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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

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