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\begin{document}
\title{Singapore Physics Olympiad Training (SPOT) - Circuits}
\author[1]{Garett Tok}%
\author[2]{Zhiming Darren TAN}%
\author[1]{guangquan.wang}%
\affil[1]{Affiliation not available}%
\affil[2]{Ministry of Education Singapore }%
\vspace{-1em}
\date{\today}
\begingroup
\let\center\flushleft
\let\endcenter\endflushleft
\maketitle
\endgroup
\sloppy
\section*{1. Direct Current (DC)
Circuits}
{\label{904379}}
We begin our analysis into circuits by analysing the most common type of
circuits we see in science experiments, and aim to gradually build up
familiarity and ease into the various circuit components.
\section*{}
{\label{904379}}
\subsection*{1.1 Open versus Closed
Circuits}
{\label{154822}}
A circuit is open if there is no possible closed path that electricity
may flow through. In such a case, no power is dissipated, although
potential difference may occur in the circuit.
\section*{}
{\label{904379}}
\subsection*{1.2 DC Sources}
{\label{577572}}
Circuits begin with some sort of input, a~\emph{source}, which are
commonly voltage sources (think batteries) or current sources. When we
talk about DC sources, what we are saying is really that the sources are
time-invariant.
\par\null\par\null
\subsection*{1.3 Kirchoff's Current Law
(KCL)}
{\label{356120}}
The sum of electrical currents flowing into a node is zero.
\[\Sigma I = 0\]
Of course, this assumes that you are consistent with your sign
convention. Notably, KCL is analogous to the continuity law for
hydrodynamics.
\par\null
\subsection*{1.4 Kirchoff's Voltage Law
(KVL)}
{\label{511552}}
Along a current loop, the sum of voltage drops over its circuit elements
is equal to the sum of electromotive forces.
\[\Sigma V = 0\]
A useful analogy for the most basic of circuits is to think of water
flowing down a channel of pipes before being pumped up again. In this
analogy, no matter the path, if you were to trace a loop, since it must
end back in its initial position, then the change in voltage
(represented in this case by the height of the pipe elements) has to be
zero, thus fulfilling Kirchoff's Voltage Law. Furthermore, no matter how
the pipes are connected, the amount of water flowing into the next set
of pipes has to be the same as the amount of water flowing into the
previous set of pipes, fulfilling Kirchoff's Current Law.
Note that the linearity of both Kirchoff's Laws allows us to perform
superposition. More specifically, if a circuit solely comprises of
voltage/current sources and resistors, we are able to analyse the
circuit with only one source (i.e. short circuiting all other voltage
sources, and open circuiting all current sources), then analysing it
with a different source, and summing the results to get the final
result. However, this may only be applied to circuits comprising solely
of independent sources.
\par\null
\subsection*{1.5 Resistors}
{\label{371443}}
The typical resistance of the resistor is denoted by the following
formula:
\[R = \frac{V}{I}\]
In this case, V refers to the voltage drop across the resistor. If,
however, the resistivity of the resistor is known, the resistance can be
obtained in a non-empirical manner by the formula:
\[R = \frac{\rho l}{A}\]
where A is the cross sectional area of the resistor.
\par\null
\subsubsection*{1.5.1 By considering voltages, derive the formula for
equivalent resistance of resistors connected in
parallel.~}
{\label{241829}}
\subsubsection*{1.5.2 By considering currents, derive the formula for
equivalent resistance of resistors connected in
series.}
{\label{506979}}
\subsubsection*{1.5.3 Determine the effective resistance of a resistor of
length l whose cross sectional area A varies with length such that A(x)
= kx/l, where k is a
constant.}
{\label{572096}}\par\null
When an electrical current passes a resistor, energy is dissipated as
heat.
\[P = I^2 R\]
\emph{Idea: In dealing with questions involving resistors, it often
helps to re-draw the diagram, emphasising parallel and series
connections}. \emph{To do this, it may often be helpful to label nodes
or resistors in order to keep track of everything.}
\par\null
\subsubsection*{1.5.4 Consider the following diagram. Determine the
relationship between the resistances A, B, C, and D such that no current
will flow through the middle resistor. This phenomenon is known as a
Wheatstone
bridge.}
{\label{744370}}\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Untitled1/Untitled1}
\caption{{Wheatstone Bridge
{\label{796099}}%
}}
\end{center}
\end{figure}
At its very core, solving problems comprising solely of resistors and
sources involves simplifying the circuits through various methods to
obtain a clean result. While the simplification techniques can be
derived, it is recommended to memorise the formulas for the olympiads as
they are often tedious in nature.
\par\null\par\null
\subsection*{1.6~\(\Delta\)-Y
Transformation}
{\label{605410}}
The Delta-Wye transformation is a problem-solving method used to
transform resistors into equivalent resistances in another arrangement
as follows:
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Untitled/Untitled}
\caption{{\(\Delta\)-Y Transformation
{\label{535837}}%
}}
\end{center}
\end{figure}
To do this, the formulas are as follows:
\[R_A = \frac{R_{AB}R_{AC}}{R_{AB} + R_{BC} + R_{AC}}\]
\[R_{AB} = \frac{R_A R_B + R_B R_C + R_A R_C}{R_C}\]
\subsection*{1.7 Thevenin's Theorem and Norton's
Theorem}
{\label{554315}}
Take any two points in a circuit, A and B. From Thevenin's Theorem, we
are able to simplify the entire circuit into a series connection of a
voltage source and a resistor. (Noting, of course, that this is only
useful if you are not concerned with what goes on within the circuit,
but rather the behaviour of the circuit when a circuit element is
attached across A and B.)
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Untitled2/Untitled2}
\caption{{Thevenin Equivalent~
{\label{349839}}%
}}
\end{center}
\end{figure}
\par\null
The equivalent voltage source is given by the potential difference
between A and B, and the equivalent resistance is given by the potential
difference between A and B divided by the short circuit current if a
wire were to link A and B directly. (This, of course, can be easily seen
from the fact that the equivalent circuit must behave like the original
circuit)
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.21\columnwidth]{figures/Untitled3/Untitled3}
\caption{{Norton Equivalent
{\label{778667}}%
}}
\end{center}
\end{figure}
Alternatively, we can look at Norton's theorem, where we simplify a
circuit into a~\emph{parallel} connection of a current source and a
resistor. The magnitude of this current source is the short circuit
current found above, and the resistance is the same as in the Thevenin
equivalent circuit.
\par\null
Thevenin's and Norton's equivalent circuits are often used to analyse
how varying a circuit element will affect the power of the circuit
(though, however, questions may ask for the current or potential
difference across a circuit element, in which case the theorems may make
the work simpler too). For drawing maximum power, the attached load has
to be equal to the Norton/Thevenin load.
\par\null
\emph{Idea: Together, the Thevenin and Norton equivalent circuits can be
used to simplify circuits by constantly changing between a voltage and a
current source then making use of voltage sources in series or current
sources in parallel to combine them.}
\par\null\par\null
\subsection*{1.8 Nodal Analysis}
{\label{273775}}
Nodal analysis is a problem-solving method by making use of KCL. The
steps are as follows:
\begin{enumerate}
\tightlist
\item
Identify every node, and group nodes with a similar voltage together.
\item
Determine the node that has the greatest number of connections and
arbitrarily ground it. (This assumes the question sets no restrictions
e.g. some other point is grounded or at a certain voltage)
\item
At each node, determine the current that goes into the node via the
other nodal voltages, and apply KCL.
\item
Repeat step 3 for every other node to obtain the necessary number of
equations.
\end{enumerate}
\par\null
\subsection*{1.9 Mesh Analysis}
{\label{862760}}
In order to apply Mesh analysis, follow the following procedure:
\begin{enumerate}
\tightlist
\item
Firstly, ensure that the circuit is planar (i.e. it can be redrawn in
a way such that no branch cross each other). The circuit should now
consist of multiple loops.
\item
Within each loop, assign a mesh current. The convention is that mesh
currents are drawn in a clockwise direction.
\item
Apply KVL in each loop, while being careful to use the algebraic sum
of mesh currents when dealing with circuit elements that border two
meshes.
\end{enumerate}
\par\null
\subsection*{1.10 Limits of Resistances}
{\label{859032}}
At times, the question may simply request for a upper and lower bound on
the resistances, rather than finding the exact resistance through two
nodes. In these cases, we make use of the fact that circuits naturally
aim to minimise the power dissipated. In particular, this implies that:
\begin{enumerate}
\tightlist
\item
If circuit branches are short circuited, then the resistance will
decrease.
\item
If circuit branches are cut, then the resistance will increase.~
\end{enumerate}
\par\null
To make sense of the conclusions above, firstly check that they comply
with the equivalent resistances of parallel and series resistances.
Next, consider a current injected at node A and removed at node B. Note
that as current moves to minimise the power dissipated, then cutting off
a possible path implies that whatever current that was passing through
the path would have to re-route, the power will increase, and therefore
the effective resistance increases. A similar argument can be put forth
for the short-circuiting of circuit branches to show that the resistance
will decrease.
\par\null\par\null
\section*{2. Resistor, Inductor, and Capacitor (RLC)
Circuits}
{\label{303070}}\par\null
\subsection*{2.1 Capacitors}
{\label{411005}}
In dealing with capacitors in circuits, we normally utilise the
formula~\(I\ =\ C\ \frac{dV}{dt}\). In dealing with RLC circuits, we commonly
come upon the term steady state, and we take that to mean when the
circuit is left unperturbed for a long period of time. In a steady state
(e.g. when everything has stabilised), the capacitor can be treated as
an~\emph{open circuit}.
\par\null
\subsubsection*{2.1.1 Consider a case where a capacitor of capacitance C
(initially uncharged) is connected to a battery with voltage V and a
resistor of resistance R. Determine the voltage on the capacitor as a
function of time. Hence, or otherwise, determine the amount of work done
on the capacitor to charge the capacitor to voltage
V.~}
{\label{595850}}\par\null
\emph{Idea: Voltage across capacitors have to be continuous} (as can be
seen from~\(\frac{dV}{dt}\)). This helps us when we are dealing with
the response of a circuit when a switch has been flipped.
\subsubsection*{}
{\label{318897}}\par\null
\subsection*{2.2 Inductors}
{\label{148138}}
We utilise the formula~\(V\ =\ L\ \frac{dI}{dt}\), and note that in the steady
state, and inductor can be treated as a~\emph{short circuit}.
\par\null
\subsubsection*{2.2.1 Consider a circuit where a capacitor of capacitance
C is fully charged, and connected only to an inductor of inductance L
that is not charged. Determine the behaviour of the circuit. This is
commonly known as a LC
circuit.}
{\label{781334}}\par\null
Idea:~\emph{Idea: Currents~through inductors are always continuous.}
\par\null
\subsection*{2.3 RLC}
{\label{256072}}
Putting them together, if you have solved 2.2.1, you would have noticed
the behaviour is very similar to oscillations. Indeed, we may further
extend this by adding a resistor into the circuit, and note that it
produces the responses of a damped oscillator.
\par\null
\subsubsection*{2.3.1 Prove that a RLC circuit functions like a damped
oscillator. Determine the responses and associated
conditions.}
{\label{778881}}\par\null
\emph{Idea: I-t or V-t functions may be found as a superposition of a
steady-state solution and the state of the circuit where all voltage
sources are replaced with a wire, and current sources replaced with open
circuits.}
\par\null
\emph{Idea: If the time in which changes are being measured is much
smaller than the time constant, consider linearising the changes.}
\par\null
Extending on this idea, the number of natural frequencies are given by
the number of independent loops minus the number of linearly independent
loops which consist only one type of element. Multiple methods can be
used to determine these natural frequencies, the most basic of which is,
by drawing on the mechanical analogue, to determine the time variance of
charge/voltage/current, and solve the differential equation.
\par\null\par\null
\section*{3. Diodes}
{\label{309727}}\par\null
\subsection*{3.1 Ideal Diodes}
{\label{502301}}
Ideal diodes can be substituted by a short circuit when there's a
forward current, and a open circuit when the current is in the reverse
direction. In solving these questions, it might be necessary to consider
and solve both cases to verify which case should be used.
\par\null
\subsection*{3.2 Non-Ideal Diode}
{\label{898725}}
Theoretical Olympiad questions may treat non-ideal diode as requiring a
minimum forward voltage, in which case the diodes can be treated as per
3.1, except with the ``addition'' of a voltage source with the
prescribed forward voltage.~
\par\null
In practical questions, however, it is not uncommon for the
voltage-current response graph of the diode to be provided, which will
have to be used against your empirical results to determine the
current/voltage pairing observed.
\par\null\par\null
\section*{4. Alternating Current (AC)
Circuits}
{\label{739229}}
\emph{Note: A firm foundation in dealing with complex numbers is
recommended.}
\par\null
Unlike DC circuits, AC circuits are excited by a time-varying source. In
particular, we wish to study the sinusoidal time-varying source, which
is both the most common type of AC that people are used to, and provides
one of the most mathematically analysable solutions.
\par\null
We note that now the source can be written as a function of time. For
example, in a voltage source, we have:
\[v(t) = A cos (\omega t + \phi)\]
In order to compare two sources, we need to ensure that both the
frequency are the same (otherwise there will be a transient response),
and that the trigonometric function used (i.e. sin/cos) are the same.
\par\null
Drawing from ideas we learned in forced oscillations, we note that the
steady-state response of a circuit will be a sinusoid of the same
frequency as the driver. For convenience when dealing with sinusoids, we
may consider replacing them with a complex number as such:
\[V(t) = Ae^{i\omega t + \phi} = Ae^\phi e^{i\omega t}\]
Where we will now use the capital letter to denote the complex
representation. Bear in mind that when we talk about measurable current
or voltage, we just have to take into account the real part of the
complex representation.
\par\null\par\null
\subsection*{\texorpdfstring{4.1 ``Simplification'' of AC
Circuits}{4.1 Simplification of AC Circuits}}
{\label{601678}}
Assuming:
\begin{enumerate}
\tightlist
\item
A linear system
\item
Sinusoidal sources
\item
All sources are of the same frequency
\item
Steady-state has been achieved
\end{enumerate}
We are able to treat an AC circuit exactly like a DC circuit, with a few
changes.
\par\null
All capacitors and inductors are treated as resistors with impedances
(denoted by \(Z\))~\(\frac{1}{j\omega C}\)
and~\(j\omega L\) respectively, and resistors have an impedance
of~\(R\). The difference between impedances and resistances
are that impedances also take into account the lead/lag between voltages
and currents based on the circuit element.~
\par\null
\subsubsection*{4.1.1 Therefore, at very low frequencies, what is the
impedance of a capacitor and an inductor? What about at a very high
frequency?}
{\label{849411}}
\selectlanguage{greek}\subsubsection*{4.1.2~ In a black box with two ports, there are three
components connected in series: a capacitor, an inductor, and a
resistor. Devise a method to determine the values of all three
components, if you have a sinusoidal voltage generator with adjustable
output frequency ν, an AC-voltmeter and an ACammeter.
~}\selectlanguage{english}
{\label{535232}}\par\null
From here on, all of Kirchoff's laws, nodal and mesh analyses, and Ohm's
law apply.~
\par\null
\subsection*{4.2 Power through an AC
Circuit}
{\label{562928}}
The dissipated power through a circuit element can be calculated as:
\[P\ =\ \frac{1}{2}\left|I\right|\ \left|V\right|\ \cos\ \phi\]
where~\(\phi\) denotes the difference in phase between the
current and voltage, and is also known as the power factor.~
\par\null
\subsubsection*{4.2.1 From the above equation, what is the power
dissipation through a inductor and a capacitor? Does it make sense?~
~}
{\label{661473}}
\subsubsection*{4.2.2 By considering the time average of the product of
real current and real voltage, derive the above
equation.~}
{\label{661473}}\par\null
Due to the above, people tend to prefer to use
the~\emph{root-mean-square (RMS)
current}~\(\left(I_{rms}\ =\ \frac{I}{\sqrt{2}}\right)\)and~\emph{voltage}~\(\left(V_{rms}\ =\ \frac{V}{\sqrt{2}}\right)\)instead
to remove the factor of half in the equation.~~
\par\null
\subsubsection*{4.2.3 By integrating the square of the current or voltage
functions over an appropriate length of time, averaging it, and taking
the square root of it, derive the above expressions for RMS
current.~}
{\label{330854}}\par\null
In extending the above, we are able to define a complex power:
\[P_c = V_{rms} I_{rms}^* = I_{rms} I_{rms}^* Z = |I_{rms}|^2 Z\]
This complex power contains all the information we need to derive the
rest of the information with regards to powers.~
\begin{enumerate}
\tightlist
\item
To obtain the apparent power (product of rms voltage and rms current),
take the magnitude of the complex power.
\item
To obtain the power factor angle, take the angle of the complex
power.~
\item
To obtain the reactive power, take the imaginary part of the complex
power.
\item
To obtain the dissipated power, take the real part of the complex
power.~
\end{enumerate}
\par\null
\subsection*{4.3 Natural Frequencies of RLC
circuits}
{\label{861443}}
We now revisit an idea first explored in section 2.3: the oscillation
within an RLC circuit.
\par\null
Let us examine what resonance means in the case of a circuit. Since
impedance is determined by the circuit elements, two sorts of resonance
can happen - current resonance and voltage resonance. The former occurs
when impedance between two points go towards zero, and the latter occurs
when the impedance between two points go towards infinity.~
\par\null
\begin{itemize}
\tightlist
\item
\textbf{\emph{Voltage Resonance}} works by choosing two points in a
circuit, and solving them for where impedance is infinity.~
\item
\textbf{\emph{Current Resonance}~}works by arbitrarily cutting the
circuit at a point. Since in the original circuit both points were
connected, the voltage between them is zero. We then set the impedance
between the two cut points to be zero (such that a current can exist)
.~
\end{itemize}
\par\null
\emph{Question: Determine the natural frequencies for the circuit
below.}
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Untitled4/Untitled4}
\caption{{LC circuit
{\label{415001}}%
}}
\end{center}
\end{figure}
First determine the number of natural frequencies. We see that there are
three independent loops (2 x~\(L_1-C\) and~\(C-C-L_2\)),
but also note that the outermost loop (\(L_{1\ }-L_2-L_1\)) consists only
inductors. Therefore, there only exists two natural frequencies.~
We will be using the current resonance method, cutting the loop at the
point shown below:
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.28\columnwidth]{figures/Untitled5/Untitled5}
\caption{{Cut LC Circuit.
{\label{978782}}%
}}
\end{center}
\end{figure}
\par\null
We note that the impedance of the circuit between the two red points can
be given by:
\[Z = \omega ( L_2(\omega^2CL_1 -1) - 2L_1)\]
Solving for zero impedance, we obtain:
\[\omega = 0 \quad or \quad \omega = \sqrt{\frac{2L_1 + L_2}{C L_1 L_2}}\]
where the first solution corresponds to the flow of current through the
outermost loop in the original circuit. Recall that we mentioned earlier
that there are two natural frequencies, but we have only derived one of
them. The reason behind is this because in doing this method, we missed
any solutions where the natural current in the original circuit is zero.
In such a case we note that current will just flow within
the~\(L_1-\ C\)~loops, giving rise to the second natural
frequency~\(\omega=\frac{1}{\sqrt{L_1C}}\).~
Although we utilised the symmetry of the question to easily obtain the
last remaining frequency, note that if the question didn't allow for
this to happen, it can still be easily obtained by selecting another
point and re-doing the analysis. Furthermore, it should be noted that in
the worst case scenario, it is still an option to solve the differential
equation, whilst making use of the natural frequency already found to
reduce the degree of the differential equation.~
\par\null
An astute learner should note that we only encountered the problem of
the missing frequency due to deliberate choosing of a symmetrical node
to cut. This was a personal preference: utilising symmetry, even if I
had to reapply the analysis, the subsequent analyses will be done on
``reduced'' circuits, which aids in my speed, and helps to reduce
complexity of both the initial and subsequent analyses.~
\subsubsection*{}
{\label{526076}}
\subsubsection*{4.3.1 Redo the analysis using the voltage resonance
method.}
{\label{526076}}
\subsection*{}
{\label{861443}}\par\null
\subsection*{4.4 Blackboxes~}
{\label{861443}}
Here, we wish to create a methodological approach to dealing with most
blackbox circuits. Although this is not fully intensive, and blackbox
problem setters are notorious to be constantly limiting the student
through the equipment provided or increasing the complexity by
increasing the number of ports, here we wish to develop a methodological
approach.
\par\null
\begin{enumerate}
\tightlist
\item
Using a voltmeter, determine presence of a battery.
\item
Determine presence of diode through ohmeter (low resistance in forward
direction, high resistance in reverse direction) or voltmeter (connect
in parallel to diode connected to DC supply)
\item
Test for capacitors by shorting all connections, then using ohmmeter
(high resistance if present), or drive voltage to charge capacitors,
then testing for decreasing current.
\item
Use AC to eliminate possible remaining configurations and obtain
inductor/capacitor values if necessary.
\end{enumerate}
\par\null
Note that this order forms only a general approach to dealing with
blackboxes, and note also the importance of following the steps
sequentially. Without knowledge of the presence of a voltage source or
diode, the conclusions of other tests will be invalid. This is also a
work in progress, please feel free to add your own procedure as you
discover other methods of dealing with a blackbox.
~
\section*{~}
{\label{624586}}
\selectlanguage{english}
\FloatBarrier
\end{document}