Part 14 - Summary

Electrical Circuits and Fields

DC

Current:

The rate of flow of electrical charge.

1A=1C1s1 A = \dfrac{1 C}{1 s} 1C=6.24 10181 C = 6.24 \cdot\ 10^{18} I(t)=dQ(t)dt [A]I(t) = \dfrac{dQ(t)}{dt}\ [A] Q(t)=t0ti(t)dt+q(t0)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=ΔEpq=Wq [V]V = \dfrac{\Delta E_p}{q} = \dfrac{W}{q}\ [V]

Resistance:

Is the opposition to the flow of current.

R=ρLA [Ω]R = \frac{\rho L}{A}\ [\Omega] ρ=resistivity of the material\rho = \text{resistivity of the material}

Ohm’s Law:

V=RIV = RI

Direction: From + to -. VabV_{ab} means aa is the positive terminal and bb negative. The same goes for IabI_{ab}

KCL:

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

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

KVL:

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

k=1n Vk=0\sum_{k = 1}^{n}\ V_k = 0

Power:

P=VI [W]P = VI\ [W]

Energy:

W=t1t2P(t)dtW = \int_{t_1}^{t_2} P(t) dt

Equivalent resistance in series:

Req=R1+R2+ RNR_{eq} = R_1 + R_2 + \ldots\ R_N

Equivalent resistance in parallel:

Req=11R1+1R2+ +1RNR_{eq} = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots\ + \dfrac{1}{R_N}} Req=R1R2R1+R2R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2} Req=R1R2R3R1R2+R2R3+R1R3R_{eq} = \dfrac{R_1 R_2 R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3}

Voltage Divider (Series):

Vk=Vtotal RkR1+R2+ RNV_{k} = V_{total} \cdot\ \dfrac{R_k}{R_1 + R_2 + \ldots\ R_N}

Current Divider (Parallel):

Ik=Itotal RotherR1+R2+ RNI_{k} = I_{total} \cdot\ \dfrac{R_{other}}{R_1 + R_2 + \ldots\ R_N}

For example, for two parallel:

I1=Itotal R2R1+R2I_{1} = I_{total} \cdot\ \dfrac{R_2}{R_1 + R_2} I2=Itotal R1R1+R2I_{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, NN, nodes we have, N1N - 1, voltages)
  4. Then we solve these using KCL on each node ( Iout= Iin\sum\ I_{out} = \sum\ I_{in})

The convention is also the following:

  • Consider iouti_{out} in resistors

  • Consider iouti_{out} as positive

  • VcurrentVadjacentV_{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=0V = 0, these become short circuits.

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

  • Add the resulting responses to find the total response.

Equivalent circuits:

  1. Replace the load, RLR_L, with open/short circuit.

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

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

  4. Then:

    • VTH=VocV_{TH} = V_{oc}

    • IN=IscI_{N} = I_{sc}

    • RTH=RN=ReqR_{TH} = R_{N} = R_{eq}

VTH=IN ReqV_{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=qVC = \dfrac{q}{V} I=CdVdtI = C \dfrac{dV}{dt} P=IV=CVdVdtP = IV = CV \dfrac{dV}{dt} W(t)=t0t P(t) dtW(t)=t0t CVdVdt dtW(t)=C t0t V dVW(t)=C2[V(t)2V(t0)2]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=0V = 0 at t0t_0 then:

W(t)=C v(t)22q=CVW(t)=v(t) q(t)2C=qVW(t)=q(t)22CV=qCW(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)=1Ct0t i(t) dt+V(t0)V(t) = \frac{1}{C} \int_{t_0}^{t}\ i(t)\ dt + V(t_0)

Capacitors in series and parallel:

Parallel:

Ceq=C1+C2+ +CNC_{eq} = C_1 + C_2 + \ldots\ + C_N

Series:

Ceq=1C1+1C2+ +1CNC_{eq} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots\ + \dfrac{1}{C_N}
  • Capacitors are open circuits to DC voltage (If VV is constant, then I=0I = 0).

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

  • Capacitors store energy (I V>0I \cdot\ V > 0), or, deliver energy (I V<0I \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)=LdIdtV(t) = L \dfrac{dI}{dt} I(t)=1Lt0t V(t)dt+I(t0)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 dIdt)=dWdtP(t) = I(t) V(t) = I(L\ \frac{dI}{dt}) = \frac{dW}{dt} W=LI22W = \frac{LI^2}{2}

Inductors in series and parallel:

Series:

Leq=L1+L2+ LNL_{eq} = L_1 + L_2 + \ldots\ L_N

Parallel:

Leq=1L1+1L2+ 1LNL_{eq} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \ldots\ \dfrac{1}{L_N}
  • Inductors are short circuits to DC voltages (If II constant, then V=0V = 0).

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

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

Time-Varying Circuits:

VC(t)=Vi etτ, where τ is:τ=RCV_{C}(t) = V_{i}\ e^{\frac{-t}{\tau}} \quad \text{, where $\tau$ is:} \newline \tau = RC I(t)=I0 eRtLτ=LRI(t) = I_{0}\ e^{\frac{-Rt}{L}} \newline \tau = \frac{L}{R}

Electrical and Magnetic Fields

Charge:

e=1.602 1019Ce^{-} = 1.602 \cdot\ 10^{-19} C
  • Repulsive if charges are the same.

  • Attractive if charges are different.

Coulomb’s Law:

F12=ke q1q20r2r12^\mathbf{\vec{F_{12}}} = k_{e}\ \frac{q_1 q_20}{r^2} \hat{r_{12}} ε0=10936π8.841 1012 [Fm] (Farads per meter)\varepsilon_{0} = \frac{10^{-9}}{36\pi} \approx 8.841 \cdot\ 10^{-12}\ \left[\dfrac{F}{m}\right]\ (\text{Farads per meter}) ke=14πε09 109 [Nm2C2]k_e = \frac{1}{4\pi\varepsilon_{0}} \approx 9 \cdot\ 10^{9}\ \left[\dfrac{Nm^2}{C^2}\right] E=ke qr2r^\mathbf{\vec{E}} = k_{e}\ \frac{q}{r^2}\hat{r} FE=q E\mathbf{\vec{F_{E}}} = q\ \mathbf{\vec{E}}

Dipoles:

p=qd\mathbf{\vec{p}} = q\mathbf{\vec{d}}

Placing a dipole in an electrical field:

τ=p×Eτ=p E sin(θ)\mathbf{\vec{\tau}} = \mathbf{\vec{p}} \times \mathbf{\vec{E}} \newline \tau = p \cdot\ E\ sin(\theta)

Electrical Flux 1D:

Φ=i=1N E n^\Phi = \sum_{i = 1}^{N}\ \mathbf{\vec{E}} \cdot\ \hat{n} Φ=L1L2 El n^ dl\Phi = \int_{L_{1}}^{L_{2}}\ \mathbf{\vec{E_{l}}} \cdot\ \hat{n} \cdot\ dl

Electrical Flux 2D:

Φ=E n^ dA=E dA cos(θ)\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:

Φ= E dA\Phi = \oiint\ \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}}

Gauss’s Law:

ΦE=E dA=qε0\Phi_{E} = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = \frac{q}{\varepsilon_{0}}

Work to move charges:

W=keq1q2RW = \dfrac{k_e q_1 q_2}{R}

Work in an electrical field:

W=qE0dW = -qE_0d

Cheat sheet:

W=F drW = - \int \mathbf{\vec{F}} \cdot\ dr E=FQ\mathbf{\vec{E}} = \dfrac{\mathbf{\vec{F}}}{Q} ΔV=WQ\Delta V = \dfrac{W}{Q} ΔV=E dr\Delta V = - \int \mathbf{\vec{E}} \cdot\ dr

Capacitors:

q=σAq = \sigma A E=qε0AE = \dfrac{q}{\varepsilon_0 A}

Energy stored in a capacitor:

W=1C Q22W = \dfrac{1}{C}\ \dfrac{Q^2}{2}

Magnetic Fields:

FB=qv×B=qvB sin(θ)\mathbf{\vec{F}_B} = q\mathbf{\vec{v}} \times \mathbf{\vec{B}} = |q|vB\ sin(\theta) B[1 T=1NA m]\mathbf{\vec{B}} \left[1\ T = 1 \dfrac{N}{A \cdot\ m}\right]

Biot-Savarts law:

dB=μ04πI ds×rr2d\mathbf{\vec{B}} = \dfrac{\mu_0}{4\pi} \dfrac{I\ d\mathbf{\vec{s}} \times \mathbf{\vec{r}}}{r^2} μ0=4π 107 1.2566 106\mu_0 = 4\pi \cdot\ 10^{-7}\ \approx 1.2566 \cdot\ 10^{-6} B=μ0I4πds×rr2\mathbf{\vec{B}} = \dfrac{\mu_0 I}{4\pi} \int \dfrac{d\mathbf{\vec{s}} \times \mathbf{\vec{r}}}{r^2} B=μ0I2πrB = \dfrac{\mu_0 I}{2\pi r} B=μ0NIl=μ0nlB = \dfrac{\mu_0 NI}{l} = \mu_0 nl

Ampere’s Law:

B ds=μ0Ienc\oint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{s}} = \mu_0 I_{enc}

Lorentz Force:

FE=qE\mathbf{\vec{F}_E} = q\mathbf{\vec{E}} FB=qv×B\mathbf{\vec{F}_B} = q\mathbf{\vec{v}} \times \mathbf{\vec{B}} F=FE+FB=q(E+v×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:

ΦB=B dA=BA cos(θ)\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:

ΦE=E dA=qμ0\Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = \dfrac{q}{\mu_0}

Gauss’s Law for magnetism:

ΦB=B dA=0\Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0

Faraday’s Law:

ε=Eds=dΦBdt\varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt}

Ampere-Maxwell Law:

ε=Bds=μ0(I+Id)=μ0(I+ε0dΦEdt)\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:

ΦE=E dA=0\Phi_E = \oiint \mathbf{\vec{E}} \cdot\ d\mathbf{\vec{A}} = 0

Gauss’s Law for magnetism:

ΦB=B dA=0\Phi_B = \oiint \mathbf{\vec{B}} \cdot\ d\mathbf{\vec{A}} = 0

Faraday’s Law:

ε=Eds=dΦBdt\varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt}

Ampere-Maxwell Law:

ε=Bds=μ0(I+Id)=μ0ε0dΦEdt\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:

EB=ωk=c\dfrac{E}{B} = \dfrac{\omega}{k} = c

EMFs in circuits:

Electrical field opposes change in voltage

C=ε0AdU=12CV2I(t)=CdVdtC = \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=μ0N2AlU=12LI2V(t)=LdIdtL = \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\cos, does not matter.

V(t)=Vmsin(ωt+θ)V(t) = V_m \sin(\omega t + \theta) VmAmplitude [V]ωAngular frequency [rads]θPhase Shift[ or rad]TPeriod [s]fFrequency [Hz]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=2πω=1fω=2πfT = \dfrac{2\pi}{\omega} = \dfrac{1}{f} \newline \omega = 2\pi f

Root Mean Square (RMS):

VRMS=1T0TV2(t)dt=Vm2IRMS=1T0TI2(t)dt=Im2V_{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}} PAvg=(VRMS)2R=(Vm2)2R=Vm22RP_{Avg} = \dfrac{(V_{RMS})^2}{R} = \dfrac{\left(\dfrac{V_m}{\sqrt{2}}\right)^2}{R} = \dfrac{V_m^2}{2R} \newline PAvg=(IRMS)2R=(Im2)2R=Im2R2P_{Avg} = (I_{RMS})^2 R = \left(\dfrac{I_m}{\sqrt{2}}\right)^2 R = \dfrac{I_m^2 R}{2}

Trigonometry:

rad=deg π180rad = deg \cdot\ \dfrac{\pi}{180} sin(ωt)=cos(ωt90)cos(ωt)=sin(ωt+90)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+jyz=z(cos(φ)+jsin(φ))z = x + jy \newline z = |z|(cos(\varphi) + j sin(\varphi))

Polar form:

z=zθz=zejθz = |z|\angle{\theta^\circ} \newline z = |z|e^{j \theta^\circ}

Polar to rectangular form:

z=rθx=rcos(θ)y=rcos(θ)z=x+jyz = 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+jyr=x2+y2θ=arctan(yx)z=rθz = 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=(bc) (de)a=(b dc+e)a = (b\angle{c^\circ}) \cdot\ (d\angle{e^\circ}) \newline a = (b \cdot\ d \angle{c^\circ + e^\circ})

Division:

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

Phasor relations:

V=RIV = RI \newline V=jωLIV = j\omega LI \newline V=1jωCIV = \dfrac{1}{j\omega C} I

Impedance:

Z=VIZ = \dfrac{V}{I}

This means that:

ZR=RZL=jωLZC=1jωC=j1ωCZ_{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)=Vmcos(ωt+θV)I(t)=Imcos(ωt+ϕI)V(t) = V_m cos(\omega t + \theta_V) \newline I(t) = I_m cos(\omega t + \phi_I) P(t)=VmImcos(ωt+θV)cos(ωt+ϕI)P(t) = V_m I_m cos(\omega t + \theta_V) cos(\omega t + \phi_I) P(t)=12VmImcos(θVϕI)+12VmImcos(2ωt+θV+ϕI)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) Pavg=VmIm2cos(θVϕI)P_{avg} = \dfrac{V_m I_m}{2} cos(\theta_V - \phi_I) Pavg=VrmsIrmscos(θVϕI)P_{avg} = V_{rms} I_{rms} cos(\theta_V - \phi_I)

Different Types:

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

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

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

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