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 2Ohms 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 AnalysisNodal 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 TheoremsSuperposition
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 InductorsCapacitors
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) = .5Cv^{2}
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) = .5Li^{2}
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 CircuitsSource 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(1e^{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 CircuitsSource 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) = Ae^{s1t} + Be^{s2t}
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(a^{2} + w0^{2})
a = 1/(2RC)
w0 = 1/sqrt(LC)
1  Overdamped (a > w0)
i(t) = Ae^{s1t} + Be^{s2t}
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 + Ae^{s1t} + Be^{s2t}
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 + Ae^{s1t} + Be^{s2t}
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 Phasorsw = 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(A^{2} + B^{2})
theta = tan^{1} (B/A)
Complex Numbers
rectangular form: z = x + jy
polar: z = r < (theta)
expolar: z = re^{j(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
voltagecurrent 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(R^{2} + X^{2})
(theta) = tan^{1}(X / R)
R = Z*cos(theta)
X = Z*sin(theta) 
Chapter 10  AC CircuitsAnalyzing 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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