EE 108 Midterm 1 Reference

Midterm 1 Cheat sheet: front back

Work

$$ \begin{gathered} \fbox{\(\omega_p = mgh\)} \end{gathered} $$

Ohm’s Law

$$ \begin{aligned} V&= I \cdot R \\[10pt] I &= \frac{V}{R} \\[10pt] R&= \frac{V}{I} \end{aligned} $$

Power

$$ \begin{aligned} & P = \frac{W_p}{t} \text{ [W, kW]} \\[10pt] & \boxed{ P = I^2R } \\[10pt] & \boxed{ P = \frac{v^2}{R} } {\scriptsize \text{ (v = voltage drop)}} \\[10pt] & 1\text{W} = 1\frac{\text{J}}{\text{s}} {\scriptsize \text{ (1 Watt = 1 Joule/s)}} \\[10pt] & 1\text{kWh} = 3.6 \times 10^6 \text{J} \end{aligned} $$

Efficient Motor (80%)

$$ \begin{aligned} & \frac{W_p}{10\text{sec}} = \frac{10,000}{10} = 1000\text{W} = P_{\text{mechanical}} \\[10pt] & 1000\text{W} = 0.8 = P_{\text{electrical}} \\[10pt] & P_E = \frac{1000}{0.8} = 1.25\text{kW} \\[25pt] & \boxed{E_c = \frac{1}{2}CV^2} \hspace{40pt} \boxed{E_L = \frac{1}{2}Li^2} \end{aligned} $$

Voltage Divider

$$
\large
\boxed{V_{\text{out}} = V_{\text{in}}\frac{R_2}{R_1 + R_2}}
$$

Parallel Resistance

$$
\large
\begin{aligned}
& R_{tot} = R_1 \parallel R_2 \\[15pt]
& \frac{1}{R_{tot}} = \frac{1}{R_1} + \frac{1}{R_2} \\[15pt]
& R_{tot} = \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1} \\[15pt]
& \boxed{ R_{tot} = \frac{R_1 R_2}{R_1 + R_2}}
\end{aligned}
$$

AC Power

$$ \large \begin{gathered} \text{Assume V}_{AC} = \text{V}_{DC}\\[10pt] \frac{V_{AC}^2}{2R} = \frac{V_{DC}^2}{R} \\[10pt] V_{AC} = \sqrt{2}V_{DC} \\[10pt] \boxed{ V_{pk} = \sqrt{2}V_{rms} }\\[10pt] \boxed{ V_{rms} = \frac{V_{pk}}{\sqrt{2}} } \end{gathered} $$

1. AC power is half of DC power under the same peak voltage.

2. Instantaneous power pulsates at 2ω. But we care about average power.

Top: voltage waveform
Bottom: AC Power waveform

Capacitance details

$$ \begin{aligned} & \varepsilon = \text{permittivity of free space } (8.854 \times 10^{-12}) \\[5pt] & A = \text{surface area } (m^2) \\[5pt] & d = \text{distance } (m) \\[20pt] \end{aligned} $$
$$ \begin{gathered} \boxed{C = \frac{\varepsilon A}{d} \text{ [F]}} \\[15pt] \end{gathered} $$
Capacitor current derivation
$$ \begin{gathered} Q = CV \\[15pt] \frac{dQ}{dt} = C\frac{dv}{dt} \\[15pt] I = C\frac{dv}{dt} \\[15pt] \text{(current leads voltage by 90}^\circ\text{)} \end{gathered} $$

First Differential Equations

$$ \begin{aligned} \text{(Inductor)} \hspace{10pt} \boxed{V_L = L \frac{di}{dt}} &\hspace{20pt} \boxed{I_C = C \frac{dv}{dt}} \hspace{10pt} \text{(Capacitor)} \\[15pt] \tau = \frac{L}{R} &\hspace{20pt} \tau = RC \end{aligned} $$

(1) Find \(I_L\) using KVL

$$ \begin{gathered} \begin{aligned} V_{dc} – V_L – V_R &= 0 \\[10pt] V_L = L \frac{di}{dt} \hspace{30pt} V_R &= I_L R \\[10pt] V_{dc} – L \frac{di}{dt} – I_L R &= 0 \\[10pt] L \frac{di}{dt} + I_L R &= V_{dc} \end{aligned} \\[15pt] \boxed{\text{(Standard form)} \hspace{15pt} \frac{di}{dt} + \frac{I_L R}{L} = \frac{V_{dc}}{L}} \end{gathered} $$

(2) Homo./Natural Solution (remove \(V_{dc}\))

$$ \begin{gathered} \frac{di}{dt} + \frac{I_L R}{L} = 0 \\[15pt] I_L(t) = A e^{\frac{t}{L/R}} \end{gathered} $$

(3) Particular/Steady Solution (remove \(\frac{di}{dt}\))

$$ \begin{gathered} \frac{I_L}{R} = \frac{V_{dc}}{R} \\[15pt] I_L = \frac{V_{dc}}{R} \longrightarrow \text{(B)} \end{gathered} $$

(4) Initial Condition

$$ \begin{gathered} I_L(0) = A + B \\[10pt] 0 = A + B \\[10pt] -B = A \end{gathered} $$

(5) Final Form (Natural + Steady state)

$$ \begin{gathered} I_L(t) = -\frac{V_{dc}}{R} e^{-\frac{t}{\tau}} + \frac{V_{dc}}{R} \\[15pt] \boxed{I_L(t) = \frac{V_{dc}}{R} \left(1 – e^{-\frac{t}{\tau}}\right)} \end{gathered} $$

General Charging and Discharging

$$ \begin{gathered} \boxed{\text{v}_c\text{(t)} = V_s (1-e^{-\frac{t}{\tau}})} \hspace{15pt} ( \text{charging} ) \hspace{15pt} \boxed{\text{i}_L\text{(t)} = \frac{V_s}{R} (1-e^{-\frac{t}{\tau}})} \\[10pt] \boxed{\text{v}_c\text{(t)} = V_s e^{-\frac{t}{\tau}}} \hspace{10pt} ( \text{discharging} ) \hspace{10pt} \boxed{\text{i}_L\text{(t)} = \frac{V_s}{R} e^{-\frac{t}{\tau}}} \end{gathered} $$
$$ \begin{gathered} 63\% \hspace{15pt} 86\% \hspace{15pt} 95\% \\[10pt] (1\tau) \hspace{16pt} (2\tau) \hspace{16pt} (3\tau) \\[10pt] 36\% \hspace{15pt} 13\% \hspace{15pt} 5\% \end{gathered} $$

Impedance

$$ \begin{gathered} \boxed{|Z_L| = \omega L \hspace{15pt} Z_L = j \omega L \hspace{15pt} \angle 90^\circ} \\[10pt] \boxed{|Z_C| = \frac{1}{\omega C} \hspace{15pt}Z_C = \underbrace{\frac{1}{j \omega C}}_{\large\substack{-j\frac{1}{\omega C}}} \hspace{15pt} \angle -90^\circ} \\[10pt] \end{gathered} $$

\(I_C\) leads voltage by 90 degrees in time.
\(I_L\) lags voltage by 90 degrees in time.

Phasors

*Notice the zero crossings of the current vs the voltage to realize leading and lagging.

Complex Impedance

$$ \Large \begin{gathered} \bar{Z}= \underbrace{R}_{\substack{\text{Resistance} \\[3pt] {|Z|cos \theta}}} + j\underbrace{X}_{\substack{\text{Reactance} \\[3pt] {|Z|sin\theta}}} \end{gathered} $$

Complex Admittance

$$ \Large \begin{gathered} \bar{Y}= \underbrace{G}_{\substack{\text{Conductance} \\[3pt] {|Y|cos \theta}}} + j\underbrace{B}_{\substack{\text{Susceptance} \\[3pt] {|Y|sin\theta}}} \end{gathered} $$
$$ \begin{aligned} &\hspace{8pt} I = \frac{V}{Z} \hspace{15pt} \small(Z = \frac{1}{Y}) \\[10pt] &\boxed{\hspace{5pt} I = V Y \hspace{5pt}} \end{aligned} $$
Parallel Case Using Admittance
$$ \begin{gathered} \begin{aligned} Z_L &= j2\Omega &\longrightarrow \quad Y_L &= \frac{1}{j2} = -j0.5\text{ S} \\[10pt] Z_C &= -j0.5\Omega &\longrightarrow \quad Y_C &= \frac{1}{-j0.5} = j2\text{ S} \\[10pt] Z_R &= 1\Omega &\longrightarrow \quad Y_R &= 1\text{ S} \end{aligned} \\[20pt] \frac{1}{Z_p} = \frac{1}{Z_1} + \frac{1}{Z_2} + \frac{1}{Z_3} \hspace{20pt} \text{But} \hspace{20pt} Y_p = Y_1 + Y_2 + Y_3 \end{gathered} $$
Series Circuit & Find admittance
$$ \begin{gathered} \begin{aligned} Z_{tot} &= 3 + j8 – j4 = 3 + j4 \quad \longrightarrow \quad \sqrt{3^2 + 4^2} = 5, \quad \arctan\left(\frac{4}{3}\right) = 53.13^\circ \\[15pt] Z_{tot} &= 5\Omega \angle 53.13^\circ \quad \longrightarrow \quad Y_{tot} = \frac{1}{5 \angle 53.13^\circ} = 0.2\text{ S} \angle -53.13^\circ \end{aligned} \end{gathered} $$

*When trying to find reciprocal of complex Z or Y, much easier to do it in polar form with angles.

(Hard) Alternate method: Reciprocal Z or Y
$$ \begin{aligned} \frac{1}{Z_p} &= \frac{1}{10} + \frac{1}{j10} – \frac{1}{j0.5} = 0.1 – j0.1 + j2 = 0.1 + j1.9 \\[15pt] Z_p &= \frac{1}{0.1 + j1.9} \cdot \frac{(0.1 – j1.9)}{(0.1 – j1.9)} = \frac{0.1 – j1.9}{0.1^2 + 1.9^2} = \frac{0.1 – j1.9}{3.62} = 0.027 – j0.52 \\[15pt] Z_{tot} &= Z_p + Z_a = 0.027 – j0.52 + 2 = 2.027 – j0.52 \\[15pt] |Z_{tot}| &= \sqrt{2.027^2 + (-0.52)^2} = 2.09\Omega \hspace{20pt} \arctan\left(\frac{-0.52}{2.027}\right) = -14.4^\circ \\[15pt] Z_{tot} &= 2.09\Omega \angle -14.4^\circ \\[15pt] I &= \frac{120 \angle 0^\circ}{2.09 \angle -14.4^\circ} = 57.4\text{ A}_{rms} \angle 14.4^\circ \text{ (leading)} \end{aligned} $$

Complex Power

Apparent Power

$$ \begin{gathered} \boxed{\bar{S} = VI^* \quad \scriptsize{\text{[VA]}}} \\[10pt] \bar{S} = P + jQ \\[10pt] |\bar{S}| = \sqrt{P^2 + Q^2} \end{gathered} $$

Power Factor

$$ \begin{gathered} \boxed{p.f = \cos(\theta)} \\[10pt] \boxed{p.f = \frac{P}{S}} \end{gathered} $$

Reactive Power

$$ \begin{gathered} \begin{aligned} &\boxed{P = S \cdot \cos(\theta)} \\[10pt] &\boxed{Q = S \cdot \sin(\theta)} \\[10pt] &S = \frac{P}{\cos(\theta)}\\[10pt] &Q = P \cdot \tan(\theta) \end{aligned} \\[15pt] \end{gathered} $$

Power \(\theta\)

$$ \begin{gathered} \theta = \cos^{-1}(\text{pf}) \\[10pt] \theta = \tan^{-1} \left(\frac{Q}{P} \right) \end{gathered} $$

Power \(\theta\) Notes:

1. Positive power \(\theta\) = current lags voltage.

2. Negative power \(\theta\) = current leads voltage.

3. If Q or \(\theta\) is > 0, circuit is more inductive.

4. If Q or \(\theta\) is < 0, circuit is more capacitive.

5. If power \(|\theta|\) > 90°, load is a generator.

Complex Current

$$ \begin{aligned} \bar{I}_{rms} &= \frac{\bar{V}_{rms}}{Z_{tot}} \\[15pt] &= \frac{120 \angle 0^\circ}{8 \angle 48^\circ} \\[15pt] &= 15 \angle -48^\circ \text{ A} \\[10pt] &\quad \downarrow \\[5pt] &\text{current lags voltage.} \end{aligned} $$

Finding Power in Parallel Circuit

$$ \begin{gathered} \text{Use} \hspace{8pt} \boxed{P = \frac{V^2}{R}} \hspace{8pt}or\hspace{8pt} \boxed{Q = \frac{V^2}{X}} \end{gathered} $$
Example: Reactive Power Compensation

Say our 3-phase complex power is \(S = 60.815\text{ kW} – j42.448\text{ kVAR}\), and we want to reduce our reactive power to zero. We need to provide more VAR. We need 3 equal capacitors and we want to find what size each are. There are two ways to do this.

First method: Find impedance (purely reactance) first using Ohm’s law \(X_C = \frac{V}{I}\). Phase current and voltage are used because we want to find the impedance of one capacitor on one of the loads on the delta configuration.

$$ \begin{gathered} X_C = \frac{V_\phi}{I_\phi} = \frac{277}{51.5} = 5.4\Omega \\[15pt] X_C = \frac{1}{\omega C} \\[15pt] \boxed{ C = \frac{1}{\omega X_C} } \\[15pt] C = \frac{1}{2\pi \cdot 60 \cdot 5.38} \hspace{20pt} (2\pi \cdot 60 = 377) \\[15pt] C = 0.0005\text{ F} \hspace{20pt} (500\mu\text{F}) \end{gathered} $$

Second method: Find impedance (purely reactance) using the power law \(Q_C = \frac{V_\phi^2}{X_C}\), where \(Q\) is reactive power. This is similar to \(P = \frac{V^2}{R}\).

$$ \begin{gathered} X_C = \frac{V_\phi^2}{\frac{1}{3}Q_C} \\[15pt] X_C = \frac{277^2}{14149} = 5.4\Omega \end{gathered} $$

Total reactive power is divided by 3 simply because we have 3 loads in delta configuration, so each capacitor will provide 1/3 of the reactive power.

Inductor Details

$$ \large \begin{gathered} L = \frac{N^2 \mu_c A_c}{l_c} \hspace{5pt} [H] \end{gathered} $$
$$ \begin{aligned} \textbf{Constants:}& \\ \mu_0 &= 4\pi \times 10^{-7}\\[15pt] \textbf{Where:}& \\ L &= \text{Inductance [H]} \\ N &= \text{Number of wire turns} \\ A_c &= \text{Cross section area of core} \\ \mu_c &= \text{Permeability of core material} \\ l_c &= \text{Length of magnetic path (} 2\pi r \text{ for toroid)} \end{aligned} $$