Part 14 - Summary

Electrical Circuits and Fields

DC

Current:

The rate of flow of electrical charge.

$$ 1 A = \dfrac{1 C}{1 s} $$

$$ 1 C = 6.24 \cdot\ 10^{18} $$

$$ I(t) = \dfrac{dQ(t)}{dt}\ [A] $$

$$ Q(t) = \int_{t_0}^{t} i(t) dt + q(t_0) $$

Voltage:

The difference in potential energy between two points, for one Coulomb of charge.

$$ V = \dfrac{\Delta E_p}{q} = \dfrac{W}{q}\ [V] $$

Resistance:

Is the opposition to the flow of current.

$$ R = \frac{\rho L}{A}\ [\Omega] $$

$$ \rho = \text{resistivity of the material} $$

Ohm’s Law: $$ V = RI $$

Direction: From + to -. $V_{ab}$ means $a$ is the positive terminal and $b$ negative. The same goes for $I_{ab}$

KCL:

The sum of current entering the node is equivalent to the sum of current leaving the node.

$$ \sum_{k = 1}^{n}\ I_{entering} = \sum_{k = 1}^{n}\ I_{leaving} $$

KVL:

The sum of voltages equals zero, for any closed loop.

$$ \sum_{k = 1}^{n}\ V_k = 0 $$

Power: $$ P = VI\ [W] $$

Energy: $$ W = \int_{t_1}^{t_2} P(t) dt $$

Equivalent resistance in series: $$ R_{eq} = R_1 + R_2 + \ldots\ R_N $$

Equivalent resistance in parallel: $$ R_{eq} = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots\ + \dfrac{1}{R_N}} $$

$$ R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2} $$

$$ R_{eq} = \dfrac{R_1 R_2 R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3} $$

Voltage Divider (Series): $$ V_{k} = V_{total} \cdot\ \dfrac{R_k}{R_1 + R_2 + \ldots\ R_N} $$

Current Divider (Parallel): $$ I_{k} = I_{total} \cdot\ \dfrac{R_{other}}{R_1 + R_2 + \ldots\ R_N} $$

For example, for two parallel: $$ I_{1} = I_{total} \cdot\ \dfrac{R_2}{R_1 + R_2} $$

$$ I_{2} = I_{total} \cdot\ \dfrac{R_1}{R_1 + R_2} $$

Node-Voltage analysis

Idea:

1. Find the nodes
2. Assign a reference node (usually, we pick the node with most connections)
3. Assign node voltages (Note, in a circuit with, $N$, nodes we have, $N - 1$, voltages)
4. Then we solve these using KCL on each node ($\sum\ I_{out} = \sum\ I_{in}$)

The convention is also the following:

• Consider $i_{out}$ in resistors

• Consider $i_{out}$ as positive

• $V_{current} - V_{adjacent}$

Mesh-Current analysis: Is the opposite of Node-voltage analysis. Therefore, we just apply KVL instead of KCL. Note that this only forks for planar circuits.

Planar circuit: It is possible to draw it in a plane without crossing wires.

Superposition: As the name suggest, it’s the principle that, given a linear system, the net response caused by two or more stimuli is the sum of these respones.

In our case, the stimuli are voltage/current sources.

So our method is:

• Leave one source ON and turn all other sources OFF.

  • Voltage sources: $V = 0$, these become short circuits.

  • Current sources: $I = 0$, these become open circuits.

• Add the resulting responses to find the total response.

Equivalent circuits:

1. Replace the load, $R_L$, with open/short circuit.

2. Find the short/open circuit current/voltage, $V_{oc} / I_{sc}$.

3. Find the equivalent resistance, $R_{eq}$, of the network with all independent sources turned off.

4. Then:

   • $V_{TH} = V_{oc}$

   • $I_{N} = I_{sc}$

   • $R_{TH} = R_{N} = R_{eq}$

$$ V_{TH} = I_N \cdot\ R_{eq} $$

Capacitors:

A capacitor is a device that stores electric charge by creating an electric field between two conductive plates separated by an insulating material.

$$ C = \dfrac{q}{V} $$

$$ I = C \dfrac{dV}{dt} $$

$$ P = IV = CV \dfrac{dV}{dt} $$

$$ W(t) = \int_{t_0}^{t}\ P(t)\ dt \newline W(t) = \int_{t_0}^{t}\ CV \frac{dV}{dt}\ dt \newline W(t) = C \cdot\ \int_{t_0}^{t}\ V\ dV \newline W(t) = \frac{C}{2} [V(t)^2 - V(t_0)^2] $$

If $V = 0$ at $t_0$ then: $$ W(t) = \frac{C \cdot\ v(t)^2}{2} \quad | \quad q = CV \newline W(t) = \frac{v(t)\ q(t)}{2} \quad | \quad C = \frac{q}{V} \newline W(t) = \frac{q(t)^2}{2C} \quad | \quad V = \frac{q}{C} $$

$$ V(t) = \frac{1}{C} \int_{t_0}^{t}\ i(t)\ dt + V(t_0) $$

Capacitors in series and parallel:

Parallel: $$ C_{eq} = C_1 + C_2 + \ldots\ + C_N $$

Series: $$ C_{eq} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots\ + \dfrac{1}{C_N} $$

• Capacitors are open circuits to DC voltage (If $V$ is constant, then $I = 0$).

• The voltage on a capacitor cannot jump (Change instantaneously, since then we would have infinite current).

• Capacitors store energy ($I \cdot\ V > 0$), or, deliver energy ($I \cdot\ V < 0$).

Inductors:

An inductor is a component in an electrical circuit that utilizes electromagnetic induction to resist changes in current flow by generating a voltage that opposes the change.

$$ V(t) = L \dfrac{dI}{dt} $$

$$ I(t) = \frac{1}{L} \int_{t_0}^{t}\ V(t) dt + I(t_0) $$

Power & Energy in Inductors $$ P(t) = I(t) V(t) = I(L\ \frac{dI}{dt}) = \frac{dW}{dt} $$

$$ W = \frac{LI^2}{2} $$

Inductors in series and parallel:

Series: $$ L_{eq} = L_1 + L_2 + \ldots\ L_N $$

Parallel: $$ L_{eq} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \ldots\ \dfrac{1}{L_N} $$

• Inductors are short circuits to DC voltages (If $I$ constant, then $V = 0$).

• The current through an inductor cannot jump (change instantaneously, otherwise we would have infinite voltage).

• Inductors store energy ($I \cdot\ V > 0$), or, deliver energy ($I \cdot\ V < 0$).

Time-Varying Circuits:

$$ V_{C}(t) = V_{i}\ e^{\frac{-t}{\tau}} \quad \text{, where $\tau$ is:} \newline \tau = RC $$

$$ I(t) = I_{0}\ e^{\frac{-Rt}{L}} \newline \tau = \frac{L}{R} $$

Electrical and Magnetic Fields

Charge: $$ e^{-} = 1.602 \cdot\ 10^{-19} C $$

• Repulsive if charges are the same.

• Attractive if charges are different.

Coulomb’s Law: $$ \mathbf{\vec{F_{12}}} = k_{e}\ \frac{q_1 q_20}{r^2} \hat{r_{12}} $$

$$ \varepsilon_{0} = \frac{10^{-9}}{36\pi} \approx 8.841 \cdot\ 10^{-12}\ \left[\dfrac{F}{m}\right]\ (\text{Farads per meter}) $$

$$ k_e = \frac{1}{4\pi\varepsilon_{0}} \approx 9 \cdot\ 10^{9}\ \left[\dfrac{Nm^2}{C^2}\right] $$

$$ \mathbf{\vec{E}} = k_{e}\ \frac{q}{r^2}\hat{r} $$

$$ \mathbf{\vec{F_{E}}} = q\ \mathbf{\vec{E}} $$

Dipoles: $$ \mathbf{\vec{p}} = q\mathbf{\vec{d}} $$

Placing a dipole in an electrical field: $$ \mathbf{\vec{\tau}} = \mathbf{\vec{p}} \times \mathbf{\vec{E}} \newline \tau = p \cdot\ E\ sin(\theta) $$

Electrical Flux 1D: $$ \Phi = \sum_{i = 1}^{N}\ \mathbf{\vec{E}} \cdot\ \hat{n} $$

$$ \Phi = \int_{L_{1}}^{L_{2}}\ \mathbf{\vec{E_{l}}} \cdot\ \hat{n} \cdot\ dl $$

Electrical Flux 2D: $$ \Phi = \iint \mathbf{\vec{E}} \cdot\ \hat{n} \cdot\ d\mathbf{\vec{A}} = \iint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}}\ cos(\theta) $$

Electrical Flux closed contour: $$ \Phi = \oiint\ \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} $$

Gauss’s Law: $$ \Phi_{E} = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = \frac{q}{\varepsilon_{0}} $$

Work to move charges: $$ W = \dfrac{k_e q_1 q_2}{R} $$

Work in an electrical field: $$ W = -qE_0d $$

Cheat sheet: $$ W = - \int \mathbf{\vec{F}} \cdot\ dr $$

$$ \mathbf{\vec{E}} = \dfrac{\mathbf{\vec{F}}}{Q} $$

$$ \Delta V = \dfrac{W}{Q} $$

$$ \Delta V = - \int \mathbf{\vec{E}} \cdot\ dr $$

Capacitors: $$ q = \sigma A $$

$$ E = \dfrac{q}{\varepsilon_0 A} $$

Energy stored in a capacitor: $$ W = \dfrac{1}{C}\ \dfrac{Q^2}{2} $$

Magnetic Fields: $$ \mathbf{\vec{F}_B} = q\mathbf{\vec{v}} \times \mathbf{\vec{B}} = |q|vB\ sin(\theta) $$

$$ \mathbf{\vec{B}} \left[1\ T = 1 \dfrac{N}{A \cdot\ m}\right] $$

Biot-Savarts law: $$ d\mathbf{\vec{B}} = \dfrac{\mu_0}{4\pi} \dfrac{I\ d\mathbf{\vec{s}} \times \mathbf{\vec{r}}}{r^2} $$

$$ \mu_0 = 4\pi \cdot\ 10^{-7}\ \approx 1.2566 \cdot\ 10^{-6} $$

$$ \mathbf{\vec{B}} = \dfrac{\mu_0 I}{4\pi} \int \dfrac{d\mathbf{\vec{s}} \times \mathbf{\vec{r}}}{r^2} $$

$$ B = \dfrac{\mu_0 I}{2\pi r} $$

$$ B = \dfrac{\mu_0 NI}{l} = \mu_0 nl $$

Ampere’s Law: $$ \oint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{s}} = \mu_0 I_{enc} $$

Lorentz Force: $$ \mathbf{\vec{F}_E} = q\mathbf{\vec{E}} $$

$$ \mathbf{\vec{F}_B} = q\mathbf{\vec{v}} \times \mathbf{\vec{B}} $$

$$ \mathbf{\vec{F} = \mathbf{\vec{F}_E} + \mathbf{\vec{F}_B} = q(\mathbf{\vec{E}} + \mathbf{\vec{v}} \times \mathbf{\vec{B}})} $$

Magnetic Flux: $$ \Phi_B = \int \int \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = BA\ cos(\theta) $$

Lenz Law:

The induced current produces a magnetic field, which oppose the change in magnetic flux that induces such currents.

Maxwell’s Equations:

Gauss’s Law for electrostatics: $$ \Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = \dfrac{q}{\mu_0} $$

Gauss’s Law for magnetism: $$ \Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0 $$

Faraday’s Law: $$ \varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt} $$

Ampere-Maxwell Law: $$ \varepsilon = \oint \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} = \mu_0(I + I_d) = \mu_0\left(I + \varepsilon_0 \dfrac{d \Phi_E}{dt}\right) $$

If q = 0 and I = 0:

Gauss’s Law for electrostatics: $$ \Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = 0 $$

Gauss’s Law for magnetism: $$ \Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0 $$

Faraday’s Law: $$ \varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt} $$

Ampere-Maxwell Law: $$ \varepsilon = \oint \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} = \mu_0(I + I_d) = \mu_0 \varepsilon_0 \dfrac{d \Phi_E}{dt} $$

Electromagnetic Waves: $$ \dfrac{E}{B} = \dfrac{\omega}{k} = c $$

EMFs in circuits:

Electrical field opposes change in voltage $$ C = \varepsilon_0 \dfrac{A}{d} \newline U = \dfrac{1}{2} C|V^2| \newline I(t) = C \dfrac{dV}{dt} $$

Magnetic field opposes change in the current $$ L = \mu_0 N^2 \dfrac{A}{l} \newline U = \dfrac{1}{2} LI^2 \newline V(t) = L \dfrac{dI}{dt} $$

AC

General:

Note: Can be $\cos$, does not matter. $$ V(t) = V_m \sin(\omega t + \theta) $$

$$ V_m - \text{Amplitude}\ [V] \newline \omega - \text{Angular frequency}\ \left[\dfrac{rad}{s}\right] \newline \theta - \text{Phase Shift} \left[^\circ\ \text{or}\ rad \right] \newline T - \text{Period}\ [s] \newline f - \text{Frequency}\ [Hz] \newline $$

$$ T = \dfrac{2\pi}{\omega} = \dfrac{1}{f} \newline \omega = 2\pi f $$

Root Mean Square (RMS): $$ V_{RMS} = \sqrt{\dfrac{1}{T} \int_0^T V^2(t) dt} = \dfrac{V_m}{\sqrt{2}} \newline I_{RMS} = \sqrt{\dfrac{1}{T} \int_0^T I^2(t) dt} = \dfrac{I_m}{\sqrt{2}} $$

$$ P_{Avg} = \dfrac{(V_{RMS})^2}{R} = \dfrac{\left(\dfrac{V_m}{\sqrt{2}}\right)^2}{R} = \dfrac{V_m^2}{2R} \newline $$

$$ P_{Avg} = (I_{RMS})^2 R = \left(\dfrac{I_m}{\sqrt{2}}\right)^2 R = \dfrac{I_m^2 R}{2} $$

Trigonometry: $$ rad = deg \cdot\ \dfrac{\pi}{180} $$

$$ sin(\omega t) = cos(\omega t - 90^\circ) \newline cos(\omega t) = sin(\omega t + 90^\circ) $$

Rules for comparing the phase of two wave functions:

• Both must be written in either sine or cosine.

• Both must be written with positive amplitude.

• Each have the same constant frequency.

Phasors:

Rectangular form: $$ z = x + jy \newline z = |z|(cos(\varphi) + j sin(\varphi)) $$

Polar form: $$ z = |z|\angle{\theta^\circ} \newline z = |z|e^{j \theta^\circ} $$

Polar to rectangular form: $$ z = r\angle{\theta^\circ} \newline x = r cos(\theta^\circ) \newline y = r cos(\theta^\circ) \newline z = x + j y $$

Rectangular to polar form: $$ z = x + jy \newline r = \sqrt{x^2 + y^2} \newline \theta = \arctan{\left(\dfrac{y}{x}\right)} \newline z = r\angle{\theta^\circ} \newline $$

Operations:

Multiplication: $$ a = (b\angle{c^\circ}) \cdot\ (d\angle{e^\circ}) \newline a = (b \cdot\ d \angle{c^\circ + e^\circ}) $$

Division:

$$ a = \dfrac{(b\angle{c^\circ})}{(d\angle{e^\circ})} \newline a = (\dfrac{b}{d} \angle{c^\circ - e^\circ}) $$

Phasor relations: $$ V = RI \newline $$ $$ V = j\omega LI \newline $$ $$ V = \dfrac{1}{j\omega C} I $$

Impedance: $$ Z = \dfrac{V}{I} $$

This means that: $$ Z_{R} = R \newline Z_{L} = j\omega L \newline Z_{C} = \dfrac{1}{j\omega C} = -j \dfrac{1}{\omega C} $$

Power in AC - general:

$$ V(t) = V_m cos(\omega t + \theta_V) \newline I(t) = I_m cos(\omega t + \phi_I) $$

$$ P(t) = V_m I_m cos(\omega t + \theta_V) cos(\omega t + \phi_I) $$

$$ P(t) = \dfrac{1}{2} V_m I_m cos(\theta_V - \phi_I) + \dfrac{1}{2} V_m I_m cos(2 \omega t + \theta_V + \phi_I) $$

$$ P_{avg} = \dfrac{V_m I_m}{2} cos(\theta_V - \phi_I) $$

$$ P_{avg} = V_{rms} I_{rms} cos(\theta_V - \phi_I) $$

Different Types:

• Real Power: $P = V_{rms} I_{rms}\ cos(\theta_V - \phi_I)\ [W]$

• Reactive Power: $Q = V_{rms} I_{rms} sin(\theta_V - \phi_I) [VAR]$

• Complex Power: $S = P + jQ$ or in polar form, $S = V_{rms} I_{rms} \angle \theta_V - \phi_I [VA]$

• Apparent Power: $|S| = V_{rms} I_{rms} [VA]$.