# Preface

This section presents examples of electric circuits and corresponding differential equations.

Introduction to Linear Algebra with Mathematica

# Electric Circuits

We show interconnection between electric circuits and differential equations used to model them in a series of examples. We start with the most simple example when resistor , inductor , and capacitor are connected in series across a voltage supply, the circuit so obtained is called series RLC circuit. A good introduction to the subject could be found on the web . The phasor diagram of series RLC circuit is drawn by combining the phasor diagram of resistor, inductor and capacitor. Before doing so, one should understand the relationship between voltage and electric current in case of resistor, capacitor and inductor.

Example 1: Consider RCL circuit in series.

The RLC circuit is a fundamental in circuits applications. In this example we will:
• Derive the time response of an RLC circuit due to a varying signal.
• Describe the individual components of the RLC circuit.
We want to set a general form for what the voltage will be. We will say it a function that varies with time with an amplitude and frequency.
V[t_] = V0*Sin[w*t]
Now what is the voltage response of the resistor, capacitor, and inductor?
• By Ohm's Law, the voltage across a resistor is the current passing through it multiplied by its resistance in ohms.
$V_R (t) = R \cdot i(t) .$
• The voltage across a capacitor is less intuitive. Physically, a basic capacitor is just two metal plates with some conductive material across it. The voltage across it is best described by this equation C*V=Q. Where the dQ/dt is the rate of charge passing through the capacitor, or the current i(t). Therefore, the voltage across a capacitor is the total charge passing through the capacitor scaled by a capacitor constant. (Helpful additional reading about the capacitor: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html )
• $V_C (t) = \frac{1}{C} \cdot \int i(t)\,{\text d}t .$
• An inductor is just a coil of wire. The coiling of the wire produces some interesting electric field properties that make the answer tot he voltage across it not so trivial. With a current source that does not vary with time, the inductor will act as a short circuit, so the voltage across it will be zero. But if current across the inductor changes, the inductor's electric field will resist the change The voltage across an inductor can be described by V=L*di/dt. In words, the voltage across an inductor is proportional to its electric field and the rate of change of current through it. (Helpful addition reading about the inductor: https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-and- calculus/)
$V_L (t) = L \cdot \frac{\text d}{{\text d}t}\, i(t) .$
By Kirchoff's voltage law, the voltage around a closed loop will be zero. Because all the components of a series RLC are connected in a closed loop, the sum of the voltages across all the individual components will equal zero. (https://www.electronics-tutorials.ws/dccircuits/kirchhoffs-voltage-law.html )
$V_{\mbox{source}} + V_{\mbox{inductor}} + V_{\mbox{capacitor}} + V_{\mbox{resistor}} = 0 .$
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Example 2: Let us consider a two-loop circuit:

 resistor1 = Graphics[{Thickness[0.01], Line[{{-30, 12}, {-25, 12}, {-23, 10}, {-23, 14}, {-19, 10}, {-19, 14}, {-15, 10}, {-15, 12}, {-10, 12}}]}, PlotLabel -> Subscript[R, 1], Ticks -> None, Axes -> False]; resistor2 = Graphics[{Thickness[0.01], Line[{{-10, 12}, {-5, 12}, {-3, 10}, {-3, 14}, {1, 10}, {1, 14}, {5, 10}, {5, 12}, {10, 12}}]}, PlotLabel -> Subscript[R, 2], Ticks -> None, Axes -> False]; coil = ParametricPlot[{2*Sin[t*3] - 10, 1*Cos[t*3 + Pi] + 1*t - 3}, {t, 0, 3*Pi}, PlotLabel -> "L", Ticks -> None, Axes -> False, ImageSize -> Tiny, PlotStyle -> {Thickness[0.01]}]; l1 = Graphics[{Thickness[0.01], Line[{{-30, -1}, {-30, -8}, {10, -8}}]}]; l2 = Graphics[{Thickness[0.01], Line[{{-30, 5}, {-30, 12}}]}]; l3 = Graphics[{Thickness[0.01], Line[{{-10, 12}, {-10, 7.5}}]}]; l4 = Graphics[{Thickness[0.01], Line[{{-10, -8}, {-10, -4}}]}]; l5 = Graphics[{Thickness[0.01], Line[{{10, 12}, {10, 3}, {8, 3}, {12, 3}}]}]; l6 = Graphics[{Thickness[0.01], Line[{{10, -8}, {10, 1}, {8, 1}, {12, 1}}]}]; c1 = Graphics[{Thickness[0.01], Circle[{-30, 2}, 3]}]; txtC = Graphics[ Text[Style["C", Bold, FontSize -> 14, Blue], {13, 2}]]; txtV = Graphics[ Text[Style["V(t)", Bold, FontSize -> 14, Blue], {-25, 2}]]; txtL = Graphics[ Text[Style["L", Bold, FontSize -> 14, Blue], {-6, 2}]]; txtR1 = Graphics[ Text[Style[Subscript[R, 1], Bold, FontSize -> 14, Blue], {-19, 8}]]; txtR2 = Graphics[ Text[Style[Subscript[R, 2], Bold, FontSize -> 14, Blue], {1, 8}]]; Show[l1, l2, l3, l4, l5, l6, c1, resistor1, resistor2, coil, txtC, txtV, txtL, txtR1, txtR2] Two-loop circuit. Mathematica code

We apply Kirchoff's voltage law (KVL) to each loop. The KVL deals with the conservation of energy around a closed circuit path. This voltage law states that for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. Since the choice of direction of current is arbitrary, we consider its direction as clockwise in every loop. As charge carriers flowing through a circuit pass though a component, they either gain or lose electrical energy, depending upon the component.

The left loop has three elements: a voltage source, V(t) = 127  sin(120πt and two passive elements: resitor R1 = 10 Ω and the coil (inductor) with inductance L = 0.1 H. The right loop consists again of the same coil, another resitor R2 = 20 Ω and capacitor with capacitance C = 1 mF = 0.001 F. Then for the left loop, we get

$V(t) = R_1 i_1 + L\,\frac{{\text d}\left( i_1 - i_2 \right)}{{\text d}t} ,$
and for the right loop, we have
$0= \frac{1}{C}\, q_2 + R_2 i_2 + L\,\frac{{\text d}\left( i_2 - i_1 \right)}{{\text d}t} .$
By adding relation between charge q2 and current i2,
$\frac{{\text d} q_2}{{\text d}t} = i_2 ,$
we obtain three differential equations of the first order for three unknowns:
$\begin{cases} \frac{{\text d} q_2}{{\text d}t} = i_2 , \\ 0.1 \, \frac{{\text d}\left( i_1 - i_2 \right)}{{\text d}t} = 127\,\sin \left( 120\pi\,t \right) - 10\,i_1 , \\ 0.1 \, \frac{{\text d}\left( i_1 - i_2 \right)}{{\text d}t} = 1000\,q_2 + 20\,i_2 . \end{cases}$
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Example 3: Consider a three-loop circuit of a first order, given that each loop will contain only one energy storage element (capacitor or inductor) in addition to resistors. The capacitor current-voltage relationship, $$i = C\,\frac{{\text d}V}{{\text d}t}$$ and inductor current-voltage relationship $$V = L\,\frac{{\text d}i}{{\text d}t}$$ are first order differential equations. Ohm’s Law, i = V R and Kirchhoff’s Voltage Law, which defines the conservation of energy around a closed loop.

connect[pointList_] := {Line[pointList], Map[Text[Style["", FontSize -> 18]] &, pointList[[{1, -1}]]]}
ap[l_ : 1] := Line[l {{{0, 0}, {1/3, 0}}, {{2/3, 0}, {1, 0}}}] resistor[l_ : 1, n_ : 3] := Line[Table[{i l/(4 n), 1/3 Sin[i Pi/2]}, {i, 0, 4 n}]]
coil[l_ : 1, n_ : 3] := Module[{ scale = l/(5/16 n + 1/2), pts = {{0, 0}, {0, 1}, {1/2, 1}, {1/2, 0}, {1/2, -1}, {5/ 16, -1}, {5/16, 0}} }, Append[Table[ BezierCurve[scale Map[{d 5/16, 0} + # &, pts]], {d, 0, n - 1}], BezierCurve[scale Map[{5/16 n, 0} + # &, pts[[1 ;; 4]]]]]]
capacitor[l_ : 1] := {gap[l],  Line[l {{{1/3, -1}, {1/3, 1}}, {{2/3, -1}, {2/3, 1}}}]}
battery[l_ : 1] := {gap[ l], {Rectangle[l {1/3, -(2/3)}, l {1/3 + 1/9, 2/3}],  Line[l {{2/3, -1}, {2/3, 1}}]}}
contact[l_ : 1] := {gap[l],  Map[{EdgeForm[Directive[Thick, Black]], FaceForm[White],  Disk[#, l/30]} &, l {{1/3, 0}, {2/3, 0}}]}
Options[display] = {Frame -> True, FrameTicks -> None,  PlotRange -> All, GridLines -> Automatic,  GridLinesStyle -> Directive[Orange, Dashed],  AspectRatio -> Automatic};
display[d_, opts : OptionsPattern[]] :=  Graphics[Style[d, Thick],  Join[FilterRules[{opts}, Options[Graphics]], Options[display]]]
at[position_, angle_ : 0][obj_] :=  GeometricTransformation[obj,  Composition[TranslationTransform[position],  RotationTransform[angle]]]
label[s_String, color_ : RGBColor[.3, .5, .8]] :=  Text@Style[s, FontColor -> color, FontFamily -> "Geneva",  FontSize -> Large]; display[{ capacitor[] // at[{0, 0}, Pi/2], connect[{{0, 1}, {0, 2}, {2, 2}}], resistor[] // at[{2, 2}], connect[{{3, 2}, {4, 2}, {4, 1}}], battery[] // at[{4, 0}, Pi/2], connect[{{4, 0}, {4, -1}, {0, -1}, {0, 0}}], connect[{{4, 2}, {6, 2}}], capacitor[] // at[{6, 2}], connect[{{7, 2}, {8, 2}}], connect[{{8, 2}, {8, 1}}], resistor[] // at[{8, 0}, Pi/2], connect[{{8, 0}, {8, -1}, {6, -1}}], resistor[] // at[{5, -1}], connect[{{5, -1}, {4, -1}}], connect[{{8, 2}, {12, 2}}], connect[{{11, 2}, {12, 2}, {12, 1}}], coil[] // at[{12, 0}, Pi/2], connect[{{12, 0}, {12, -1}, {8, -1}}]}]
For the circuit configuration, a system of five equations can be derived using Kirchhoff’s Voltage Law, the capacitor current-voltage relations and Ohm’s Law defined above. Three of the equations are differential equations while the other two are algebraic equations.    ■
 Unit Name Ubit Symbol Quantity Ampere (amp) A Electric current (I) Volt V Voltage (V, E) Electromotive force (E) Potential difference (Δφ) Ohm Ω Resistance (R) Watt W Electric power (P) Farad F Capacitance (C) Henry H Inductance (L) Coulomb C Electric charge (Q) Joule J Energy (E) Tesla T Magnetic field (B) Weber Wb Magnetic flux (Φm) Hertz Hz Frequency (f)