Terminal into which electrons flow in
Terminal out of which electrons flow out
Scalar quantity \(q\) representing the physical property in material's matter that causes a force when exposed to an electric field. An electron holds a charge \(-1e\), while a proton holds a charge \(+1e\)
Denoted as \(e\), represent the smallest possible charge, equivalent to \(1e=1.6 \times 10^{-19}C\) in Coulombs
Scalar quantity \(I\) representing the rate at which charge flows in a material from the anode to the cathode
\( I = \frac{dq}{dt} \)
Current is the flow of electrons from a place of lower charge (anode) to a place of higher charge (cathode)
Arbitrarily defining the flow of current from cathode to anode for notational purposes
Scalar quantity \( \Delta V, V, U\) representing the difference of electric potential energy of two points, occuring due to unequal charge between the pair of points
Since like charges repel and dislike charges attract, the difference in charge causes current to occur.
\( \Delta V_{0,1} = V_{1} - V_{0} \)
Scalar quantity \( \Delta V, V, U\) representing the work required to move a single charge from a reference point to a certain point in an electric field, occuring due to unequal charge between the pair of points
\( V_{\vec{E}} = - \int_{\mathcal{C}} \vec{E} \cdot d\ell \)
Since energy is conserved by the Maxwell equation \(\nabla \times \vec{E} = 0\), the electric potential is dependent purely on the endpoints of the path \(\mathcal{C}\)
Scalar quantity \(R\) representing a material's resistance to current
\(R = \frac{\rho \ell}{A}\)
Scalar quantity \(G\) representing a material's propensity to conduct changes throughout itself
\(G = \frac{1}{R}\)
When resistance is constant, voltage and current are proportional
\(V=IR\)
Scalar quantity \(E\) representing energy created by current that can be transferred into work
\( E=qV \)
Scalar quantity \(P\) representing rate at which energy is transferred
\( P = \frac{dE}{dt}\)
\( P= \frac{dE}{dt}= \frac{dq}{dt}V=IV \)
\( P=IV=I^{2}R=\frac{V^2}{R} \)
\( E= \int_{t_1}^{t_2} P(t) dt \)
A closed path for electrons to move across to produce current
Collection of points on a circuit graph that mathematically represent a specifc node
Circuit such that there is a single path from anode to cathode for electrons to flow though. The following characteristic of series circuits is due to KCL (to be covered later):
\(I = I_n\)
Circuit such that there is are multiple path from anode to cathode for electrons to flow though. The following characteristic of series circuits is due to KVL (to be covered later):
\(V = V_m\)
Law asserting conservation of charge in a circuit; charge cannot be lost so therefore all current flowing into a node must flow out of the node
\( \sum I_{n} = 0 \)
Law asserting conservation of energy in a circuit; energy cannot be lost so therefore the voltage drops around a closed circuit sum to the voltage of the closed circuit
This means voltage is uniform in parallel from the electrically common points
\(\sum V_{m} = 0\)
\( R_{eq} = \sum R_{m}\)
\( V = \sum V_{m} = I(\sum R_{m}) = I R_{eq} \)
\( \frac{1}{R_{eq}} = \sum \frac{1}{R_{m}}\)
\( I = \sum I_{n} = V(\sum \frac{1}{R_{n}}) = V(\frac{1}{R_{eq}}) \)
In parallel, the current is shared between each branch according to KCL. Using KCL, Ohm's law, and the fact that voltage across a parallel circuit is uniformi:
\( I_{m} =\frac{R_{eq}}{R_{m}}I \)
\(I_{m} =\frac{V}{R_{m}} = \frac{IR_{eq}}{R_{m}}\)
In series, the voltage is shared between each element. Using KVL, and Ohm's law, and the fact that current across a series circuit is uniform:
\( V_{m} = \frac{R_{m}}{R_{eq}}V \)
\(V_{m} =IR_{m} = \frac{V}{R_{eq}}R_{m}\)
Graph theory concept that a circuit can be constructed in some form without crossing an edge over an edge. The utilities problem is an example of the failure to draw a planar graph
A closed path where the first and last node are the same
Loop that does not contain any smaller loop within it
The idea of classifying an ideal current that circulates only the perimeter of its respective mesh. You can choose this current to circulate either clockwise or counterclockwise (so long as you're consisent), at UTS we do clockwise. To denote this in a circuit diagram, you can draw a circular clockwise arrow that almost closes in on itself with a label for the mesh current (such as \(i_1\)) (\(i\) is a mesh current, while \(I\) is a current)
Therefore we can think of any branch in one mesh to have the current \(i_{m}\) and a branch shared by two meshes as \(i_{a} - i_{b}\)
\( \begin{bmatrix} R & R & R \\ R & R & R \\ R & R & R\end{bmatrix} \begin{bmatrix} I \\ I \\ I \end{bmatrix} = \begin{bmatrix} V \\ V \\ V \end{bmatrix} \)
When a current source is included, it gives us some information but makes formulation a little difficult. So do the following:
\( \begin{bmatrix} G & G & G \\ G & G & G \\ G & G & G\end{bmatrix} \begin{bmatrix} V \\ V \\ V \end{bmatrix} = \begin{bmatrix} I \\ I \\ I \end{bmatrix} \)
When a voltage source is included, it gives us some information but makes formulation a little difficult. So do the following:
Electrical element where the input voltage is linear (proportional) to the output.
Circuit of only linear elements, meaning if you measure the voltage between some branch \(V_m\) and the voltage has source \(V_s\), then \( V_s \propto V_m \)
With linear circuits, we can analyse a circuit in two parts: shorting the voltage sources, and opening the current sources. Then we add the respective results together for the values of the full circuit. This can be done since linear circuits obey the law \(f(V_{i1} + V_{i2})=V_{o1} + V_{o2}\)
A voltage source which delivers a static number as its voltage, however in reality this doesn't happen, for instace, if I shorted an ideal 9V battery, Ohm's law says \(V_s =9 \land V=IR \implies I=\frac{9}{0}\), which is impossible as it means infinite power.
In reality, we pair \(V_s\) with a \(R_{sv}\) in series to represent the internal resistance of the voltage source
With current, we pair \(I_s\) with a \(R_{si}\) in parallel to represent the internal resistance of the current source
You can switch a practical voltage source to a practical current source and vice versa given that in your transformation:
Any linear circuit can be represented as one of the following:
Any linear circuit can be represented as one of the following:
To construct a Thevenin circuit, you must:
By finding the circuit's total resistance, then by shorting the open terminals and finding the current across the branch with the largest current in which it splits, you have the current and resistance to make a thevenin equivalent
Given a practical voltage source, the maximum power through a load is when the resistance of a load \(R_{L}\) is equal to the source's resistance \(R_{sv}\), so when \(R_{sv}=R_{L}\)
\(P_{L}=I_{L}^{2}R_{L} = V_{s}^2 \frac{R_{L}}{(R_{sv}+R_{L})^2} \) (Generalising a circuit based on Thevenin's theorem)
\(f(x)= c \frac{x}{(k+x)^2} \) (Generalise this into a function)
\(f'(x)= \frac{c(1-2x(k+1)^{-1})}{(k+x)^2}=0 \) (Maximum occurs when function is not decreasing or increasing)
\(\implies \frac{1-2x}{k+x}=0 \) (numerator must be zero)
\(\implies 2x=k+x \)
\(\therefore k=x \implies f'(x)=0 \)
\(R_{sv}=R_{L} \)
\( P_{L} = \frac{V_{s}^2}{4R_{L}} \)
\(P_{L} = R_{L}I_{L}^2= \frac{R_{L}V_{s}^2}{(R_{s}+R_{L})^2}\)
\( \because R_{L}=R_{s} \text{ at maximum power }, P_{L} = \frac{R_{L}V_{s}^2}{(2R_{L})^2}\)
\( \therefore P_{L} = \frac{V_{s}^2}{4R_{L}}\)
Component with conductive plates on each side with an insulator internally. This allows for capacitors to accumulate charges on its plates. As the charge \(q\) on the plates increases, the electric field \(E\) increases and has its own voltage \(V_{c}\). Accumulation of charge stops when \(V_{c}=V_{s}\)
Scalar quantity \(C\) representing material's ability to store electric charges per volt
\( C = \frac{q}{V}\)
\( C = \frac{\varepsilon A}{\ell}\)
\( i(t)= C\frac{dV(t)}{dt}\) where:
\( V(t) = \frac{1}{C} \int_{t_{0}}^{t} I(t) dt + V(t_{0}) \) where:
\( W_{c}(t) = \frac{CV^2(t)}{2} \) where:
Quantity \(\varepsilon\) that describes how easily a material polarises with an applied voltage, how easily polarizing is permitted to occur
Permittivity of a vacuumm denoted \(\varepsilon_0\)
\(\varepsilon_{0} = 8.8541878128(13) \times 10^{-12} F/m\)
\( \frac{1}{C_{eq}} = \sum \frac{1}{C_{n}}\)
\( C_{eq} = \sum C_{n}\)
\( I_{m} = C_{m} \frac{dv_{c}}{dt} \)
\( I_{T} = \frac{dv_{c}}{dt} \sum C_{m} \) (KCL)
\( \therefore C_{eq} = \sum C_{n}\)
When conducting transient analysis on capacitors, the coefficient for time can be deduced using knowlegde of differential equations and the nature of circuits to be the following:
\(\tau = RC \)
The following shows how fast the current in an inductor changes with respect to time
\(I_{c} = \frac{V_{b}}{R} (e^{-\frac{t}{\tau}}) \)
\(V_{c} = V_{b} (1-e^{-\frac{t}{\tau}}) \) where:
And by using the fact \(I = C\frac{dV}{dt}\), we see how the voltage changes in respect to time
This means during charging, current fades and voltage accumulates in a capcitor
\(I_{c} = -\frac{V_{Cmax}}{R} e^{-\frac{t}{\tau}} \)
\(V_{c} = V_{Cmax}e^{-\frac{t}{\tau}} \) where:
And by using the fact \(V = L\frac{dI}{dt}\), we see how the voltage changes in respect to time
This means during discharging, current in the opposite direction is jolted and fades and voltage fades in a capcitor
A component composed of a twisted wire through which a current going through it will produce a magnetic field that is ideally contained within the center of the wire coiling
Scalar quantity \(\Phi_{B}\) representing the total magnetic field through the surface area of a material
\(\Phi_{B}\ = \vec{B} \cdot \vec{S} \)
Scalar quantity \(\lambda\) representing the aggregate magnetic field of a series of turns in an inductor's wire
\( \lambda = N\Phi \) where:
Scalar quantity \(L\) representing material's ability to form a magnetic field per ampere
\( L = \frac{\lambda}{I} \)
\( L \equiv N^2 \frac{\mu A}{\ell}\)
\( V = L\frac{dI(t)}{dt}\) where:
\( I(t) = \frac{1}{L} \int_{t_{0}}^{t} V(t) dt + I(t_{0}) \) where:
\( W_{L}(t) = \frac{Li^2(t)}{2} \) where:
Law predicting the effect of a magnetic field on EMF (electromotive force), stating that the change in a magnetic field produces an electric field of opposite polarity to the change due to Lenz law
\(\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \) where:
When a current source is removed from a circuit, inductors release an Electromagnetic frequency that creates a bit of voltage to oppose this change. This makes
Quantity \(\mu \)that describes the amount of magnetisation in an object when a magnetic field is applied to it, how much a magnetic field permeates in a material
Permeability of a vacuumm denoted \(\mu_0\)
\(\mu_{0} = 4\pi \times 10^{-7} H/m\)
\( L_{eq} = \sum L_{n}\)
\( \frac{1}{L_{eq}} = \sum \frac{1}{L_{n}}\)
When conducting transient analysis on inductors, the coefficient for time can be deduced using knowlegde of differential equations and the nature of circuits to be the following:
\(\tau = \frac{L}{R} \)
The following shows how fast the current in an inductor changes with respect to time
\(I_{L} = \frac{V_{b}}{R} (1-e^{-\frac{t}{\tau}}) \) where:
\(V_{L} = V_{b} e^{-\frac{t}{\tau}} \) where:
And by using the fact \(V = L\frac{dI}{dt}\), we see how the voltage changes in respect to time
\(I_{L} = I_{Lmax} e^{-\frac{t}{\tau}} \)
\(V_{L} = I_{Lmax} R e^{-\frac{t}{\tau}} \) where:
And by using the fact \(V = L\frac{dI}{dt}\), we see how the voltage changes in respect to time
A component that allows for current to flow in one direction only. This component is non-linear as the VI characteristic is not linear. Assume \(V_B \equiv 0.7\) as the diode's "voltage barrier" and \(V_D\) as the voltage applied to the diode. \(V_Z\) is the Zener voltage
Diodes are non-linear, making things annoying. To find the voltage of a point of a circuit with diodes, make an assumption for each case of the diode being on or off (combinatoric says this is \(2^n\) assumptions where n is amount of diodes)
Using a single diode to ignore the negative phase of AC (since diodes allow single directional current)
When using a capacitor to create a pulsating effect, \(V_r = \frac{I_L T}{C}\)
Using a diode bridge so that the negative phase is caught and its absolute value used. The idea is that on a positive phase, \(D_1\) permits it onto the circuit and the current end up at parallel diodes \(D_3\) and \(D_4\), which both permit the current to flow back into the circuit. On a negative phase, \(D_2\) accepts the negative current and directs it through the circuit and to the parallel \(D_3\) and \(D_4\)
When using a capacitor to create a pulsating effect, \(V_r = \frac{I_L T}{2C}\)
By puting a capacitor in parallel with a component in rectifier circuits, the voltage can have a pulsating effect called a ripple. This occurs because the discharging of the capacitor releases charge and creates potential difference
\( V(t) = k\)
Wave function that can represent oscillating values
\( V(t) = \alpha \cos ( \omega (t - \varphi))\) where;
Horizontal translation of a sinusoid; said to be in-phase if the translation is the same \(\varphi_f = \varphi_g\) and out-of-phase \(| \varphi_f - \varphi_g|= \pi\)
To find the average over some interval \([T_0,T_1]\), \( \text{AV} = \frac{1}{T_{1}-T_{0}} \int_{T_{0}}^{T_{1}} V(t) dt \)
To find the whole sinusoidal average \( \text{AV} = \frac{V_{p+} + V_{p-}}{2} \)
An average where each element of the set is squared, and then the average of these square rooted. This is done in electrophysics since integrating under a sinusoidal curve in an interval of \(2 \pi+k\) will get you zero. T
\( \mu = \sqrt{\frac{\sum_{k=1}^{n} x_{k}^{2}}{n}} \)
For some AC or DC voltage...
\( \text{RMS} = \sqrt{ \frac{1}{T_{1}-T_{0}} \int_{T_{0}}^{T_{1}} V(t)^2 dt } = \frac{V_m}{\sqrt{2}} \)
\( V(t) = V_m \cos (\omega t + \theta) \implies X = V_m e^{j\theta}\)
Since adding trigonometric functions can sometimes be tiresome, it is usually easier to take the phasor for each function, and add the phasors together to get the phase of the new AC voltage comprised of both voltages in synthesis
Measure of opposition against an AC voltage or current due to resistance and reactance
Complex scalar quantity \(Z\) representing resistance and reactance in an AC circuit
\(Z_{eq} = \sum Z\)
\( \frac{1}{Z_{eq}} = \sum \frac{1}{Z}\)
Given a practical voltage source, the maximum power through a load is when the impedance of a load \(Z_{L}\) is equal to the source's conjugate impedance \(\overline{Z_{sv}}\), so when \(\overline{Z_{sv}}=Z_{L}\). If the resistance can't be complex, then it occurs at the modulus of the source's conjugate impedance (which is just equal to the source's impedance)\(Z_L=|Z_{sv}|\)
Resistors often have bands that give details on the type of transistor. Each band will have a colour that when read correctly display the resistance of the resistor and the tolerance (the percent of inaccuracy from the given resistance). The bands on the left are digits that make up the resistance of the resistor, with the last band on the left being the multiplier. The band on the right tells the tolerance of the resistor