Electrical Circuits and Fields
DC
Recall (Current) The rate of flow of electrical charge.
1 A = 1 C 1 s 1 \ \mathrm{A} = \dfrac{1 \ \mathrm{C}}{1 \ \mathrm{s}} 1 A = 1 s 1 C 1 C = 6.24 ⋅ 10 18 1 \ \mathrm{C} = 6.24 \cdot 10^{18} 1 C = 6.24 ⋅ 1 0 18 I ( t ) = d Q ( t ) d t [ A ] I(t) = \dfrac{dQ(t)}{dt} \ [\mathrm{A}] I ( t ) = d t d Q ( t ) [ A ] Q ( t ) = ∫ t 0 t i ( t ) d t + q ( t 0 ) Q(t) = \int_{t_0}^{t} i(t) \, dt + q(t_0) Q ( t ) = ∫ t 0 t i ( t ) d t + q ( t 0 ) Recall (Voltage) The difference in potential energy between two points, for one Coulomb of charge.
V = Δ E p q = W q [ V ] V = \dfrac{\Delta E_p}{q} = \dfrac{W}{q} \ [\mathrm{V}] V = q Δ E p = q W [ V ] Recall (Resistance) The opposition to the flow of current.
R = ρ L A [ Ω ] R = \frac{\rho L}{A} \ [\Omega] R = A ρ L [ Ω ] ρ = resistivity of the material \rho = \text{resistivity of the material} ρ = resistivity of the material Ohm’s Law:
V = R I V = RI V = R I Direction:
From + to -. V a b V_{ab} V ab a a a b b b I a b I_{ab} I ab
Recall (KCL) The sum of current entering the node is equivalent to the sum of current leaving the node.
∑ k = 1 n I entering = ∑ k = 1 n I leaving \sum_{k = 1}^{n} I_{\text{entering}} = \sum_{k = 1}^{n} I_{\text{leaving}} k = 1 ∑ n I entering = k = 1 ∑ n I leaving Recall (KVL) The sum of voltages equals zero, for any closed loop.
∑ k = 1 n V k = 0 \sum_{k = 1}^{n} V_k = 0 k = 1 ∑ n V k = 0 Power:
P = V I [ W ] P = VI \ [\mathrm{W}] P = V I [ W ] Energy:
W = ∫ t 1 t 2 P ( t ) d t W = \int_{t_1}^{t_2} P(t) \, dt W = ∫ t 1 t 2 P ( t ) d t Equivalent resistance in series:
R e q = R 1 + R 2 + … + R N R_{eq} = R_1 + R_2 + \ldots + R_N R e q = R 1 + R 2 + … + R N Equivalent resistance in parallel:
R e q = 1 1 R 1 + 1 R 2 + … + 1 R N R_{eq} = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots + \dfrac{1}{R_N}} R e q = R 1 1 + R 2 1 + … + R N 1 1 R e q = R 1 R 2 R 1 + R 2 R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2} R e q = R 1 + R 2 R 1 R 2 R e q = R 1 R 2 R 3 R 1 R 2 + R 2 R 3 + R 1 R 3 R_{eq} = \dfrac{R_1 R_2 R_3}{R_1 R_2 + R_2 R_3 + R_1 R_3} R e q = R 1 R 2 + R 2 R 3 + R 1 R 3 R 1 R 2 R 3 Voltage Divider (Series):
V k = V total ⋅ R k R 1 + R 2 + … + R N V_{k} = V_{\text{total}} \cdot \dfrac{R_k}{R_1 + R_2 + \ldots + R_N} V k = V total ⋅ R 1 + R 2 + … + R N R k Current Divider (Parallel):
I k = I total ⋅ R other R 1 + R 2 + … + R N I_{k} = I_{\text{total}} \cdot \dfrac{R_{\text{other}}}{R_1 + R_2 + \ldots + R_N} I k = I total ⋅ R 1 + R 2 + … + R N R other For example, for two parallel:
I 1 = I total ⋅ R 2 R 1 + R 2 I_{1} = I_{\text{total}} \cdot \dfrac{R_2}{R_1 + R_2} I 1 = I total ⋅ R 1 + R 2 R 2 I 2 = I total ⋅ R 1 R 1 + R 2 I_{2} = I_{\text{total}} \cdot \dfrac{R_1}{R_1 + R_2} I 2 = I total ⋅ R 1 + R 2 R 1 Node-Voltage analysis
Idea:
Algorithm 1: Node-voltage analysis
Find the nodes
Assign a reference node (usually, we pick the node with most connections)
Assign node voltages (Note, in a circuit with, N N N N − 1 N - 1 N − 1
Then we solve these using KCL on each node (∑ I o u t = ∑ I i n \sum I_{out} = \sum I_{in} ∑ I o u t = ∑ I in
The convention is also the following:
Consider i o u t i_{out} i o u t
Consider i o u t i_{out} i o u t
V current − V adjacent V_{\text{current}} - V_{\text{adjacent}} 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 works for planar circuits .
Recall (Planar circuit) It is possible to draw it in a plane without crossing wires.
Superposition:
As the name suggests, it’s the principle that, given a linear system, the net response caused by two or more stimuli is the sum of these responses.
In our case, the stimuli are voltage/current sources.
So our method is:
Algorithm 2: Superposition
Leave one source ON and turn all other sources OFF.
Voltage sources: V = 0 V = 0 V = 0 short circuits .
Current sources: I = 0 I = 0 I = 0 open circuits .
Add the resulting responses to find the total response.
Equivalent circuits:
Algorithm 3: Thévenin and Norton equivalents
Replace the load, R L R_L R L
Find the short/open circuit current/voltage, V o c / I s c V_{oc} / I_{sc} V oc / I sc
Find the equivalent resistance, R e q R_{eq} R e q
Then:
V T H = V o c V_{TH} = V_{oc} V T H = V oc I N = I s c I_{N} = I_{sc} I N = I sc R T H = R N = R e q R_{TH} = R_{N} = R_{eq} R T H = R N = R e q
V T H = I N ⋅ R e q V_{TH} = I_N \cdot R_{eq} V T H = I N ⋅ R e q Capacitors:
Recall (Capacitor) A capacitor is a device that stores electric charge by creating an electric field between two conductive plates separated by an insulating material.
C = q V C = \dfrac{q}{V} C = V q I = C d V d t I = C \dfrac{dV}{dt} I = C d t d V P = I V = C V d V d t P = IV = CV \dfrac{dV}{dt} P = I V = C V d t d V W ( t ) = ∫ t 0 t P ( t ) d t = ∫ t 0 t C V d V d t d t = C ∫ t 0 t V d V = C 2 [ V ( t ) 2 − V ( t 0 ) 2 ] \begin{align*}
W(t) & = \int_{t_0}^{t} P(t) \, dt \newline
& = \int_{t_0}^{t} CV \frac{dV}{dt} \, dt \newline
& = C \int_{t_0}^{t} V \, dV \newline
& = \frac{C}{2} [V(t)^2 - V(t_0)^2]
\end{align*} W ( t ) = ∫ t 0 t P ( t ) d t = ∫ t 0 t C V d t d V d t = C ∫ t 0 t V d V = 2 C [ V ( t ) 2 − V ( t 0 ) 2 ] If V = 0 V = 0 V = 0 t 0 t_0 t 0
W ( t ) = C v ( t ) 2 2 ∥ q = C V W ( t ) = v ( t ) q ( t ) 2 ∥ C = q V W ( t ) = q ( t ) 2 2 C ∥ V = q C \begin{align*}
W(t) & = \frac{C v(t)^2}{2} & & \| \quad q = CV \newline
W(t) & = \frac{v(t) q(t)}{2} & & \| \quad C = \frac{q}{V} \newline
W(t) & = \frac{q(t)^2}{2C} & & \| \quad V = \frac{q}{C}
\end{align*} W ( t ) W ( t ) W ( t ) = 2 C v ( t ) 2 = 2 v ( t ) q ( t ) = 2 C q ( t ) 2 ∥ q = C V ∥ C = V q ∥ V = C q V ( t ) = 1 C ∫ t 0 t i ( t ) d t + V ( t 0 ) V(t) = \frac{1}{C} \int_{t_0}^{t} i(t) \, dt + V(t_0) V ( t ) = C 1 ∫ t 0 t i ( t ) d t + V ( t 0 ) Capacitors in series and parallel:
Parallel:
C e q = C 1 + C 2 + … + C N C_{eq} = C_1 + C_2 + \ldots + C_N C e q = C 1 + C 2 + … + C N Series:
1 C e q = 1 C 1 + 1 C 2 + … + 1 C N \dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots + \dfrac{1}{C_N} C e q 1 = C 1 1 + C 2 1 + … + C N 1
Capacitors are open circuits to DC voltage (If V V V I = 0 I = 0 I = 0
The voltage on a capacitor cannot jump (Change instantaneously, since then we would have infinite current).
Capacitors store energy (I ⋅ V > 0 I \cdot V > 0 I ⋅ V > 0 I ⋅ V < 0 I \cdot V < 0 I ⋅ V < 0
Inductors:
Recall (Inductor) 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 d I d t V(t) = L \dfrac{dI}{dt} V ( t ) = L d t d I I ( t ) = 1 L ∫ t 0 t V ( t ) d t + I ( t 0 ) I(t) = \frac{1}{L} \int_{t_0}^{t} V(t) \, dt + I(t_0) I ( t ) = L 1 ∫ t 0 t V ( t ) d t + I ( t 0 ) Power & Energy in Inductors
P ( t ) = I ( t ) V ( t ) = I ( L d I d t ) = d W d t P(t) = I(t) V(t) = I \left(L \frac{dI}{dt}\right) = \frac{dW}{dt} P ( t ) = I ( t ) V ( t ) = I ( L d t d I ) = d t d W W = L I 2 2 W = \frac{LI^2}{2} W = 2 L I 2 Inductors in series and parallel:
Series:
L e q = L 1 + L 2 + … + L N L_{eq} = L_1 + L_2 + \ldots + L_N L e q = L 1 + L 2 + … + L N Parallel:
1 L e q = 1 L 1 + 1 L 2 + … + 1 L N \dfrac{1}{L_{eq}} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \ldots + \dfrac{1}{L_N} L e q 1 = L 1 1 + L 2 1 + … + L N 1
Inductors are short circuits to DC voltages (If I I I V = 0 V = 0 V = 0
The current through an inductor cannot jump (change instantaneously, otherwise we would have infinite voltage).
Inductors store energy (I ⋅ V > 0 I \cdot V > 0 I ⋅ V > 0 I ⋅ V < 0 I \cdot V < 0 I ⋅ V < 0
Time-Varying Circuits:
V C ( t ) = V i e − t / τ , where τ = R C V_{C}(t) = V_{i} e^{-t / \tau}, \quad \text{where } \tau = RC V C ( t ) = V i e − t / τ , where τ = R C I ( t ) = I 0 e − R t / L , where τ = L R I(t) = I_{0} e^{-Rt / L}, \quad \text{where } \tau = \frac{L}{R} I ( t ) = I 0 e − R t / L , where τ = R L Electrical and Magnetic Fields
Charge:
e − = 1.602 ⋅ 10 − 19 C e^{-} = 1.602 \cdot 10^{-19} \ \mathrm{C} e − = 1.602 ⋅ 1 0 − 19 C
Repulsive if charges are the same .Attractive if charges are different .
Coulomb’s Law:
F ⃗ 12 = k e q 1 q 2 r 2 r ^ 12 \mathbf{\vec{F}}_{12} = k_{e} \frac{q_1 q_2}{r^2} \hat{r}_{12} F 12 = k e r 2 q 1 q 2 r ^ 12 ε 0 = 10 − 9 36 π ≈ 8.841 ⋅ 10 − 12 [ F / m ] ( Farads per meter ) \varepsilon_{0} = \frac{10^{-9}}{36\pi} \approx 8.841 \cdot 10^{-12} \ [\mathrm{F/m}] \ (\text{Farads per meter}) ε 0 = 36 π 1 0 − 9 ≈ 8.841 ⋅ 1 0 − 12 [ F/m ] ( Farads per meter ) k e = 1 4 π ε 0 ≈ 9 ⋅ 10 9 [ N m 2 / C 2 ] k_e = \frac{1}{4\pi\varepsilon_{0}} \approx 9 \cdot 10^{9} \ [\mathrm{N\,m^2/C^2}] k e = 4 π ε 0 1 ≈ 9 ⋅ 1 0 9 [ N m 2 / C 2 ] E ⃗ = k e q r 2 r ^ \mathbf{\vec{E}} = k_{e} \frac{q}{r^2}\hat{r} E = k e r 2 q r ^ F ⃗ E = q E ⃗ \mathbf{\vec{F}}_{E} = q \mathbf{\vec{E}} F E = q E Dipoles:
p ⃗ = q d ⃗ \mathbf{\vec{p}} = q\mathbf{\vec{d}} p = q d Placing a dipole in an electrical field:
τ ⃗ = p ⃗ × E ⃗ , τ = p E sin ( θ ) . \begin{align*}
\mathbf{\vec{\tau}} & = \mathbf{\vec{p}} \times \mathbf{\vec{E}}, \newline
\tau & = p E \sin(\theta).
\end{align*} τ τ = p × E , = pE sin ( θ ) . Electrical Flux 1D:
Φ = ∑ i = 1 N E ⃗ ⋅ n ^ \Phi = \sum_{i = 1}^{N} \mathbf{\vec{E}} \cdot \hat{n} Φ = i = 1 ∑ N E ⋅ n ^ Φ = ∫ L 1 L 2 E ⃗ l ⋅ n ^ d l \Phi = \int_{L_{1}}^{L_{2}} \mathbf{\vec{E}}_{l} \cdot \hat{n} \, dl Φ = ∫ L 1 L 2 E l ⋅ n ^ d l Electrical Flux 2D:
Φ = ∬ E ⃗ ⋅ n ^ d A = ∬ E d A cos ( θ ) \Phi = \iint \mathbf{\vec{E}} \cdot \hat{n} \, dA = \iint E \, dA \cos(\theta) Φ = ∬ E ⋅ n ^ d A = ∬ E d A cos ( θ ) Electrical Flux closed contour:
Φ = ∯ E ⃗ ⋅ d A ⃗ \Phi = \oiint \mathbf{\vec{E}} \cdot d\mathbf{\vec{A}} Φ = ∬ E ⋅ d A Gauss’s Law:
Φ E = ∯ E ⃗ ⋅ d A ⃗ = q ε 0 \Phi_{E} = \oiint \mathbf{\vec{E}} \cdot d\mathbf{\vec{A}} = \frac{q}{\varepsilon_{0}} Φ E = ∬ E ⋅ d A = ε 0 q Work to move charges:
W = k e q 1 q 2 R W = \dfrac{k_e q_1 q_2}{R} W = R k e q 1 q 2 Work in an electrical field:
W = − q E 0 d W = -qE_0d W = − q E 0 d Cheat sheet:
W = − ∫ F ⃗ ⋅ d r ⃗ W = -\int \mathbf{\vec{F}} \cdot d\mathbf{\vec{r}} W = − ∫ F ⋅ d r E ⃗ = F ⃗ Q \mathbf{\vec{E}} = \dfrac{\mathbf{\vec{F}}}{Q} E = Q F Δ V = W Q \Delta V = \dfrac{W}{Q} Δ V = Q W Δ V = − ∫ E ⃗ ⋅ d r ⃗ \Delta V = -\int \mathbf{\vec{E}} \cdot d\mathbf{\vec{r}} Δ V = − ∫ E ⋅ d r Capacitors:
q = σ A q = \sigma A q = σ A E = q ε 0 A E = \dfrac{q}{\varepsilon_0 A} E = ε 0 A q Energy stored in a capacitor:
W = 1 C Q 2 2 W = \dfrac{1}{C} \dfrac{Q^2}{2} W = C 1 2 Q 2 Magnetic Fields:
F ⃗ B = q v ⃗ × B ⃗ , ∣ F ⃗ B ∣ = ∣ q ∣ v B sin ( θ ) \mathbf{\vec{F}}_B = q\mathbf{\vec{v}} \times \mathbf{\vec{B}}, \quad |\mathbf{\vec{F}}_B| = |q|vB \sin(\theta) F B = q v × B , ∣ F B ∣ = ∣ q ∣ v B sin ( θ ) B ⃗ [ 1 T = 1 N / ( A ⋅ m ) ] \mathbf{\vec{B}} \ \left[1 \ \mathrm{T} = 1 \ \mathrm{N/(A \cdot m)}\right] B [ 1 T = 1 N/ ( A ⋅ m ) ] Biot-Savart’s law:
d B ⃗ = μ 0 4 π I d s ⃗ × r ^ r 2 d\mathbf{\vec{B}} = \dfrac{\mu_0}{4\pi} \dfrac{I \, d\mathbf{\vec{s}} \times \hat{r}}{r^2} d B = 4 π μ 0 r 2 I d s × r ^ μ 0 = 4 π ⋅ 10 − 7 ≈ 1.2566 ⋅ 10 − 6 \mu_0 = 4\pi \cdot 10^{-7} \approx 1.2566 \cdot 10^{-6} μ 0 = 4 π ⋅ 1 0 − 7 ≈ 1.2566 ⋅ 1 0 − 6 B ⃗ = μ 0 I 4 π ∫ d s ⃗ × r ^ r 2 \mathbf{\vec{B}} = \dfrac{\mu_0 I}{4\pi} \int \dfrac{d\mathbf{\vec{s}} \times \hat{r}}{r^2} B = 4 π μ 0 I ∫ r 2 d s × r ^ B = μ 0 I 2 π r B = \dfrac{\mu_0 I}{2\pi r} B = 2 π r μ 0 I B = μ 0 N I l = μ 0 n I B = \dfrac{\mu_0 NI}{l} = \mu_0 nI B = l μ 0 N I = μ 0 n I Ampere’s Law:
∮ B ⃗ ⋅ d s ⃗ = μ 0 I enc \oint \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} = \mu_0 I_{\text{enc}} ∮ B ⋅ d s = μ 0 I enc Lorentz Force:
F ⃗ E = q E ⃗ \mathbf{\vec{F}}_E = q\mathbf{\vec{E}} F E = q E F ⃗ B = q v ⃗ × B ⃗ \mathbf{\vec{F}}_B = q\mathbf{\vec{v}} \times \mathbf{\vec{B}} F B = q v × B F ⃗ = F ⃗ E + F ⃗ B = 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}}) F = F E + F B = q ( E + v × B ) Magnetic Flux:
Φ B = ∬ B ⃗ ⋅ d A ⃗ = B A cos ( θ ) \Phi_B = \iint \mathbf{\vec{B}} \cdot d\mathbf{\vec{A}} = BA \cos(\theta) Φ B = ∬ B ⋅ d A = B A cos ( θ ) Lenz’s Law:
Recall (Lenz’s law) The induced current produces a magnetic field, which opposes the change in magnetic flux that induces such currents.
Maxwell’s Equations:
Gauss’s Law for electrostatics :
Φ E = ∯ E ⃗ ⋅ d A ⃗ = q ε 0 \Phi_E = \oiint \mathbf{\vec{E}} \cdot d\mathbf{\vec{A}} = \dfrac{q}{\varepsilon_0} Φ E = ∬ E ⋅ d A = ε 0 q Gauss’s Law for magnetism :
Φ B = ∯ B ⃗ ⋅ d A ⃗ = 0 \Phi_B = \oiint \mathbf{\vec{B}} \cdot d\mathbf{\vec{A}} = 0 Φ B = ∬ B ⋅ d A = 0 Faraday’s Law :
ε = ∮ E ⃗ ⋅ d s ⃗ = − d Φ B d t \varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt} ε = ∮ E ⋅ d s = − d t d Φ B Ampere-Maxwell Law :
∮ B ⃗ ⋅ d s ⃗ = μ 0 ( I + I d ) = μ 0 ( I + ε 0 d Φ E d t ) \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) ∮ B ⋅ d s = μ 0 ( I + I d ) = μ 0 ( I + ε 0 d t d Φ E ) If q = 0 q = 0 q = 0 I = 0 I = 0 I = 0
Gauss’s Law for electrostatics :
Φ E = ∯ E ⃗ ⋅ d A ⃗ = 0 \Phi_E = \oiint \mathbf{\vec{E}} \cdot d\mathbf{\vec{A}} = 0 Φ E = ∬ E ⋅ d A = 0 Gauss’s Law for magnetism :
Φ B = ∯ B ⃗ ⋅ d A ⃗ = 0 \Phi_B = \oiint \mathbf{\vec{B}} \cdot d\mathbf{\vec{A}} = 0 Φ B = ∬ B ⋅ d A = 0 Faraday’s Law :
ε = ∮ E ⃗ ⋅ d s ⃗ = − d Φ B d t \varepsilon = \oint \mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} = -\dfrac{d \Phi_B}{dt} ε = ∮ E ⋅ d s = − d t d Φ B Ampere-Maxwell Law :
∮ B ⃗ ⋅ d s ⃗ = μ 0 ( I + I d ) = μ 0 ε 0 d Φ E d t \oint \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} = \mu_0(I + I_d) = \mu_0 \varepsilon_0 \dfrac{d \Phi_E}{dt} ∮ B ⋅ d s = μ 0 ( I + I d ) = μ 0 ε 0 d t d Φ E Electromagnetic Waves:
E B = ω k = c \dfrac{E}{B} = \dfrac{\omega}{k} = c B E = k ω = c EMFs in circuits:
Recall Electrical field opposes change in voltage.
C = ε 0 A d , U = 1 2 C ∣ V ∣ 2 , I ( t ) = C d V d t . \begin{align*}
C & = \varepsilon_0 \dfrac{A}{d}, \newline
U & = \dfrac{1}{2} C|V|^2, \newline
I(t) & = C \dfrac{dV}{dt}.
\end{align*} C U I ( t ) = ε 0 d A , = 2 1 C ∣ V ∣ 2 , = C d t d V . Recall Magnetic field opposes change in the current.
L = μ 0 N 2 A l , U = 1 2 L I 2 , V ( t ) = L d I d t . \begin{align*}
L & = \mu_0 N^2 \dfrac{A}{l}, \newline
U & = \dfrac{1}{2} LI^2, \newline
V(t) & = L \dfrac{dI}{dt}.
\end{align*} L U V ( t ) = μ 0 N 2 l A , = 2 1 L I 2 , = L d t d I . AC
General:
Note: Can be cos \cos cos
V ( t ) = V m sin ( ω t + θ ) V(t) = V_m \sin(\omega t + \theta) V ( t ) = V m sin ( ω t + θ ) V m Amplitude [ V ] ω Angular frequency [ r a d / s ] θ Phase Shift [ ∘ or r a d ] T Period [ s ] f Frequency [ H z ] \begin{align*}
V_m & \quad \text{Amplitude} \ [\mathrm{V}] \newline
\omega & \quad \text{Angular frequency} \ [\mathrm{rad/s}] \newline
\theta & \quad \text{Phase Shift} \ [^\circ \text{ or } \mathrm{rad}] \newline
T & \quad \text{Period} \ [\mathrm{s}] \newline
f & \quad \text{Frequency} \ [\mathrm{Hz}]
\end{align*} V m ω θ T f Amplitude [ V ] Angular frequency [ rad/s ] Phase Shift [ ∘ or rad ] Period [ s ] Frequency [ Hz ] T = 2 π ω = 1 f , ω = 2 π f . \begin{align*}
T & = \dfrac{2\pi}{\omega} = \dfrac{1}{f}, \newline
\omega & = 2\pi f.
\end{align*} T ω = ω 2 π = f 1 , = 2 π f . Root Mean Square (RMS):
V RMS = 1 T ∫ 0 T V 2 ( t ) d t = V m 2 , I RMS = 1 T ∫ 0 T I 2 ( t ) d t = I m 2 . \begin{align*}
V_{\text{RMS}} & = \sqrt{\dfrac{1}{T} \int_0^T V^2(t) \, dt} = \dfrac{V_m}{\sqrt{2}}, \newline
I_{\text{RMS}} & = \sqrt{\dfrac{1}{T} \int_0^T I^2(t) \, dt} = \dfrac{I_m}{\sqrt{2}}.
\end{align*} V RMS I RMS = T 1 ∫ 0 T V 2 ( t ) d t = 2 V m , = T 1 ∫ 0 T I 2 ( t ) d t = 2 I m . P avg = ( V RMS ) 2 R = ( V m 2 ) 2 R = V m 2 2 R P_{\text{avg}} = \dfrac{(V_{\text{RMS}})^2}{R} = \dfrac{\left(\dfrac{V_m}{\sqrt{2}}\right)^2}{R} = \dfrac{V_m^2}{2R} P avg = R ( V RMS ) 2 = R ( 2 V m ) 2 = 2 R V m 2 P avg = ( I RMS ) 2 R = ( I m 2 ) 2 R = I m 2 R 2 P_{\text{avg}} = (I_{\text{RMS}})^2 R = \left(\dfrac{I_m}{\sqrt{2}}\right)^2 R = \dfrac{I_m^2 R}{2} P avg = ( I RMS ) 2 R = ( 2 I m ) 2 R = 2 I m 2 R Trigonometry:
rad = deg ⋅ π 180 \text{rad} = \text{deg} \cdot \dfrac{\pi}{180} rad = deg ⋅ 180 π sin ( ω t ) = cos ( ω t − 90 ∘ ) , cos ( ω t ) = sin ( ω t + 90 ∘ ) . \begin{align*}
\sin(\omega t) & = \cos(\omega t - 90^\circ), \newline
\cos(\omega t) & = \sin(\omega t + 90^\circ).
\end{align*} sin ( ω t ) cos ( ω t ) = cos ( ω t − 9 0 ∘ ) , = sin ( ω t + 9 0 ∘ ) . 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 has the same constant frequency.
Phasors:
Rectangular form:
z = x + j y = ∣ z ∣ ( cos ( φ ) + j sin ( φ ) ) \begin{align*}
z & = x + jy \newline
& = |z|(\cos(\varphi) + j \sin(\varphi))
\end{align*} z = x + j y = ∣ z ∣ ( cos ( φ ) + j sin ( φ )) Polar form:
z = ∣ z ∣ ∠ θ ∘ = ∣ z ∣ e j θ \begin{align*}
z & = |z|\angle{\theta^\circ} \newline
& = |z|e^{j \theta}
\end{align*} z = ∣ z ∣∠ θ ∘ = ∣ z ∣ e j θ Polar to rectangular form:
z = r ∠ θ ∘ , x = r cos ( θ ∘ ) , y = r sin ( θ ∘ ) , z = x + j y . \begin{align*}
z & = r\angle{\theta^\circ}, \newline
x & = r \cos(\theta^\circ), \newline
y & = r \sin(\theta^\circ), \newline
z & = x + j y.
\end{align*} z x y z = r ∠ θ ∘ , = r cos ( θ ∘ ) , = r sin ( θ ∘ ) , = x + j y . Rectangular to polar form:
z = x + j y , r = x 2 + y 2 , θ = arctan ( y x ) , z = r ∠ θ ∘ . \begin{align*}
z & = x + jy, \newline
r & = \sqrt{x^2 + y^2}, \newline
\theta & = \arctan{\left(\dfrac{y}{x}\right)}, \newline
z & = r\angle{\theta^\circ}.
\end{align*} z r θ z = x + j y , = x 2 + y 2 , = arctan ( x y ) , = r ∠ θ ∘ . Operations:
Multiplication:
a = ( b ∠ c ∘ ) ⋅ ( d ∠ e ∘ ) = b d ∠ ( c ∘ + e ∘ ) \begin{align*}
a & = (b\angle{c^\circ}) \cdot (d\angle{e^\circ}) \newline
& = b d \angle{(c^\circ + e^\circ)}
\end{align*} a = ( b ∠ c ∘ ) ⋅ ( d ∠ e ∘ ) = b d ∠ ( c ∘ + e ∘ ) Division:
a = b ∠ c ∘ d ∠ e ∘ = b d ∠ ( c ∘ − e ∘ ) \begin{align*}
a & = \dfrac{b\angle{c^\circ}}{d\angle{e^\circ}} \newline
& = \dfrac{b}{d} \angle{(c^\circ - e^\circ)}
\end{align*} a = d ∠ e ∘ b ∠ c ∘ = d b ∠ ( c ∘ − e ∘ ) Phasor relations:
V = R I , V = j ω L I , V = 1 j ω C I . \begin{align*}
V & = RI, \newline
V & = j\omega LI, \newline
V & = \dfrac{1}{j\omega C} I.
\end{align*} V V V = R I , = j ω L I , = j ω C 1 I . Impedance:
Z = V I Z = \dfrac{V}{I} Z = I V This means that:
Z R = R , Z L = j ω L , Z C = 1 j ω C = − j 1 ω C . \begin{align*}
Z_{R} & = R, \newline
Z_{L} & = j\omega L, \newline
Z_{C} & = \dfrac{1}{j\omega C} = -j \dfrac{1}{\omega C}.
\end{align*} Z R Z L Z C = R , = j ω L , = j ω C 1 = − j ω C 1 . Power in AC - general:
V ( t ) = V m cos ( ω t + θ V ) , I ( t ) = I m cos ( ω t + ϕ I ) . \begin{align*}
V(t) & = V_m \cos(\omega t + \theta_V), \newline
I(t) & = I_m \cos(\omega t + \phi_I).
\end{align*} V ( t ) I ( t ) = V m cos ( ω t + θ V ) , = I m cos ( ω t + ϕ I ) . P ( t ) = V m I m cos ( ω t + θ V ) cos ( ω t + ϕ I ) P(t) = V_m I_m \cos(\omega t + \theta_V) \cos(\omega t + \phi_I) P ( t ) = V m I m cos ( ω t + θ V ) cos ( ω t + ϕ I ) P ( t ) = 1 2 V m I m cos ( θ V − ϕ I ) + 1 2 V m I m cos ( 2 ω t + θ V + ϕ 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 ( t ) = 2 1 V m I m cos ( θ V − ϕ I ) + 2 1 V m I m cos ( 2 ω t + θ V + ϕ I ) P avg = V m I m 2 cos ( θ V − ϕ I ) P_{\text{avg}} = \dfrac{V_m I_m}{2} \cos(\theta_V - \phi_I) P avg = 2 V m I m cos ( θ V − ϕ I ) P avg = V rms I rms cos ( θ V − ϕ I ) P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\theta_V - \phi_I) P avg = V rms I rms cos ( θ V − ϕ I ) Different Types:
Real Power: P = V rms I rms cos ( θ V − ϕ I ) [ W ] P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_V - \phi_I) \ [\mathrm{W}] P = V rms I rms cos ( θ V − ϕ I ) [ W ]
Reactive Power: Q = V rms I rms sin ( θ V − ϕ I ) [ V A R ] Q = V_{\text{rms}} I_{\text{rms}} \sin(\theta_V - \phi_I) \ [\mathrm{VAR}] Q = V rms I rms sin ( θ V − ϕ I ) [ VAR ]
Complex Power: S = P + j Q S = P + jQ S = P + j Q S = V rms I rms ∠ ( θ V − ϕ I ) [ V A ] S = V_{\text{rms}} I_{\text{rms}} \angle (\theta_V - \phi_I) \ [\mathrm{VA}] S = V rms I rms ∠ ( θ V − ϕ I ) [ VA ]
Apparent Power: ∣ S ∣ = V rms I rms [ V A ] |S| = V_{\text{rms}} I_{\text{rms}} \ [\mathrm{VA}] ∣ S ∣ = V rms I rms [ VA ]