Cheatography

# CircuitsFinal2014 Cheat Sheet by Lini

Formulas and Notes for ECE2004

### Chapter 1 - Basics

 - Electric current = (i): time rate of change of charge, measured in amperes (A). - Charge = (q): integral of i - Voltage (or potential differ­ence) = (V): energy required to move a unit charge through an element - Power = (W): vi = (i^2)R - Passive sign conven­tion: when the current enters through the positive terminal of an element (p = +vi) Remember: +Power absorbed = -Power supplied --> sum of power in a circuit = 0 - Energy (J) = integral of P

### Chapter 2

 Ohms Law: v=iR Conduc­tance (G) = 1/R = i/v Branch: single element such as a voltage source or a resistor. Node: point of connection between two or more branches Loop: any closed path in a circuit. Kirchh­off’s current law (KCL): algebraic sum of currents entering a node (or a closed boundary) is zero. Kirchh­off’s voltage law (KVL): algebraic sum of all voltages around a closed path (or loop) is zero. Voltage D: v1 = ((R1) / (R1 + R2)) * v Voltage D: v2 = ((R2 / (R1 + R2)) * v Current D: i1 = (R2 * i) / (R1 + R2) Current D: i2 = (R1 * i) / (R1 + R2)

### Chapter 3 - Methods of Analysis

 Nodal Analysis: want to fine the node voltages Step 1: select reference node - assign voltages v1 --> vn to remaining nodes Step 2: apply KCL to each node - want to express branch currents in terms of voltage Step 3: solve for unknowns Important: current flows from high to low (+ ==> -) SuperNode Properties 1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages 2. Supernode had no voltage of its own 3. Supernode requires the applic­ation of both KCL and KVL Mesh Analysis Step 1: Assign mesh currents or loops Step 2: Apply KVL - use OHMS LAW to express voltages in terms of the mesh current Step 3: Solve for the unknown Supermesh - when two meshes have an indepe­ndent or dependent CURRENT source between them

### Chapter 4 - Circuit Theorems

 Superp­osition principal states that the VOLTAGE ACROSS or CURRENT THROUGH an element in a linear circuit is the SUM of the VOLTAGES OR CURRENTS that are caused after solving for each INDEPE­NDENT source separately How to solve a superp­osition circuit Step 1: Turn OFF ALL indepe­ndent sources except for ONE ==> find voltage or current Step 2: Repeat above for all other indepe­ndent sources Step 3: Add all voltag­es/­cur­rents together to find final value Thevenin's Theorem V(th) = V(oc) circuit with Load: I(L) = V(th) / (R(th) + R(L)) ==> V(L) = R(L) I (L) ==> (R(L) / ((R(th) + R(L)) V(th)) Norton's Theorem R(n) = R(th) I(n) = i(sc) ==> (sc) = short circuit I(n) = V(th) / R(th) Maximum Power Transfer max power is transf­erred to the LOAD RESISTOR when the LOAD RESISTOR is EQUAL to the THEVENIN RESIST­ANCE: R(L) = R(th) p(max) = V(th)2 / 4R(th)

### Chapter 6 - Capacitors and Inductors

 Capacitors q = C * v capaci­tance: ratio of the charge on one plate to the voltage difference between the two plates i(t) = C(dv/dt) v(t) = 1/C [Integral: i(T)dT + v(t0))] T = time constant energy (w) = .5Cv2 Important: VOLTAGE of a capacitor cannot change instan­tan­eously Capacitors in Series: 1 / Ceq = 1/C1 + 1/C2 + 1/Cn Capacitors in Parallel: Ceq = C1 + C2 + Cn Inductors v = L(di / dt) i = (1/L) [Integral: (v(T)dT + i(t0)] energy (w) = .5Li2 Important: CURRENT through an inductor cannot change instan­tan­eously Inductors in Series: Leq = L1 + L2 + Ln Inductors in Parallel: 1/Leq = 1/L1 + 1/L2 + 1/Ln

### Chapter 7 - First Order Circuits

 Source Free RC Circuits v(t) = V0 * e-t/T ==> T = RC How to Solve SOURCE FREE RC CIRCUITS Step 1: Find v0 = V0 across the capacitor Step 2: Find T (time constant) Source Free RL Circuits i(t) = I0 * e-t/T ==> T = L / R vr(t) = iR = I0 * Re-t/T How to Solve SOURCE FREE RL CIRCUITS Step 1: Find i(0) = I0 through the inductor Step 2: Find T (time constant) Step response of an RC circuit v(t) = V0 when t < 0 v(t) = Vs + (V0 - Vs)e-t/T when t > 0 v = vn + vf ==> vn = V0e-t/T, vf = Vs(1-e-t/T) OR v(t) = v(infi­nity) + [( v(0) - v(infi­nity)]e-t/T How to solve a STEP RESPONSE OF AN RC CIRCUIT Step 1: Find initial capacitor voltage v0 (t < 0) Step 2: Find final capacitor voltage v(in) (t > 0) Step 3: Find T (time constant) (t > 0) Step response of an RL circuit i(t) = i(infi­niti) + [ i(0) - i(infi­nity)]e-t/T How to solve a STEP RESPONSE OF AN RL CIRCUIT Step 1: Find initial inductor current i0 (t = 0) Step 2: Find final final inductor current i(inf) ==> (t > 0) Step 3: Find T (time constant) (t > 0)

### Chapter 8 - Second Order Circuits

 Source Free RLC Circuits v(0) = 1/C [integral ( idt = v0 ) from 0 to -infinity] i(0) = I(0) Determ­ining Dampness (alpha) = R / (2L) (omega w0) = 1 / sqrt(LC) 1 - Overdamped (a > w0) i(t) = Aes1t + Bes2t 2 - Critically Damped (a = w0) s1 = s2 = a i(t) = (A + Bt)e-at 3 - Underd­amped (a < w0) i(t) = e-at(Acos(w0t) + Bsin(w0t)) Source Free Parallel Circuits roots of charac­ter­istic euqation s1,2 = -a (+-) sqrt(a2 + w02) a = 1/(2RC) w0 = 1/sqrt(LC) 1 - Overdamped (a > w0) i(t) = Aes1t + Bes2t 2 - Critically Damped (a = w0) s1 = s2 = a i(t) = (A + Bt)e-at 3 - Underd­amped (a < w0) i(t) = e-at(Acos(­wd(t)) + Bsin(w­d(t))) Step Response of a SERIES RLC Circuit 1 - Overdamped (a > w0) v(t) = Vs + Aes1t + Bes2t 2 - Critically Damped (a = w0) s1 = s2 = a v(t) = Vs + (A + Bt)e-at 3 - Underd­amped (a < w0) v(t) = Vs + e-at(Acos(­wd(t)) + Bsin(w­d(t))) Step Response of a PARALLEL RLC Circuit 1 - Overdamped (a > w0) i(t) = Is + Aes1t + Bes2t 2 - Critically Damped (a = w0) s1 = s2 = a i(t) = Is + (A + Bt)e-at 3 - Underd­amped (a < w0) i(t) = Is + e-at(Acos(­wd(t)) + Bsin(w­d(t)))

### Chapter 9 - Sinusoids and Phasors

 w = omega T = 2*pie / w freq = 1 / T (Hertz) v(t) = v(m)*s­in(wt + theta) v1(t) = v(m)*s­in(wt) v2(t) = v(m)*s­in(wt + theta) sin(A +- B) = sinAcosB +- cosAsinB cos(A +- B) = cosAcosB +- sinAsinB Acos(wt) + Bsin(wt) = C*cos(wt - theta) C = sqrt(A2 + B2) theta = tan-1 (B/A) Complex Numbers rectan­gular form: z = x + jy polar: z = r < (theta) expolar: z = rej(theta) sin: r (cos(t­heta) + j*sin(­theta)) z = x + jy z1 = x1 + jy1 == r1 < (theta)1 z2 = x2 + jy2 == r2 < (theta)2 operations addition: z1 + z2 == (x1 + x2) + j*(y1 + y2) subtra­ction: z1 - z2 == (x1 - x2) + j*(y1 - y2) multip­lic­ation: z1z2 == r1r2 < ((theta)1 + (theta)2) division: z1/z2 == r1/r2 < ((theta)1 - (theta)2) recipr­ocal: 1/z = 1/r < -(theta) square: sqrt(z) = sqrt(r) < (theta)/2 complex conjugate: z* = x - jy = r < -(theta) = re-j(theta) real vs. imaginary e+-j(theta) = cos(theta) +- j*sin(­theta) cos(theta) = REAL jsin(t­heta) = IMAGINARY voltag­e-c­urrent relati­onship R v = Ri (time domain) v = RI (frequency domain) L v = L(di/dt) (time) v = jwLI C i = C(dv/dt) (time) V = I / jwC Impedance vs. admittance R Z = R (imped­ance) Y = 1 / R I Z = jwL Y = 1 / jwL C Z = 1 / jwC Y = jwC Complex Numbers with Impedance Z = R + jx = |Z| < (theta) |Z| = sqrt(R2 + X2) (theta) = tan-1(X / R) R = |Z|*co­s(t­heta) X = |Z|*si­n(t­heta)

### Chapter 10 - AC Circuits

 Analyzing AC Circuits Step 1: Transform circuit to phasor or frequency domain Step 2: Solve Using Circuit Techniques Step 3: Transform phasor ==> time domain