) + {[} i(0) - i(infinity){]}e\textasciicircum{}-t/T\textasciicircum{} \newline % Row Count 26 (+ 1)
{\emph{How to solve a STEP RESPONSE OF AN RL CIRCUIT}} \newline % Row Count 27 (+ 1)
Step 1: Find initial inductor current i0 (t = 0) \newline % Row Count 28 (+ 1)
Step 2: Find final final inductor current i(inf) ==\textgreater{} (t \textgreater{} 0) \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 7 - First Order Circuits (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{Step 3: Find T (time constant) (t \textgreater{} 0)% Row Count 1 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 8 - Second Order Circuits}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Source Free RLC Circuits}} \newline % Row Count 1 (+ 1)
v(0) = 1/C {[}integral ( idt = v0 ) from 0 to -infinity{]} \newline % Row Count 3 (+ 2)
i(0) = I(0) \newline % Row Count 4 (+ 1)
{\bf{Determining Dampness}} \newline % Row Count 5 (+ 1)
{\emph{(alpha) = R / (2L)}}   \newline % Row Count 6 (+ 1)
{\emph{(omega w0) = 1 / sqrt(LC)}} \newline % Row Count 7 (+ 1)
{\bf{1 - Overdamped (a \textgreater{} w0)}} \newline % Row Count 8 (+ 1)
i(t) = Ae\textasciicircum{}s1t\textasciicircum{} + Be\textasciicircum{}s2t\textasciicircum{} \newline % Row Count 9 (+ 1)
{\bf{2 - Critically Damped (a = w0)}} \newline % Row Count 10 (+ 1)
s1 = s2 = a \newline % Row Count 11 (+ 1)
i(t) = (A + Bt)e\textasciicircum{}-at\textasciicircum{} \newline % Row Count 12 (+ 1)
{\bf{3 - Underdamped (a \textless{} w0)}} \newline % Row Count 13 (+ 1)
i(t) = e\textasciicircum{}-at\textasciicircum{}(Acos(w0t) + Bsin(w0t)) \newline % Row Count 14 (+ 1)
{\bf{Source Free Parallel Circuits}} \newline % Row Count 15 (+ 1)
roots of characteristic euqation \newline % Row Count 16 (+ 1)
s1,2 = -a (+-) sqrt(a\textasciicircum{}2\textasciicircum{} + w0\textasciicircum{}2\textasciicircum{}) \newline % Row Count 17 (+ 1)
a = 1/(2RC) \newline % Row Count 18 (+ 1)
w0 = 1/sqrt(LC) \newline % Row Count 19 (+ 1)
{\bf{1 - Overdamped (a \textgreater{} w0)}} \newline % Row Count 20 (+ 1)
i(t) = Ae\textasciicircum{}s1t\textasciicircum{} + Be\textasciicircum{}s2t\textasciicircum{} \newline % Row Count 21 (+ 1)
{\bf{2 - Critically Damped (a = w0)}} \newline % Row Count 22 (+ 1)
s1 = s2 = a \newline % Row Count 23 (+ 1)
i(t) = (A + Bt)e\textasciicircum{}-at\textasciicircum{} \newline % Row Count 24 (+ 1)
{\bf{3 - Underdamped (a \textless{} w0)}} \newline % Row Count 25 (+ 1)
i(t) = e\textasciicircum{}-at\textasciicircum{}(Acos(wd(t)) + Bsin(wd(t))) \newline % Row Count 26 (+ 1)
{\bf{Step Response of a SERIES RLC Circuit}} \newline % Row Count 27 (+ 1)
{\bf{1 - Overdamped (a \textgreater{} w0)}} \newline % Row Count 28 (+ 1)
v(t) = Vs + Ae\textasciicircum{}s1t\textasciicircum{} + Be\textasciicircum{}s2t\textasciicircum{} \newline % Row Count 29 (+ 1)
{\bf{2 - Critically Damped (a = w0)}} \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
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\vfill
\columnbreak
\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 8 - Second Order Circuits (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{s1 = s2 = a \newline % Row Count 1 (+ 1)
v(t) = Vs + (A + Bt)e\textasciicircum{}-at\textasciicircum{} \newline % Row Count 2 (+ 1)
{\bf{3 - Underdamped (a \textless{} w0)}} \newline % Row Count 3 (+ 1)
v(t) = Vs + e\textasciicircum{}-at\textasciicircum{}(Acos(wd(t)) + Bsin(wd(t))) \newline % Row Count 4 (+ 1)
{\bf{Step Response of a PARALLEL RLC Circuit}} \newline % Row Count 5 (+ 1)
{\bf{1 - Overdamped (a \textgreater{} w0)}} \newline % Row Count 6 (+ 1)
i(t) = Is + Ae\textasciicircum{}s1t\textasciicircum{} + Be\textasciicircum{}s2t\textasciicircum{} \newline % Row Count 7 (+ 1)
{\bf{2 - Critically Damped (a = w0)}} \newline % Row Count 8 (+ 1)
s1 = s2 = a \newline % Row Count 9 (+ 1)
i(t) = Is + (A + Bt)e\textasciicircum{}-at\textasciicircum{} \newline % Row Count 10 (+ 1)
{\bf{3 - Underdamped (a \textless{} w0)}} \newline % Row Count 11 (+ 1)
i(t) = Is + e\textasciicircum{}-at\textasciicircum{}(Acos(wd(t)) + Bsin(wd(t)))% Row Count 12 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 9 - Sinusoids and Phasors}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{w = omega \newline % Row Count 1 (+ 1)
T = 2*pie / w \newline % Row Count 2 (+ 1)
freq = 1 / T (Hertz) \newline % Row Count 3 (+ 1)
v(t) = v(m)*sin(wt + theta) \newline % Row Count 4 (+ 1)
v1(t) = v(m)*sin(wt) \newline % Row Count 5 (+ 1)
v2(t) = v(m)*sin(wt + theta) \newline % Row Count 6 (+ 1)
sin(A +- B) = sinAcosB +- cosAsinB \newline % Row Count 7 (+ 1)
cos(A +- B) = cosAcosB +- sinAsinB \newline % Row Count 8 (+ 1)
Acos(wt) + Bsin(wt) = C*cos(wt - theta) \newline % Row Count 9 (+ 1)
C = sqrt(A\textasciicircum{}2\textasciicircum{} + B\textasciicircum{}2\textasciicircum{}) \newline % Row Count 10 (+ 1)
theta = tan\textasciicircum{}-1\textasciicircum{} (B/A) \newline % Row Count 11 (+ 1)
{\bf{Complex Numbers}} \newline % Row Count 12 (+ 1)
rectangular form: z = x + jy \newline % Row Count 13 (+ 1)
polar: z = r \textless{} (theta) \newline % Row Count 14 (+ 1)
expolar: z = re\textasciicircum{}j(theta)\textasciicircum{} \newline % Row Count 15 (+ 1)
sin: r (cos(theta) + j*sin(theta)) \newline % Row Count 16 (+ 1)
z = x + jy \newline % Row Count 17 (+ 1)
z1 = x1 + jy1 == r1 \textless{} (theta)1 \newline % Row Count 18 (+ 1)
z2 = x2 + jy2 == r2 \textless{} (theta)2 \newline % Row Count 19 (+ 1)
{\bf{operations}} \newline % Row Count 20 (+ 1)
addition: z1 + z2 == (x1 + x2) + j*(y1 + y2) \newline % Row Count 21 (+ 1)
subtraction: z1 - z2 == (x1 - x2) + j*(y1 - y2) \newline % Row Count 22 (+ 1)
multiplication: z1{\emph{z2 == r1}}r2 \textless{} ((theta)1 + (theta)2) \newline % Row Count 24 (+ 2)
division: z1/z2 == r1/r2 \textless{} ((theta)1 - (theta)2) \newline % Row Count 25 (+ 1)
reciprocal: 1/z = 1/r \textless{} -(theta) \newline % Row Count 26 (+ 1)
square: sqrt(z) = sqrt(r) \textless{} (theta)/2  \newline % Row Count 27 (+ 1)
complex conjugate: z* = x - jy = r \textless{} -(theta) = re\textasciicircum{}-j(theta)\textasciicircum{} \newline % Row Count 29 (+ 2)
{\bf{real vs. imaginary}} \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 9 - Sinusoids and Phasors (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{e\textasciicircum{}+-j(theta)\textasciicircum{} = cos(theta) +- j*sin(theta) \newline % Row Count 1 (+ 1)
cos(theta) = REAL \newline % Row Count 2 (+ 1)
jsin(theta) = IMAGINARY \newline % Row Count 3 (+ 1)
{\bf{voltage-current relationship}} \newline % Row Count 4 (+ 1)
R    v = Ri (time domain)    v = RI (frequency domain) \newline % Row Count 6 (+ 2)
L    v = L(di/dt) (time)         v = jwLI \newline % Row Count 7 (+ 1)
C    i = C(dv/dt) (time)        V = I / jwC \newline % Row Count 8 (+ 1)
{\bf{Impedance vs. admittance}} \newline % Row Count 9 (+ 1)
R    Z = R (impedance)    Y = 1 / R \newline % Row Count 10 (+ 1)
I     Z = jwL                     Y = 1 / jwL \newline % Row Count 11 (+ 1)
C    Z = 1 / jwC               Y = jwC \newline % Row Count 12 (+ 1)
{\bf{Complex Numbers with Impedance}} \newline % Row Count 13 (+ 1)
Z = R + jx = |Z| \textless{} (theta) \newline % Row Count 14 (+ 1)
|Z| = sqrt(R\textasciicircum{}2\textasciicircum{} + X\textasciicircum{}2\textasciicircum{}) \newline % Row Count 15 (+ 1)
(theta) = tan\textasciicircum{}-1\textasciicircum{}(X / R) \newline % Row Count 16 (+ 1)
R = |Z|*cos(theta) \newline % Row Count 17 (+ 1)
X = |Z|*sin(theta)% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Chapter 10 - AC Circuits}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Analyzing AC Circuits}} \newline % Row Count 1 (+ 1)
Step 1: Transform circuit to phasor or frequency domain \newline % Row Count 3 (+ 2)
Step 2: Solve Using Circuit Techniques \newline % Row Count 4 (+ 1)
Step 3: Transform phasor ==\textgreater{} time domain% Row Count 5 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


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

\end{document}