# Examples of Circuits

One of the basic fundamentals to simplifying and solving a circuit is to focus on multiple resistors that have connections with one another. Depending upon the configuration of the ladder network, the resistors in series can be added up to simplify the resistance in that certain region of the circuit.

In addition, resistors in parallel can be combined by adding the inverse of all of the parallel resistances.

gap[l_: 1] := Line[l {{{0, 0}, {1/3, 0}}, {{2/3, 0}, {1, 0}}}]

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}}]}}

resistor[l_: 1, n_: 3] := Line[Table[{i l/(4 n), 1/3 Sin[i Pi/2]}, {i, 0, 4 n}]]

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[{battery[] // at[{0, 0}, Pi/2], connect[{{0, 1}, {0, 2}, {3, 2}, {3, 1}}], resistor[] // at[{3, 0}, Pi/2], connect[{{3, 2}, {6, 2}, {6, 1}}], resistor[] // at[{6, 0}, Pi/2], connect[{{6, 0}, {6, -1}, {0, -1}, {0, 0}}], connect[{{3, 0}, {3, -1}}]}]

label[s_String, color_: RGBColor[.3, .5, .8]] := Text@Style[s, FontColor -> color, FontFamily -> "Geneva", FontSize -> Large];

display[{battery[] // at[{0, 0}, Pi/2], connect[{{0, 1}, {0, 2}, {3, 2}, {3, 1}}], resistor[] // at[{3, 0}, Pi/2], connect[{{3, 0}, {3, -1}, {0, -1}, {0, 0}}]}]

After the simplification of combining series and parallel resistors, there are operations in electrical engineering that can be used to solve for other elements in the circuit. For instance, having resistors in series enables the use of voltage division. Voltage division is a procedure that is used to solve for the voltage of a resistor (when that particular resistor is in series with another resistor in the circuit).

Moreover, resistors in parallel allows the use of an operation known as current division. Current division is a function that is used to solve for the current that passes through a resistor (when that particular resistor is connected in parallel with another resistor in the circuit).

The input voltage (*v*_{1}) and the input current (*i*_{1}) in an electrical
circuit is set up in a matrix form that looks like \( \begin{bmatrix} v_1 \\ i_1
\end{bmatrix} . \) Meanwhile, the output voltage (*v*_{2}) and the output current
(*i*_{2}) is set up in a column vector form that equates to \( \begin{bmatrix} v_2 \\ i_2
\end{bmatrix} . \) Taking a standard electrical circuit with a linear transformation is known
as a transfer matrix, which is written as \( \begin{bmatrix} v_2 \\ i_2
\end{bmatrix} = {\bf A}\, \begin{bmatrix} v_1 \\ i_1
\end{bmatrix} . \)

It is entirely possible that an electrical engineer could find themselves faced with solving a ladder
network, which links two circuits together that are connected in series with one another. The output
of the first circuit in the network is subsequently used as the input to the next circuit that is in
the network. In the figure pictured below, the circuit with resistor *R*_{1} is recognized as a series
circuit and the circuit with resistor *R*_{2} is recognized as a shunt circuit.

Given both of these circuits, the transfer matrices for each of the respective circuits are written as
\( {\bf A}_1 = \begin{bmatrix} 1& -R_1 \\ 0&1 \end{bmatrix} \) (transfer matrix of
series circuit) and
\( {\bf A}_2 = \begin{bmatrix} 1& 0 \\ -1/R_2&1 \end{bmatrix} \) (transfer matrix of
shunt circuit). Evidently, it is possible to solve through the transfer matrix (**A**_{1}
for the series circuit and **A**_{2} for the shunt circuit) that epitomizes the
figure of the ladder network that was just shown. After using an input vector **x**, the
transfer matrix is illustrated as

*R*

_{1}and

*R*

_{2}from the ladder network that is depicted on the previous page, the ladder network would be equal to \( \begin{bmatrix} 1& -R_1 \\ -1/R_2&1+ R_1 / R_2 \end{bmatrix} = \begin{bmatrix} 1& -8 \\ -0.5&5 \end{bmatrix} . \) Essentially, one is a constant value from the previous transfer matrices,

*R*

_{1}is equal to eight ohms, -1/

*R*

_{2}is equal to -.5 ohms and 1+

*R*

_{1}/

*R*

_{2}is equal to five ohms. Based off of this newly designed ladder network,

*R*

_{2}is equal to two ohms because one divided by .5 is equal to two ohms.

Just like in this scenario, an electrical engineer must first find out if a network like the ladder network in figure four can be built. Assuming that the network can be created, the next step is to use matrix factorization on the transfer matrix in order to find the matrices that are connected to smaller circuits. The smaller circuits can then be used to compose the network or the smaller circuits may already be a part of a configuration such as a ladder network. In some cases, the transfer matrix can contain values that contain complex numbers. Ultimately, the goal is to try to build the network that the electrical engineer needs while expending the least number of electrical elements.

For the more daring electrical engineer, matrices can help simplify the most complex circuits that are non-linear, like a delta-delta transformer. A transformer is an electrical device that is designed on the basis of the concept of magnetic coupling. It uses magnetically coupled coils to transfer energy from one circuit to another. The transformer can be used for “stepping up” or “stepping down” AC voltages or currents. AC stands for alternating current and the voltage that was dealt with before is DC (directed current). Delta configuration represents the order of components, usually two grounded components, and one connection between them. Stating a delta-delta transformation essentially states that they are connected together.

gap[l_: 1] := Line[l {{{0, 0}, {1/3, 0}}, {{2/3, 0}, {1, 0}}}]

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}}]}}

resistor[l_: 1, n_: 3] := Line[Table[{i l/(4 n), 1/3 Sin[i Pi/2]}, {i, 0, 4 n}]]

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[{connect[{{0, 1}, {3.5, 1}}], resistor[] // at[{3.5, 1}], connect[{{4.5, 1}, {8, 1}}], connect[{{2.75, -0.75}, {4, -2}}], resistor[] // at[{2, 0}, 7 Pi/4], connect[{{1, 1}, {2, 0}}], connect[{{4, -2}, {5.25, -0.75}}], resistor[] // at[{5.25, -0.75}, Pi/4], connect[{{6, 0}, {7, 1}}], connect[{{0, -2}, {8, -2}}]}]

label[s_String, color_: RGBColor[.3, .5, .8]] := Text@Style[s, FontColor -> color, FontFamily -> "Geneva", FontSize -> Large];

display[{connect[{{0, 1}, {3.5, 1}}], resistor[] // at[{3.5, 1}], connect[{{4.5, 1}, {8, 1}}], connect[{{1, -1}, {1, -2}}], connect[{{1, 1}, {1, 0}}], connect[{{0, -2}, {8, -2}}], resistor[] // at[{1, 0}, 3 Pi/2], connect[{{7, 1}, {7, 0}}], connect[{{7, -1}, {7, -2}}], resistor[] // at[{7, 0}, 3 Pi/2]}]

Not all electrical engineering problems have numbers that can be solved, however that does not stop someone from making an equation for future use, once other values are known. In reality, this is actually a common practice for electrical engineers when the desired input and output is unknown.

As stated before, a transformer's transfer energy comes from one circuit. Due to power conservation, the energy that may be lost or gain is still maintained within a ratio before the circuits, as shown below

The following formula was derived from the voltages on the two inductors multiple by the top inductor’s reactance

Now merge the above equations, and drive the V' and V'' from the circuit based on power conservation.

Plug and reduce, Therefore \( I_\alpha = \alpha^{-1} \left( \alpha^{-1} V_\alpha - \beta^{-1} V_\beta \right) = \alpha^{-2} V_\alpha L - (\alpha \beta )^{-1} V_\beta L . \)

Now from \( I_\beta \), it followsbecause

Therefore

div class = "math"> \[ -I_\beta = \beta^{-2} \left(V_\beta L \right) + \beta\alpha^{-1} \left(V_\alpha L\right) \]