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FUNDAMENTALS OF ELECTRICAL CIRCUITS
ING. DELL'AUTOVEICOLO
A.A. 2023/24
DOC. FABIO FRESCHI
SARA PASTINE
CONTENTS
- Definitions
- Equations
- Components
- Connection of Components
- Method to Solve Circuits
- Controlled Generators
- Nodal Analysis
- Dynamic Circuits
- Transient Analysis
- Complex Numbers
- AC Circuits
- Power Balance
- Formulary
- Definition Summary
Components
- Passive component: component that always absorbs positive energy (resistors, capacitors, inductors)
- Active component: may deliver positive energy (voltage and current generators)
Resistors
- [R]-resistance-Ω=m²·Ω
- [G]-conductance-siemens-S
- [ρ]-resistivity-Ω·m = m²·Ω m
- [γ]-conductivity-S/m
v(t)=i(t) R
i(t)=v(t): R(t)= σ(t)
R= L / S
1 / R=G
p(t)=v(t)·i(t)=Ri²+v² / R=v² / R+Gv²
Voltage Generator
v(t)=e(t)
i(t) depends only on the circuit
p(t)=v(t)·i(t)=e(t)·i(t) ≥0
e:electromotive force
Current Generator
i(t)=α(t)
v(t) depends only on the circuit
Short Circuit
v(t)=0 iL, vL = resistor with R=0.Ω
= voltage generator with e(t)=0V
= ideal connection
Open Circuit
i(t)=0 vL vL = current generator with α(t)=0V
= resistor with R=^∞^Ω or G=0S
- Switch =
- open circuit if open
- short circuit if closed
- THEVENIN ⟶ NORTON
RTh = RN JTh = RNJN
EXAMPLE
RN = RAB • RB//RS
JN1 = e1/RB JS1 = e2/RS iN = J3 + J5 = e1/RB + e2/RS
iN = RN/(RN + RL) iN
- MILLIAN'S THEOREM
I) Assume JTh as known and calculate the branch currents
e1 - Rii1 - JThN = 0 ⟶ i1 = (e1 - JTh)/R4
-e2 - Rii2 - JThN = 0 ⟶ i2 = (-e2 - JTh)/R3
JTh - Rii3 = 0 ⟶ i3 = JTh/R3
i4 = Qu i5 = Q5
II) Apply KCL to node A
i1 + i2 + iS + i4 + i5 = 0
(e1 - JTh)/R4 + (-e2 + JTh)/R3 + Qu + (Qu5)/R0) = 0
III) Find JN
JN = (e1/R2 + Q5 + The)/The
JN = e2/R2 + e1/R1 + e1/R2
NODAL ANALYSIS
- Label nodes from 0 (n)
- Define node potentials (φi)
- Assign φ0=0 to 3 unknown potentials
Write KCL equations at nodes (e excluded) assuming '+' for the i flowing outside the node using the rules below
g = Rjk/Rjk = GJrK(φq - φj)iJ = isrNODE 1: -@1 + G2(φ2 - φ1) + G1(φ1 - φ3 + φ3 = 0)NODE 2: -G2(φ3 - φ2) = 0NODE 3: G3(φ3 - φq) + G4(φ4 - φy) + G1(φ3 - φ2)
- GJJ: sum of the conductances connected to node 3
- GJR: the conductance that connects nodes 1-K (Symmetric matrix)
- φJ: impressed current write '+' when current enters node 'J' and '-' when leaves node J
- The matrix is symmetric
- The sums of the rows are the G connected to node φ
- The sum of the result matrix is the total current leaving the node φ
CIRCUITS WITH VOLTAGE GENERATORS
- Generator in series with resistor
- Generator between 2 nodes
Norton's transformation
ix is unknown
φk - φj - e
t>0
RB=RAB=(RA+R2)/RB
IN=(R1/R4+R2)A
IR(t)=k et/τ
MAGIC FORMULA
IR(t) = (I0 - I∞) e(-t/τ) +I∞
τ = L/Req
[XC] - capacitive reactance
- Generalized Ohm's Law
Resistor V=RI Capacitor VI=jXCI Inductor V=jXLI
Re = jX = Z = V/Icomplex operator
Resistors, capacitors and inductors can all be transformed into impedances. Impedances in circuits can be simplified with series and parallel rules of resistors.
- RLC Circuits
R ZR=R C ZC=jXC=j (1/wC) L ZL=jXL=jwL
VR=ZRI VC=ZCI VL=ZLI
Zeq=ZR+ZC+ZL V=VR+VC+VL=ZRI+ZCI+ZLI=(ZR+ZC+ZL)I=ZeqI
Zeq(w)=ZR+ZC+ZL=R+jXC+jXL=R+j(jXC+jXL)=R+j(1/(jwC)+jwL)
=R-j(1/wC)+jwL=R+j (R+j (1/wC)+jwL)=R+j (Xeq(w))
limw→0Xeq(w)-limw→0(1/wC) → -∞→Xeqacts like C
limw→∞Xeq(w)-limw→∞(wL)→∞→Xeqacts like L
wsℓws+Xeq<0→capacitive reactance w=0(w=ws)+Xeq=0→inductive reactance ws=wsubeq subeq =−1/p w2L⇔sub eq
Xeq=limw→∞Xeq(w)-1/(wL)-1/(2π)(wsS) → -1/(wL(wsR))
L0=(wc-m)/2π→1/2π wL2mT(L/C)Resonance Frequency