path: root/master.tex

\documentclass{general}
\usepackage{macros}

\title{ELEC 342 Lab 1\\AC/DC Circuits and Basic Measurements}
\author{Manav Khangura \and Warrick Lo \and Yuvraj Chadha}
\date{}

\DeclareSIUnit{\VAR}{VAR}

\begin{document}

\vfill

\maketitle

\vfill

\begin{centering}
	\Large

	Lab Experiment: 1 \\
	Section: L2C \\
	Bench: \#7

	\vfill

	\begin{tabular}{|c|c|c|c|}
		\hline
		Partners & Student ID & Participation & Signatures \\
		\hline
		\hline
		Manav Khangura & 16725475 & \qty{33.3}{\%} & \includegraphics[height=20pt]{images/manav.png} \\
		Warrick Lo & 47938733 & \qty{33.3}{\%} & \includegraphics[height=20pt]{images/warrick.png} \\
		Yuvraj Chadha & 23005697 & \qty{33.3}{\%} & \includegraphics[height=20pt]{images/yuvraj.png} \\
		\hline
	\end{tabular}

	\vfill

	Date(s) Performed: 30 January, 13 February \\
	Date Submitted: 20 February 2026 \\
\end{centering}

\vfill

\newpage

\tableofcontents

\listoftables
\listoffigures

\newpage

\phantomsection
\addcontentsline{toc}{section}{Pre-Lab}
\section*{Pre-Lab}

\textbf{Warrick Lo, 09 February 2026}

Using $S = P + \j Q = VI$, we get
\begin{align*}
	S &= I^2 Z = I^2 \left(R + \j X\right) \quad \text{and} \\
	S &= \frac{V^2}{Z} = \frac{V^2}{R + \j X}.
\end{align*}
Thus, we can solve for P and Q when the impedances are all either purely
resistive or reactive:
\begin{subequations}
	\begin{align}
		P &= I^2 R, \label{eq:p:ser} \\
		&= \frac{V^2}{R}, \quad \text{and} \label{eq:p:par} \\
		Q &= I^2 X, \label{eq:q:ser} \\
		&= \frac{V^2}{X}. \label{eq:q:par}
	\end{align}
\end{subequations}

Using the power triangle and the definition of the power factor angle,
\begin{align*}
	\cos\phi = \frac{P}{S}, \quad \tan\phi = \frac{Q}{P},
\end{align*}
and thus,
\begin{align}
	Q &= P \tan\phi. \label{eq:q}
\end{align}

\textbf{Resistance}

\begin{subequations}
	\begin{alignat}{3}
		\intertext{Series resistance:}
		R &= \frac{P}{I^2} & \quad\quad\quad &
			\text{Using equations \eqref{eq:p:ser} and \eqref{eq:q}} \\
		\intertext{Parallel resistance:}
		R &= \frac{V^2}{P} & \quad\quad\quad &
			\text{Using equations \eqref{eq:p:par} and \eqref{eq:q}}
	\end{alignat}
\end{subequations}

\textbf{Inductance}

\begin{align}
	\j X_L &= \j\omega L \notag \\
	L &= \frac{X_L}{\omega} \label{eq:l}
\end{align}

\begin{subequations}
	\begin{alignat}{3}
		\intertext{Series inductance:}
		L &= \frac{Q}{\omega I^2} & \quad\quad\quad &
			\text{Using equations \eqref{eq:q:ser} and \eqref{eq:l}} \notag \\
		&= \frac{P \tan\phi}{\omega I^2} & \quad\quad\quad &
			\text{Using equation \eqref{eq:q}} \notag \\
		&= \frac{P \tan\phi}{2\uppi f I^2} \\
		\intertext{Parallel inductance:}
		L &= \frac{V^2}{\omega Q} & \quad\quad\quad &
			\text{Using equations \eqref{eq:q:par} and \eqref{eq:l}} \notag \\
		&= \frac{V^2}{\omega P \tan\phi} & \quad\quad\quad &
			\text{Using equation \eqref{eq:q}} \notag \\
		&= \frac{V^2}{2\uppi f P \tan\phi}
	\end{alignat}
\end{subequations}

\textbf{Capacitance}

\begin{align}
	-\j X_C &= \frac{1}{\j \omega C} = \frac{-\j}{\omega C} \notag \\
	C &= \frac{1}{\omega X_C} \label{eq:c}
\end{align}

\begin{subequations}
	\begin{alignat}{3}
		\intertext{Series capacitance:}
		C &= \frac{I^2}{\omega Q} & \quad\quad\quad &
			\text{Using equations \eqref{eq:q:ser} and \eqref{eq:c}} \notag \\
		&= \frac{I^2}{\omega P \tan\phi} & \quad\quad\quad &
			\text{Using equation \eqref{eq:q}} \notag \\
		&= \frac{I^2}{2\uppi f P \tan\phi} \\
		\intertext{Parallel capacitance:}
		L &= \frac{Q}{\omega V^2} & \quad\quad\quad &
			\text{Using equations \eqref{eq:q:par} and \eqref{eq:c}} \notag \\
		&= \frac{P \tan\phi}{\omega V^2} & \quad\quad\quad &
			\text{Using equation \eqref{eq:q}} \notag \\
		&= \frac{P \tan\phi}{2\uppi f V^2}
	\end{alignat}
\end{subequations}

\textbf{Instantaneous Power}

\begin{align}
	\cos\alpha \cos\beta
		= \frac{1}{2} \left( \cos(\alpha+\beta) + \cos(\alpha-\beta) \right)
		\label{eq:cos}
\end{align}

\begin{align*}
	v(t) &= \sqrt{2} V_\text{RMS} \cos\left(\omega t + \phi_v\right) \\
	i(t) &= \sqrt{2} I_\text{RMS} \cos\left(\omega t + \phi_i\right)
\end{align*}

\begin{align}
	p(t) &= v(t) i(t)
		\notag \\
	&= 2 V_\text{RMS} I_\text{RMS} \cos\left(\omega t\right)
		\cos\left(\omega t + \phi \right)
		\notag \\
	&= V_\text{RMS} I_\text{RMS}
		\left( \cos\phi + \cos(2\omega t + \phi) \right)
\end{align}

\textbf{Complex Power}

\begin{subequations}
	\begin{align}
		S &= P + \j Q
			\notag \\
		&= VI \cos\phi + \j VI \sin\phi
			\notag \\
		P &= VI \cos\phi \\
		Q &= VI \sin\phi
	\end{align}
\end{subequations}

\textbf{Three-Phase Power}

\begin{align*}
	V_\text{line} &= \sqrt{3} V_\text{phase}
\end{align*}

\begin{subequations}
	\begin{align}
		P &= 3 V_\text{phase} I_\text{phase} \cos\phi \\
		&= \sqrt{3} V_\text{line} I_\text{line} \cos\phi
	\end{align}
\end{subequations}

\begin{subequations}
	\begin{align}
		Q &= 3 V_\text{phase} I_\text{phase} \sin\phi \\
		&= \sqrt{3} V_\text{line} I_\text{line} \sin\phi
	\end{align}
\end{subequations}

\newpage

\phantomsection
\addcontentsline{toc}{section}{Lab}
\addcontentsline{toc}{subsection}{Task 1}
\section*{Task 1A. Measurements with RL Load Box}

\begin{table}[H]
	\centering
	\begin{tabular}{*{8}{c}}
		\toprule
		Measurement & $V_\text{1(RMS)}$ (\si{\V}) & $I_\text{1(RMS)}$ (\si{\A})
			& $\phi_1$ (\si{\degree}) & $P_\text{1(AVG)}$ (\si{\W}) & $Z_\text{eq}$ (\si{\ohm})
			& $R_\text{eq(series)}$ (\si{\ohm}) & $L_\text{eq}$ (\si{\mH}) \\
		\midrule
		One inductor & 14.11 & 0.64 & 85.4 & 0.72 & 22.18 & 1.758 & 57.95 \\
		Three inductors & 14.07 & 1.98 & 85.2 & 2.33 & 7.11 & 0.594 & 18.77 \\
		\bottomrule
	\end{tabular}
	\caption[Single-phase parallel inductors]{Single-phase parallel inductors.}
	\label{tab:1a}
\end{table}

\begin{itemize}
	\item \textbf{How is this inductor different from an ideal inductor element?} \\
		The inductor has many parasitic elements, including physical resistance
		from the metal conductor, self-capacitance, magnetic flux leakage,
		radiation losses at high frequencies, and core losses if it's not an
		air-core inductor.
	\item \textbf{Are the calculated values of inductance close to the expected value?
		What is the difference?} \\
		The nominal value for each inductor, as indicated on the load box, is \qty{40}{\mH}.
		With three inductors in parallel, this gives us an equivalent inductance of
		\qty{13.33}{\mH}. The measured values are higher, at \qty{57.95}{\mH} for the
		single inductor and \qty{18.77}{\mH} for the parallel inductors, giving a deviation of
		\qty{45}{\%} and \qty{41}{\%}, respectively. This is a relatively high amount
		of deviation and can be attributed to the parasitic properties of the
		inductor described above.
\end{itemize}

\section*{Task 1B. Measurements with RC Load Box}

\begin{table}[H]
	\centering
	\begin{tabular}{*{8}{c}}
		\toprule
		Measurement & $V_\text{1(RMS)}$ (\si{\V}) & $I_\text{1(RMS)}$ (\si{\A})
			& $\phi_1$ (\si{\degree}) & $P_\text{1(AVG)}$ (\si{\W}) & $Z_\text{eq}$ (\si{\ohm})
			& $R_\text{eq(parallel)}$ (\si{\ohm}) & $C_\text{eq}$ (\si{\uF}) \\
		\midrule
		One capacitor & 14.07 & 0.22 & -89.6 & 0.022 & 62.82 & 0.4545 & 40.74 \\
		Three capacitors & 14.12 & 0.68 & -89.8 & 0.034 & 20.47 & 0.0735 & 125.9 \\
		\bottomrule
	\end{tabular}
	\caption[Single-phase parallel capacitors]{Single-phase parallel capacitors.}
	\label{tab:1b}
\end{table}

\begin{itemize}
	\item \textbf{How is this capacitor different from an ideal capacitor element?} \\
		The capacitor has physical resistance from the metal conductor, self-inductance,
		fringing field losses, leakage current through the dielectric, and the skin effect
		at high frequencies, among other losses.
	\item \textbf{Are the calculated values of capacitance close to the expected value?
		What is the difference?} \\
		The nominal value for each capacitor, as indicated on the load box, is \qty{40}{\uF}.
		With three capacitors in parallel, the equivalence capacitance is \qty{120}{\uF}.
		The measured values are slightly higher, at \qty{40.74}{\uF} for the single capacitor
		and \qty{125.9}{\uF} for the parallel capacitors. This gives us a deviation of
		\qty{1.9}{\%} and \qty{4.9}{\%}, respectively, which is a reasonably low amount
		of deviation that can be attributed to the real-world properties of the
		capacitor described earlier.
\end{itemize}

\section*{Task 1C. Parallel Connection of RLC Components}

\begin{table}[H]
	\centering
	\begin{tabular}{*{8}{c}}
		\toprule
		Measurement & $I_\text{1(RMS)}$ (\si{\A}) & $\phi_1$ (\si{\degree})
			& $I_\text{2(RMS)}$ (\si{\A}) & $\phi_2$ (\si{\degree})
			& $I_\text{3(RMS)}$ (\si{\A}) & $\phi_3$ (\si{\degree})
			& $P_\text{1(AVG)}$ (\si{\W}) \\
		\midrule
		$R_1$ & 0.71 & 0.2 & --- & --- & --- & --- & 10.0 \\
		$R_1$ and $L_1$ & 0.99 & 39.8 & 0.64 & 85.2 & --- & --- & 10.7 \\
		$R_1$, $L_1$, $C_1$, $C_2$, $C_3$ & 0.79 & -1.0 & 0.63 & 85.1 & 0.67 & -89.5 & 10.7 \\
		\bottomrule
	\end{tabular}
	\caption[Parallel connection of RLC components]
		{Parallel connection of RLC components.
		Supply voltage $V_1 = 14.16 \angle -1.7 \si{\degree}$ \si{\V} RMS.}
	\label{tab:1c}
\end{table}

\begin{itemize}
	\item \textbf{How has the real power changed when you connected inductors
		and capacitors in parallel?} \\
		In the ideal scenario, adding inductors and capacitors to the circuit wouldn't
		change the real power as the components are purely reactive. However due to the
		parasitic resistances of the components, they consume small amounts of real power.
		The total real power of the circuit is then simply the sum of the resistor's
		real power and the reactive components' real power from losses.
	\item \textbf{How are the magnitudes of the currents $I_1$, $I_2$, and $I_3$ related
		to each other when all elements are connected in parallel?} \\
		The current $I_1$ is the sum of all three branch currents as phasors:
		\begin{align*}
			\tilde{I}_1 &= \tilde{I}_R + \tilde{I}_L + \tilde{I}_C.
		\end{align*}
		In the ideal model, the inductor and capacitor currents are, respectively,
		\ang{90} and \ang{-90} out of phase from the resistor current. Since they are
		\ang{180} apart, the magnitudes will subtract from each other. Thus the magnitude
		of the current $I_1$ can be given by
		\begin{align}
			\left|\tilde{I}_1\right| &= \sqrt{\left|\tilde{I}_R\right|^2
				+ \left(\left|\tilde{I}_2\right|
				- \left|\tilde{I}_3\right|\right)^2}.
			\label{eq:1c}
		\end{align}
	\item \textbf{Can either $I_2$ and/or $I_3$ have a magnitude larger than the
		magnitude of $I_1$? Explain.} \\
		Yes, if the resistor current is small enough, or equally if both inductor and
		capacitor currents are large enough. We can see this in equation \eqref{eq:1c};
		the reactive components' currents ``nearly cancel'' each other out, giving only
		a small increase to $I_1$, but they can each be larger than $I_1$ individually.
\end{itemize}

\section*{Task 1D. Series Connection of RLC Components}

\begin{table}[H]
	\centering
	\begin{tabular}{*{8}{c}}
		\toprule
		Measurement & $I_\text{1(RMS)}$ (\si{\A}) & $\phi_1$ (\si{\degree})
			& $V_\text{2(RMS)}$ (\si{\V}) & $\phi_2$ (\si{\degree})
			& $V_\text{3(RMS)}$ (\si{\V}) & $\phi_3$ (\si{\degree})
			& $P_\text{1(AVG)}$ (\si{\W}) \\
		\midrule
		$R_1$, $L_1$, $C_1$ & 0.32 & 55.3 & 8.65 & 218.3 & 20.58 & 31.2 & 2.3 \\
		$R_1$, $L_1$, $C_1$, $C_2$, $C_3$ & 0.64 & 0.8 & 14.25 & 80.6 & 13.97 & 90.5 & 9.0 \\
		$R_1$, $R_2$, $R_3$, $L_1$, $C_1$, $C_2$, $C_3$
			& 1.56 & 25.3 & 27.17 & 245.0 & 34.08 & 60.6 & 19.5 \\
		\bottomrule
	\end{tabular}
	\caption[Series connection of RLC components]
		{Series connection of RLC components.
		Supply voltage $V_1 = 14.14 \angle -1.6 \si{\degree}$ \si{\V} RMS.}
	\label{tab:1d}
\end{table}

\begin{itemize}
	\item \textbf{How has the real power changed when you connected capacitors in parallel
		or resistors in parallel?} \\
		When the capacitors are connected in parallel, the equivalent impedance of
		the capacitors is significantly lowered. This matches with how see $V_3$ being
		lower afterwards. Thus, more current is flowing through the circuit and more
		power is being dissipated by the resistor. When the resistors are connected
		in parallel, more current is again going through the circuit and thus the
		real power increases as well.
	\item \textbf{How are the magnitudes of voltages $V_1$, $V_2$, and $V_3$ related
		to each other when all elements are connected in series?} \\
		The voltage $V_1$ is the sum of all three components' voltages as phasors:
		\begin{align*}
			\tilde{V}_1 = \tilde{V}_R + \tilde{V}_L + \tilde{V}_C.
		\end{align*}
		In the ideal model, the inductor and capacitor voltages are, respectively,
		\ang{90} and \ang{-90} out of phase from the resistor voltage. Since they are
		\ang{180} apart, the magnitudes will subtract from each other. Thus the magnitude
		of the voltage $V_1$ can be given by
		\begin{align}
			\left|\tilde{V}_1\right| = \sqrt{\left|\tilde{V}_R\right|^2
				+ \left(\left|\tilde{V}_2\right|
				- \left|\tilde{V}_3\right|\right)^2}.
			\label{eq:1d}
		\end{align}
	\item \textbf{Can either $V_2$ and/or $V_3$ have a magnitude larger than the
		magnitude of $V_1$? Explain.} \\
		Yes, if the resistor voltage is small enough, or equally if both inductor and
		capacitor voltages are large enough. Similar to Task 1C, we can see how according to
		equation \eqref{eq:1d}, the reactive components' voltages ``nearly cancel'' each
		other out, which allows $V_2$ and $V_3$ to be each larger than $V_1$ individually.
	\item \textbf{Support your answer with the phasor diagram corresponding to the
		measurement Task 1D, step 5.} \\
		See figure \ref{fig:1d:pha}.
\end{itemize}

\begin{figure}
	\centering
	\includegraphics[width=0.75\linewidth]{images/1d-3-phasor.png}
	\caption[Phasor view of circuit with three resistors, one inductor, and three capacitors]
		{Phasor view of circuit with three resistors, one inductor, and three capacitors.}
	\label{fig:1d:pha}
\end{figure}

\phantomsection
\addcontentsline{toc}{subsection}{Task 2}
\section*{Task 2A. Three-Phase Measurements with Wye-Connected RL Load Box}

\begin{table}[H]
\begin{adjustwidth}{-15mm}{-15mm}
	\centering
	\begin{tabular}{l*{7}{c}}
		\toprule
		& \multicolumn{2}{c}{Phase 1 (RMS)} & \multicolumn{2}{c}{Phase 2 (RMS)}
			& \multicolumn{2}{c}{Phase 3 (RMS)} & Total \\
		\cmidrule(lr){2-3} \cmidrule(lr){4-5} \cmidrule(lr){6-7} \cmidrule(l){8-8}
		\multicolumn{1}{c}{Measurement}
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $P_\text{AVG}$ (\si{\W}) \\
		\midrule
		RL balanced load,
			& 10.17 $\angle$ 4.5 & 0.27 $\angle$ -46.3
			& 10.15 $\angle$ -238.0 & 0.29 $\angle$ 73.3
			& 9.79 $\angle$ -115.2 & 0.29 $\angle$ -163.6
			& 5.6 \\
		\; neutrals disconnected \\[0.5em]
		RL unbalanced load,
			& 10.06 $\angle$ 4.4 & 0.38 $\angle$ -80.0
			& 12.91 $\angle$ -251.8 & 0.39 $\angle$ 60.8
			& 8.94 $\angle$ -139.9 & 0.26 $\angle$ -189.0
			& 5.2 \\
		\; shorted resistor, \\
		\; neutrals disconnected \\[0.5em]
		RL unbalanced load,
			& 8.88 $\angle$ 0.9 & 0.45 $\angle$ 0.6
			& 9.15 $\angle$ -199.3 & 0.26 $\angle$ -247.8
			& 13.71 $\angle$ -98.4 & 0.43 $\angle$ -145.0
			& 9.9 \\
		\; shorted inductor, \\
		\; neutrals disconnected \\[0.5em]
		RL unbalanced load,
			& 10.16 $\angle$ -1.5 & 0.4 $\angle$ -85.7
			& 10.31 $\angle$ -241.4 & 0.3 $\angle$ 70.2
			& 10.19 $\angle$ -120.6 & 0.30 $\angle$ -168.8
			& 4.6 \\
		\; shorted resistor, \\
		\; neutrals connected \\[0.5em]
		RL unbalanced load,
			& 10.09 $\angle$ -1.8 & 0.51 $\angle$ -2.0
			& 10.27 $\angle$ -241.5 & 0.30 $\angle$ 70.3
			& 10.19 $\angle$ -120.8 & 0.30 $\angle$ -168.9
			& 9.6 \\
		\; shorted inductor, \\
		\; neutrals connected \\
		\bottomrule
	\end{tabular}
	\caption[Balanced and unbalanced wye-connected RL three-phase load]
		{Balanced and unbalanced wye-connected RL three-phase load.}
	\label{tab:2a}
\end{adjustwidth}
\end{table}

\begin{itemize}
	\item \textbf{How close are your measurements to the ideal case when the
		RL load is balanced?} \\
		In the balanced load circuit, the voltages had phase differences that were
		calculated to be \ang{120 \pm 2.8} and maximum voltage spread of less than
		\qty{\pm 0.38}{\V}. Similarly, the currents had phase differences of
		\ang{120 \pm 3.1} and variations in magnitude of \qty{\pm 0.02}{\A}.
		These can be attributed to the tolerances in individual components,
		manufacturing variations, and parasitic impedances, among other things.
	\item \textbf{What happens to the phase voltages and currents when the
		resistor is shorted in one of the phases?} \\
		Looking at the phase current through the shorted load, the magnitude will
		increase due to the lower impedance from the short. That current will also lag
		the voltage by around \ang{90}, due to it being a (nearly) pure inductive load.
		Without the neutral line connected, the phase voltages drift from the target
		of \qty{10}{\V} and the phases deviate from their nominal \ang{120} separation.
	\item \textbf{What happens to the phase voltages and currents when the
		inductor is shorted in one of the phases?} \\
		Looking at the phase current through the shorted load, the magnitude will
		again increase due to the lower impedance from the short. That current will
		be roughly in phase with the voltage, due to it being purely resistive.
		Without the neutral line connected, the phase voltages also drift from the target
		of \qty{10}{\V} and the phases again deviate from their nominal \ang{120} separation.
	\item \textbf{How does connecting or disconnecting the neutral wire affect the
		phase voltages and currents when the load is unbalanced?} \\
		The neutral line allows the unbalanced current to flow back to the source.
		This stabilises the phase voltages even when the load is unbalanced,
		ensuring that the magnitudes stay around \qty{10}{\V} and the phase differences
		are around \ang{120}.
\end{itemize}

\section*{Task 2B. Three-Phase Measurements with Delta-Connected RL Load Box}

\begin{table}[H]
\begin{adjustwidth}{-15mm}{-15mm}
	\centering
	\begin{tabular}{l*{7}{c}}
		\toprule
		& \multicolumn{2}{c}{Phase 1 (RMS)} & \multicolumn{2}{c}{Phase 2 (RMS)}
			& \multicolumn{2}{c}{Phase 3 (RMS)} & Total \\
		\cmidrule(lr){2-3} \cmidrule(lr){4-5} \cmidrule(lr){6-7} \cmidrule(l){8-8}
		\multicolumn{1}{c}{Measurement}
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $P_\text{AVG}$ (\si{\W}) \\
		\midrule
		RL balanced load
			& 9.98 $\angle$ -1.6 & 0.93 $\angle$ -47.0
			& 10.06 $\angle$ -241.9 & 0.95 $\angle$ 72.0
			& 9.93 $\angle$ -121.3 & 0.96 $\angle$ -167.1
			& 19.7 \\[0.5em]
		RL unbalanced load,
			& 10.10 $\angle$ -1.5 & 0.94 $\angle$ -81.0
			& 10.14 $\angle$ -241.7 & 1.39 $\angle$ 55.0
			& 10.02 $\angle$ -121.3 & 0.97 $\angle$ -167.2
			& 14.6 \\
		\; shorted resistor \\[0.5em]
		RL unbalanced load,
			& 9.97 $\angle$ -1.5 & 1.41 $\angle$ -26.2
			& 10.15 $\angle$ -242.3 & 0.91 $\angle$ -248.7
			& 10.03 $\angle$ -120.9 & 0.97 $\angle$ -166.6
			& 28.8 \\
		\; shorted inductor \\
		\bottomrule
	\end{tabular}
	\caption[Balanced and unbalanced delta-connected RL three-phase load]
		{Balanced and unbalanced delta-connected RL three-phase load.}
	\label{tab:2b}
\end{adjustwidth}
\end{table}

\begin{itemize}
	\item \textbf{How close are your measurements to the ideal case when the
		RL load is balanced?} \\
		In the balanced load circuit, the voltages had phase differences that were
		calculated to be \ang{120 \pm 0.6} and maximum voltage spread of less than
		\qty{\pm 0.13}{\V}. Similarly, the currents had phase differences of
		\ang{120 \pm 1} and variations in magnitude of \qty{\pm 0.03}{\A}.
		These can be attributed to the tolerances in individual components,
		manufacturing variations, and parasitic impedances, among other things.
	\item \textbf{How do the current and power measurements (values) compare with
		the previous case when the balanced load was wye-connected?}
		The average power increases by a factor of \num{3}, compared to the
		wye-connected load. This is because the power per phase in a wye-connected
		load is given by $\left(V_\text{line}/\sqrt{3}\right)^2/Z \cdot \cos\phi$,
		whereas the delta-connected load's power per phase is given by
		$V_\text{line}^2/Z \cdot \cos\phi$. Similarly, the line current will increase
		by a factor of \num{3}, since in wye and delta connections the line current
		is given by $V_\text{line}/\left(\sqrt{3}Z\right)$ and $\sqrt{3} V_\text{line}/Z$,
		respectively.
	\item \textbf{What happens to the phase voltages and currents when the
		resistor is shorted in one of the phases?}
		The phase voltages do not change, staying approximately \ang{120} apart
		and having a maximum voltage spread of \qty{0.12}{\V}. Looking at the
		phase current through the shorted load, the magnitude will increase due to
		the lower impedance from the short. That current will also lag the voltage
		by around \ang{90}, due to it being a (nearly) pure inductive load.
	\item \textbf{What happens to the phase voltages and currents when the
		inductor is shorted in one of the phases?}
		The phase voltages do not change, staying approximately \ang{120} apart
		and having a maximum voltage spread of \qty{0.18}{\V}. Looking at the
		phase current through the shorted load, the magnitude will again increase
		due to the lower impedance from the short. That current will be in phase
		with the voltage, due to it being a purely resistive load.
\end{itemize}

\section*{Task 2C. Two-Wattmeter Measurement Method with Delta-Connected RL Load Box}

\begin{table}[H]
	\centering
	\begin{tabular}{l*{7}{c}}
		\toprule
		& \multicolumn{2}{c}{Phase 1 (RMS)} & \multicolumn{2}{c}{Phase 2 (RMS)}
			& \multicolumn{3}{c}{Average Power} \\
		\cmidrule(lr){2-3} \cmidrule(lr){4-5} \cmidrule(l){6-8}
		\multicolumn{1}{c}{Measurement}
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $V \angle \phi_v$ (\si{\V}) & $I \angle \phi_i$ (\si{\A})
			& $P_1$ (\si{\W}) & $P_2$ (\si{\W}) & $P_\text{total}$ (\si{\W}) \\
		\midrule
		RL balanced load
			& 17.34 $\angle$ -0.7 & 0.94 $\angle$ -76.2
			& 17.58 $\angle$ 59.4 & 0.96 $\angle$ 42.6
			& 4.1 & 15.6 & 19.8 \\[0.5em]
		RL unbalanced load,
			& 17.41 $\angle$ -0.7 & 0.94 $\angle$ -110.3
			& 17.53 $\angle$ 59.5 & 1.40 $\angle$ 25.7
			& -5.5 & 19.7 & 14.2 \\
		\; shorted resistor \\[0.5em]
		RL unbalanced load,
			& 17.37 $\angle$ -0.6 & 1.42 $\angle$ -55.6
			& 17.67 $\angle$ 59.1 & 0.92 $\angle$ 81.8
			& 14.3 & 14.5 & 28.7 \\
		\; shorted inductor \\
		\bottomrule
	\end{tabular}
	\caption[Two-wattmeter measurements with delta-connected RL three-phase load]
		{Two-wattmeter measurements with delta-connected RL three-phase load.}
	\label{tab:2c}
\end{table}

\begin{itemize}
	\item \textbf{The load circuit in Task 2B and Task 2C is identical in each step.
		Compare the measurements of currents and voltages of Task 2B and Task 2C.
		Explain the results.} \\
		The voltages measured here are greater compared to Task 2B because we are
		measuring the line-to-line voltage, as opposed to the phase voltage like
		what was done previously. As the line and phase voltages are related to
		each other by $V_\text{line} = \sqrt{3} V_\text{phase} \angle \ang{30}$,
		the voltage supply is giving the same output voltage in both Task 2B
		and Task 2C (as $\sqrt{3} \cdot \qty{10}{\V} \cong \qty{17.32}{\V}$).
		We also see this through our measurement of the current; the magnitude
		remains relatively unchanged from Task 2B to Task 2C, while the current
		shows a \ang{-30}  phase shift because of the change in our reference
		(the angle of $V_1$).
	\item \textbf{Compare the measurements of the real power of Task 2B and Task 2C.
		Explain the results.} \\
		The total real power remains relatively the same, as expected, since no part
		of the physical circuit changed. We see that when the resistor is shorted
		(the load is purely inductive), $P_1$ is negative. This can be explained by
		the power factor angle being more than \ang{90}, which will make the power,
		given by $P = V_\text{line} I_\text{phase} \cos\phi$, negative.
\end{itemize}

\phantomsection
\addcontentsline{toc}{subsection}{Task 3}
\section*{Task 3A. Single-Phase Full-Wave Rectifier}

\begin{table}[H]
\begin{adjustwidth}{-15mm}{-15mm}
	\centering
	\begin{tabular}{*{9}{c}}
		\toprule
		& $V_\text{1(RMS)}$ (\si{\V}) & $I_\text{1(RMS)}$ (\si{\A})
			& $P_\text{1(in)}$ (\si{\W}) & $V_\text{3(RMS)}$ (\si{\V})
			& $V_\text{3(PP)}$ (\si{\V}) & $I_\text{3(RMS)}$ (\si{\A})
			& $I_\text{3(PP)}$ (\si{\A}) & $P_\text{3(out)}$ (\si{\W}) \\
		\midrule
		No load & 15.05 $\angle$ -1.7 & 0.08 $\angle$ 36.2 & 0.8
			& 14.19 $\angle$ -93.8 & 20.5
			& 0.11 $\angle$ -224.5 & 0.4 & 1.0 \\
		Light load & 14.87 $\angle$ -1.8 & 0.84 $\angle$ -2.4 & 12.5
			& 13.34 $\angle$ -93.9 & 20.6
			& 0.88 $\angle$ -96.2 & 0.9 & 11.7 \\
		Heavy load & 14.77 $\angle$ -1.8 & 1.50 $\angle$ -2.0 & 22.0
			& 13.09 $\angle$ -94.1 & 20.2
			& 1.53 $\angle$ -95.6 & 2.6 & 19.9 \\
		\bottomrule
	\end{tabular}
	\caption[Single-phase full-wave rectifier, without a capacitor filter]
		{Single-phase full-wave rectifier, without a capacitor filter.}
	\label{tab:3a}
\end{adjustwidth}
\end{table}

\begin{itemize}
	\item \textbf{What relationship between the voltage and current
		peak/ripple do you observe?} \\
		The current follows the same waveform shape as the voltage. Since the
		load is resistive and there is no capacitor, the current peak is in phase
		with the voltage peak. As the load resistance decreases (heavier load),
		the peak to peak current increases while the voltage peak to peak
		remains about constant.
\end{itemize}

\begin{table}[H]
\begin{adjustwidth}{-15mm}{-15mm}
	\centering
	\begin{tabular}{*{10}{c}}
		\toprule
		& $V_\text{1(RMS)}$ (\si{\V}) & $I_\text{1(RMS)}$ (\si{\A})
			& $I_\text{1(PP)}$ (\si{\A}) & $P_\text{1(in)}$ (\si{\W})
			& $V_\text{3(RMS)}$ (\si{\V}) & $V_\text{3(PP)}$ (\si{\V})
			& $I_\text{3(RMS)}$ (\si{\A}) & $I_\text{3(PP)}$ (\si{\A})
			& $P_\text{3(out)}$ (\si{\W}) \\
		\midrule
		No load & 15.04 $\angle$ -1.7 & 0.34 $\angle$ -8.6 & 2.8 & 2.2
			& 19.90 $\angle$ -176.3 & 0.29
			& 0.17 $\angle$ 13.1 & 0.16 & 2.8 \\
		Light load & 14.73 $\angle$ -1.8 & 2.5 $\angle$ -3.9 & 13.5 & 24.8
			& 17.73 $\angle$ -165.8 & 1.58
			& 1.23 $\angle$ 203.1 & 0.23 & 21.7 \\
		Heavy load & 14.54 $\angle$ -2.1 & 3.80 $\angle$ -2.3 & 18.5 & 39.4
			& 16.73 $\angle$ -161.9 & 2.30
			& 2.01 $\angle$ -159.5 & 0.36 & 33.6 \\
		\bottomrule
	\end{tabular}
	\caption[Single-phase full-wave rectifier, with a capacitor filter]
		{Single-phase full-wave rectifier, with a capacitor filter.}
	\label{tab:3b}
\end{adjustwidth}
\end{table}

\begin{itemize}
	\item \textbf{How do the output voltage and current ripples compare with
		the previous case without the capacitor filter?} \\
		Compared to the no-capacitor case, adding the capacitor filter reduces both
		the output voltage and current ripple, because the capacitor supplies current
		between rectified peaks. Ripple increases with heavier load due to
		faster discharge, but it is smaller than without the filter.
	\item \textbf{How does the presence of a capacitor impact the input current peak?} \\
		The input current peak is increased when the capacitor is present.
		Current flows in brief, high-amplitude pulses close to the AC peaks
		rather than continuously throughout the cycle because the capacitor charges
		only when the input voltage is greater than the capacitor voltage.
		In comparison to the scenario in which the capacitor filter is not used,
		this leads to significantly greater input current peaks.
	\item \textbf{How does the peak of the input current change with the load current?} \\
		The peak input current rises noticeably as the load current does.
		A higher recharge current is needed over a brief interval when the load
		is heavier because the capacitor discharges more between AC peaks.
		Higher amplitude current pulses at the input are the outcome of this.
\end{itemize}

\phantomsection
\addcontentsline{toc}{subsection}{Task 4}
\section*{Task 4A. Three-Phase Full-Wave Rectifier}

\begin{table}[H]
	\centering
	\begin{tabular}{*{7}{c}}
		\toprule
		& $I_\text{1(RMS)}$ (\si{\A}) & $V_\text{2(RMS)}$ (\si{\V})
			& $V_\text{2(PP)}$ (\si{\V}) & $I_\text{2(RMS)}$ (\si{\A})
			& $I_\text{2(PP)}$ (\si{\A}) & $P_\text{3(out)}$ (\si{\W}) \\
		\midrule
		No load & 0.09 $\angle$ 33.3 & 12.18 $\angle$ -62.2 & 1.5
			& 0.16 $\angle$ -148.3 & 0.14 & 1.5 \\
		Light load & 0.65 $\angle$ 33.3 & 11.99 $\angle$ -50.1 & 1.5
			& 0.86 $\angle$ -36.7 & 0.22 & 9.9 \\
		Heavy load & 1.10 $\angle$ 34.2 & 11.86 $\angle$ -40.6 & 1.5
			& 1.43 $\angle$ -32.2 & 0.30 & 16.2 \\
		\bottomrule
	\end{tabular}
	\caption[Three-phase full-wave rectifier, without a capacitor filter]
		{Three-phase full-wave rectifier, without a capacitor filter.}
	\label{tab:4a}
\end{table}

\begin{itemize}
	\item \textbf{What relationship between the voltage and current peak/ripple
		do you observe?} \\
		The ripples are the same shape and in phase as the load is purely resistive.
		The voltage peak to peak remains constant while the current peak to increases.
	\item \textbf{How does the input current differ from that of the single-phase rectifier?} \\
		The input current of a three-phase rectifier is smoother and more continuous
		than that of a single-phase rectifier. In the single-phase case, current flows
		in large pulses and drops to zero, in the three-phase case, conduction overlaps
		between phases, reducing ripple.
\end{itemize}

\begin{table}[H]
\begin{adjustwidth}{-15mm}{-15mm}
	\centering
	\begin{tabular}{*{10}{c}}
		\toprule
		& $I_\text{1(RMS)}$ (\si{\A}) & $I_\text{1(PP)}$ (\si{\A})
			& $I_\text{2(RMS)}$ (\si{\A}) & $I_\text{2(PP)}$ (\si{\A})
			& $V_\text{3(RMS)}$ (\si{\V}) & $V_\text{3(PP)}$ (\si{\V})
			& $I_\text{3(RMS)}$ (\si{\A}) & $I_\text{3(PP)}$ (\si{\A})
			& $P_\text{3(out)}$ (\si{\W}) \\
		\midrule
		No load & 0.08 $\angle$ 22.4 & 0.5 & 0.20 $\angle$ -31.0 & 0.6
			& 12.58 $\angle$ -236.0 & 0.1 & 0.12 $\angle$ -218.5 & 0.1 & 1.6 \\
		Light load & 0.70 $\angle$ 28.6 & 2.7 & 1.02 $\angle$ -54.3 & 2.0
			& 11.91 $\angle$ -95.9 & 0.3 & 0.83 $\angle$ 96.0 & 0.1 & 10.0 \\
		Heavy load & 1.17 $\angle$ 29.0 & 4.2 & 1.73 $\angle$ -54.4 & 2.7
			& 11.63 $\angle$ -96.0 & 0.4 & 1.40 $\angle$ -183.2 & 0.1 & 16.3 \\
		\bottomrule
	\end{tabular}
	\caption[Three-phase full-wave rectifier, with a capacitor filter]
		{Three-phase full-wave rectifier, with a capacitor filter.}
	\label{tab:4b}
\end{adjustwidth}
\end{table}

\begin{itemize}
	\item \textbf{How do the output voltage and current ripples compare with
		the previous case without the capacitor filter?} \\
		With the capacitor filter, both output voltage and current become smoother.
		Ripple still increases with heavier load but it remains smaller than
		the case without the capacitor.
	\item \textbf{How does the presence of a capacitor impact the input current peak?} \\
		The input current peak is increased when the capacitor is present.
		Current only flows when the input voltage is higher than the capacitor
		voltage because the capacitor keeps the output voltage close to its
		maximum value. This causes short pulses as opposed to continuous conduction,
		which results in larger peak input currents.
	\item \textbf{How do the output voltage and current ripples compare with the
		single-phase rectifier?} \\
		Compared to the single-phase rectifier, the three-phase rectifier produces
		significantly smaller output voltage and current ripple. This is because
		the three-phase rectifier has overlapping output between phases,
		which reduces the voltage dips between peaks. As a result,
		the output is smoother.
	\item \textbf{How does the peak of the input current change with the load current?} \\
		The peak input current increases as the load current increases.
		A heavier load causes the capacitor to discharge more current between peaks,
		requiring a larger recharge current over a short time. As a result,
		the input current peaks become taller under heavier load conditions.
\end{itemize}


\phantomsection
\addcontentsline{toc}{subsection}{Task 5}
\section*{Task 5A. Instantaneous, Real, and Reactive Power in Parallel RLC Circuit}

The instantaneous power is given by
\begin{align}
	p(t) = v(t) i(t),
	\label{eq:inst-power}
\end{align}
where $v(t)$ and $i(t)$ are the instantaneous voltages and currents,
respectively. The average power is
\begin{align}
	P = \frac{1}{T} \int_{\langle T \rangle} v(t) i(t) \odif{t}.
	\label{eq:power}
\end{align}

The measured RMS voltages and currents are slightly different than the ones
calculated by MATLAB, which uses the discrete version of equation \eqref{eq:power}
(see the appendix). We will use the values calculated by MATLAB here.

For a sinusoidal signal, $P = V_\text{RMS} I_\text{RMS} \cos\phi$ and
$Q = V_\text{RMS} I_\text{RMS} \sin\phi$. We will assume that the signals
are ``sinusoidal enough''.

\begin{table}[H]
	\centering
	\begin{tabular}{ccc}
		\toprule
		& Real Power, $P$ (\si{\W}) & Reactive Power, $Q$ (\si{\VAR}) \\
		\midrule
		Task 3C, step 3 & 9.98 & 0.03 \\
		Task 3C, step 4 & 10.68 & 8.90 \\
		Task 3C, step 5 & 11.25 & -0.20 \\
		\bottomrule
	\end{tabular}
	\caption[Calculations for the instantaneous, real, and reactive power
		in a parallel RLC circuit]
		{Calculations for the instantaneous, real, and reactive power in a
		parallel RLC circuit. The equations for a sinusoidal signal,
		shown above, were used. RMS values are calculated from MATLAB.}
\end{table}

The instantaneous power gives the power absorbed at any given moment in time,
while the real power is the average power absorbed or dissipated over
an interval. The reactive power is power that is temporarily stored in the
magnetic and electric fields of inductors and capacitors, respectively.

By adding inductors, the reactive power increases since they have positive reactance
and will increase the power factor angle. Capacitors, on the other hand, decrease
the reactive power since they have negative reactive and will decrease the
power factor angle.

\begin{figure}[H]
	\centering
	\includegraphics[width=0.5\linewidth]{images/5a-phasor.png}
	\caption[Phasor plot of the voltage and currents for the parallel RLC circuit]
		{Phasor plot of the applied voltage $V_1$, input current $I_1$,
		inductor current $I_2$, and capacitor current $I_3$ for the parallel
		RLC circuit. Here the voltage $V_1$ is normalised to $1$. Notice how
		the phasors $I_2$ and $I_3$ effectively cancel each other.}
\end{figure}

\begin{figure}[H]
	\centering
	\begin{subfigure}{0.75\linewidth}
		\includegraphics[width=\linewidth]{images/5a-1-plot.png}
		\caption{Plot of circuit with one resistor.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.75\linewidth}
		\includegraphics[width=\linewidth]{images/5a-2-plot.png}
		\caption{Plot of circuit with one resistor and one inductor.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.75\linewidth}
		\includegraphics[width=\linewidth]{images/5a-3-plot.png}
		\caption{Plot of circuit with one resistor, one inductor,
			and three capacitors.}
	\end{subfigure}

	\caption[Plots of instantaneous, real, and reactive power in the parallel RLC circuits]
		{Plots of instantaneous, real, and reactive power in the parallel RLC circuits.}
\end{figure}

\section*{Task 5B. Single-Phase Phasor Diagram for the Series RLC Circuit}

We see that in the phasor diagrams, the inductor and capacitor voltages are
always \ang{180} apart from each other, as expected since each of them have
purely reactive loads. We see that with one resistor, the voltages $V_2$
and $V_3$ are almost exactly \ang{\pm 90} from $V_1$. This means that the
power factor angle is near the minimum (\ang{0}) and almost all of the
power delivered is real.

When we add the other resistors in parallel and decrease the series resistance,
the current starts to lead the voltage, meaning that less of the total power
delivered is real and more of it is reactive. Because the voltages $V_2$ and $V_3$
have to be \ang{90} apart from the current $I_1$, we see that the shift in the
current also causes the angles of the reactive components to shift as well.

Finally we observe that the difference between $V_1$ and $V_2+V_3$ should be
the voltage drop across the resistor(s).

\begin{figure}[H]
	\centering
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5b-1-phasor.png}
		\caption{Phasor diagram of circuit with one resistor, one inductor,
			and three capacitors.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5b-2-phasor.png}
		\caption{Phasor diagram of circuit with three resistors, one inductor,
			and three capacitors.}
	\end{subfigure}
	\caption[Phasor plots of the voltages and current for the series RLC circuit]
		{Phasor plots of the applied voltage $V_1$, inductor voltage $V_2$,
		capacitor voltage $V_3$, and input current $I_1$ for the series RLC circuit.}
\end{figure}

\section*{Task 5C. Three-Phase Phasor Diagrams for Wye-Connected RL Load}

We see that without the neutral line connected, shorting one of the load elements
will shift the phase voltage angles away from \ang{120}. Shorting the resistor
caused the angle of the unbalanced load's voltage to increase, and shorting the
inductor caused it to decrease.

With the neutral line connected, the unbalanced current now has a return path
back to the source, ensuring that the phase voltages stay relatively close to
\ang{120} apart from each other.

Looking at our measurements in table \ref{tab:2b}, we see that the unbalanced
loads wouldn't affect the phase voltages, since each point of connection in the
delta configuration is directly connected to a voltage source. As expected from
having unbalanced impedances, we will see differences between the phases in
current.

\begin{figure}[H]
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5c-1-phasor.png}
		\caption{Phasor diagram of circuit with RL balanced load, \\
			neutrals disconnected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5c-2-phasor.png}
		\caption{Phasor diagram of circuit with RL unbalanced load, \\
			shorted resistor, neutrals disconnected.}
	\end{subfigure}
	\caption[Phasor plots of the applied phase voltages and input currents
		for the wye-connected RL load]
		{Phasor plots of the applied phase voltages, $V_1$, $V_2$, $V_3$,
		and input currents, $I_1$, $I_2$, $I_3$, for the wye-connected RL load.
		Note that all current magnitudes have been scaled up by $10$.}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5c-3-phasor.png}
		\caption{Phasor diagram of circuit with RL unbalanced load, \\
			shorted inductor, neutrals disconnected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5c-4-phasor.png}
		\caption{Phasor diagram of circuit with RL unbalanced load, \\
			shorted resistor, neutrals connected.}
	\end{subfigure}

	\caption[Phasor diagrams of the applied phase voltages and input currents
		for the wye-connected RL load]
		{Phasor diagrams of the applied phase voltages, $V_1$, $V_2$, $V_3$,
		and input currents, $I_1$, $I_2$, $I_3$, for the wye-connected RL load.
		Note that all current magnitudes have been scaled up by $10$.}
\end{figure}

\begin{figure}[H]
	\ContinuedFloat
	\centering
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/5c-5-phasor.png}
		\caption{Phasor diagram of circuit with RL unbalanced load, \\
			shorted inductor, neutrals connected.}
	\end{subfigure}

	\captionsetup{list=no}
	\caption{Phasor diagrams of the applied phase voltages, $V_1$, $V_2$, $V_3$,
		and input currents, $I_1$, $I_2$, $I_3$, for the wye-connected RL load.
		Note that all current magnitudes have been scaled up by $10$.}
	\captionsetup{list=yes}
\end{figure}

\section*{Task 5D. Harmonics in AC Mains}

\textit{For this section, we will use $V_1$ from Task 2B, with a balanced load.}

Harmonics are distortions that occur at integer multiples of the fundamental
frequency. For AC mains in North America, this is \qty{60}{\Hz}. These are caused
by non-linear loads. In our lab, these harmonics may be caused by the power supply's
internal components being non-linear. Since our load is purely ohmic, the load should
not be contributing to the harmonic content.

\begin{figure}[H]
	\centering
	\includegraphics[width=0.75\linewidth]{images/5d-plot.png}
	\caption[Frequency spectrum of a sinusoidal voltage signal]
		{Frequency spectrum of $V_1$ from Task 2B using MATLAB's \texttt{fft()}
		function (see the appendix).}
\end{figure}

As expected, we see a large spike at \qty{60}{\Hz} as that is our power supply's
frequency. We also notice distortions at multiples the fundamental frequency.

\section*{Task 5E. Single-Phase AC/DC}

\textit{For this section, we will use $I_1$ from Task 3A,
	under a full load with and without a capacitor filter.}

We see that the output current without a capacitor filter does look close to a
sinusoidal wave. However, the peaks are sharper than a sine wave, resembling
more of a sawtooth wave.

The output current with a capacitor filter, apart from the non-zero parts
of the current, does not resemble a sinusoidal wave.

As expected from our initial observations on the waveforms, the frequency spectrum
of the current without a capacitor filter has a single large spike at \qty{60}{\Hz},
with a small amount of distortion at higher frequencies. On the other hand,
the current with a capacitor filter has strong harmonic content.

\begin{figure}[H]
	\centering

	\begin{subfigure}{0.75\linewidth}
		\includegraphics[width=\linewidth]{images/5e-1-plot.png}
		\caption{Waveform and frequency spectrum under a full load without
			a capacitor filter.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.75\linewidth}
		\includegraphics[width=\linewidth]{images/5e-2-plot.png}
		\caption{Waveform and frequency spectrum under a full load with
			a capacitor filter.}
	\end{subfigure}

	\caption[Waveform and frequency spectrum of a rectified current]
		{Waveform and frequency spectrum of $I_1$ from Task 3A using MATLAB's
		\texttt{fft()} function (see the appendix).}
\end{figure}

\section*{Task 5F. Three-Phase AC/DC Rectifiers}

\textit{For this section, we will use $I_1$, $I_2$, and $I_3$ from
	Task 4A, under a full load with a capacitor filter.}

The current $I_2$ is the output of the current from the full-bridge rectifier.
The diodes ensure that all of the current flows in one single direction. For $I_3$,
the capacitor helps maintain a steady voltage and current, making it comparable
to an ideal DC current.

\begin{figure}[H]
	\centering
	\includegraphics[width=0.75\linewidth]{images/5f-plot.png}
	\caption[Waveform and frequency spectrum of the current in a
		three-phase rectifier]
		{Waveform and frequency spectrum of $I_1$, $I_2$, and $I_3$ from Task 4A
		using MATLAB's \texttt{fft()} function (see the appendix).}
\end{figure}

The $I_1$ currents in Task 5E are much closer to an ideal sinusoidal wave,
compared to here. However, we also observe that the three-phase input current
has much less harmonic content compared to the single-phase currents.

\phantomsection
\addcontentsline{toc}{subsection}{Conclusion}
\section*{Conclusion}

In this laboratory experiment, we were able to measure and calculate different
parameters of single-phase and three-phase circuits. We looked into real-world
applications and ideal versus non-ideal components, such as inductors, capacitors,
and rectifiers. We also saw the difference between balanced and unbalanced loads,
and their different effects in the wye and delta configurations. In addition, this
lab showed us the difference between single-phase and three-phase rectifiers.
Finally, we looked into the harmonic content of AC voltages and currents.

\newpage

\appendix

\phantomsection
\addcontentsline{toc}{section}{Appendix}

\addcontentsline{toc}{subsection}{Code Listings}
\section*{Appendix A. Code Listings}

\lstinputlisting[
	style=Matlab-editor,
	basicstyle=\ttfamily,
	numberstyle=\ttfamily\small\color{gray},
	keywordstyle=\color{blue},
	commentstyle=\color{gray},
	stringstyle=\color{red},
	caption={Frequency spectrum analysis code.},
	captionpos=b
]{matlab/specan.m}

\newpage

\lstinputlisting[
	style=Matlab-editor,
	basicstyle=\ttfamily,
	numberstyle=\ttfamily\small\color{gray},
	keywordstyle=\color{blue},
	commentstyle=\color{gray},
	stringstyle=\color{red},
	caption={Phasor plot code.},
	captionpos=b
]{matlab/poweran.m}

\newpage

\lstinputlisting[
	style=Matlab-editor,
	basicstyle=\ttfamily,
	numberstyle=\ttfamily\small\color{gray},
	keywordstyle=\color{blue},
	commentstyle=\color{gray},
	stringstyle=\color{red},
	caption={Phasor plot code.},
	captionpos=b
]{matlab/phasor.m}

\lstinputlisting[
	language=bash,
	basicstyle=\ttfamily,
	numberstyle=\ttfamily\small\color{gray},
	keywordstyle=\color{blue},
	commentstyle=\color{gray},
	stringstyle=\color{red},
	caption={Shell script for preparing data.},
	captionpos=b
]{matlab/prep.sh}

\newpage

\phantomsection
\addcontentsline{toc}{subsection}{Images}
\section*{Appendix B. Images}

\begin{figure}[H]
	\centering
	\includegraphics[width=0.75\linewidth]{images/1a-phasor.png}
	\caption[Phasor measurements for the RL load box with three inductors]
		{Phasor measurements for the RL load box with three inductors (Task 1A).}
	\label{fig:1a}
\end{figure}

\begin{figure}[H]
	\centering
	\includegraphics[width=0.75\linewidth]{images/1b-phasor.png}
	\caption[Phasor measurements for the RC load box with three capacitors]
		{Phasor measurements for the RC load box with three capacitors (Task 1B).}
	\label{fig:1b}
\end{figure}

\begin{figure}[H]
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1c-1-wave.png}
		\caption{Waveform of circuit with one resistor.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1c-2-wave.png}
		\caption{Waveform of circuit with one resistor and one inductor.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1c-3-wave.png}
		\caption{Waveform of circuit with one resistor,
			one inductor, \\ and three capacitors.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1c-3-phasor.png}
		\caption{Phasor view of circuit with one resistor,
			one inductor, \\ and three capacitors.}
	\end{subfigure}
	\caption[Measurements of parallel connections of RLC components]
		{Measurements of parallel connections of RLC components (Task 1C).}
	\label{fig:1c}
\end{figure}

\begin{figure}[H]
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1d-1-wave.png}
		\caption{Waveform of circuit with one resistor,
			one inductor, \\ and one capacitor.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1d-2-wave.png}
		\caption{Waveform of circuit with one resistor,
			one inductor, \\ and three capacitors.}
	\end{subfigure}
	\caption[Measurements of series connections of RLC components]
		{Measurements of series connections of RLC components (Task 1D).}
	\label{fig:1d}
\end{figure}

% Task 1D continued.

\begin{figure}[H]
	\ContinuedFloat
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1d-3-wave.png}
		\caption{Waveform of circuit with three resistors,
			one inductor, \\ and three capacitors.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/1d-3-phasor.png}
		\caption{Phasor view of circuit with three resistors,
			one inductor, \\ and three capacitors.}
	\end{subfigure}
	\captionsetup{list=no}
	\caption{Measurements of series connections of RLC components (Task 1D).}
	\captionsetup{list=yes}
\end{figure}

\begin{figure}[H]
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-1-wave.png}
		\caption{Waveform of circuit with RL balanced load, \\ neutrals disconnected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-1-phasor.png}
		\caption{Phasor view of circuit with RL balanced load, \\ neutrals disconnected.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-2-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\
			shorted resistor, neutrals disconnected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-2-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\
			shorted resistor, neutrals disconnected.}
	\end{subfigure}

	\caption[Three-phase measurements with wye-connected RL load box]
		{Three-phase measurements with wye-connected RL load box (Task 2A).}
	\label{fig:2a}
\end{figure}

% Task 2A continued.

\begin{figure}
	\ContinuedFloat
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-3-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\
			shorted inductor, neutrals disconnected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-3-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\
			shorted inductor, neutrals disconnected.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-4-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\
			shorted resistor, neutrals connected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-4-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\
			shorted resistor, neutrals connected.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-5-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\
			shorted inductor, neutrals connected.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2a-5-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\
			shorted inductor, neutrals connected.}
	\end{subfigure}

	\captionsetup{list=no}
	\caption{Three-phase measurements with wye-connected RL load box (Task 2A).}
	\captionsetup{list=yes}
\end{figure}

\begin{figure}
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2b-1-wave.png}
		\caption{Waveform of circuit with RL balanced load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2b-1-phasor.png}
		\caption{Phasor view of circuit with RL balanced load.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2b-2-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\ shorted resistor.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2b-2-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\ shorted resistor.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2b-3-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\ shorted inductor.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2b-3-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\ shorted inductor.}
	\end{subfigure}

	\caption[Three-phase measurements with delta-connected RL load box]
		{Three-phase measurements with delta-connected RL load box (Task 2B).}
	\label{fig:2b}
\end{figure}

\begin{figure}
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2c-1-wave.png}
		\caption{Waveform of circuit with RL balanced load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2c-1-phasor.png}
		\caption{Phasor view of circuit with RL balanced load.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2c-2-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\ shorted resistor.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2c-2-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\ shorted resistor.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2c-3-wave.png}
		\caption{Waveform of circuit with RL unbalanced load, \\ shorted inductor.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/2c-3-phasor.png}
		\caption{Phasor view of circuit with RL unbalanced load, \\ shorted inductor.}
	\end{subfigure}

	\caption[Two-wattmeter measurements with delta-connected RL load box]
		{Two-wattmeter measurements with delta-connected RL load box (Task 2C).}
	\label{fig:2c}
\end{figure}

\begin{figure}
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/3a-1-wave.png}
		\caption{Waveform of circuit without a capacitor filter \\ under no load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/3a-2-wave.png}
		\caption{Waveform of circuit without a capacitor filter \\ under half load.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/3a-3-wave.png}
		\caption{Waveform of circuit without a capacitor filter \\ under full load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/3b-1-wave.png}
		\caption{Waveform of circuit with a capacitor filter \\ under no load.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/3b-2-wave.png}
		\caption{Waveform of circuit with a capacitor filter \\ under half load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/3b-3-wave.png}
		\caption{Waveform of circuit with a capacitor filter \\ under full load.}
	\end{subfigure}

	\caption[Single-phase full-wave rectifier, without and with capacitor filter]
		{Single-phase full-wave rectifier, without and with capacitor filter (Task 3A).}
	\label{fig:3a}
\end{figure}

\begin{figure}
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/4a-1-wave.png}
		\caption{Waveform of circuit without a capacitor filter \\ under no load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/4a-2-wave.png}
		\caption{Waveform of circuit without a capacitor filter \\ under half load.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/4a-3-wave.png}
		\caption{Waveform of circuit without a capacitor filter \\ under full load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/4b-1-wave.png}
		\caption{Waveform of circuit with a capacitor filter \\ under no load.}
	\end{subfigure}

	\medskip

	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/4b-2-wave.png}
		\caption{Waveform of circuit with a capacitor filter \\ under half load.}
	\end{subfigure}
	\hfill
	\begin{subfigure}{0.45\linewidth}
		\includegraphics[width=\linewidth]{images/4b-3-wave.png}
		\caption{Waveform of circuit with a capacitor filter \\ under full load.}
	\end{subfigure}

	\caption[Three-phase full-wave rectifier, without and with capacitor filter]
		{Three-phase full-wave rectifier, without and with capacitor filter (Task 4A).}
	\label{fig:4a}
\end{figure}

\end{document}