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Intro to Electrical Engineering
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Volt
Work per unit charge - (difference in potential energy)
V = 1J/C = (1 kg·m
^{2}
/s
^{2}
)/C
V = IR
Coulomb
Measure of charge
1C = 1 A*s
Joule
Work to move 1 Newton 1 meter.
1J = 1 kg·m2/s2
Resistance
Measure opposition to the passage of current.
R=l/σA
Power (P)
P = VI
= I
^{2}
R
= V
^{2}
/R
Charge for Electron
q
_{e}
= -1.602*10
^{-19}
C
Charge for Proton
qe = 1.602*10
^{-19}
C
Electric Current
i = Δq/ΔT
units: 1C/s
Kirchoff's Current Law
charge is conserved
i = i
_{0}
+ i
_{1 }
+ i
_{n}
Ohm's Law
V = IR
Conductance
I=GV G: element
Equivalent series resistance
Resistors appear as a single equivalent resistance of value R
_{eq}
.
R
_{eq}
= R
_{1}
+ R
_{2}
+ R
_{3}
Voltage Divider
When source voltage is divided among the resistors.
V
_{n}
= (R
_{n}
/(R
_{1}
+ R
_{2}
+ R
_{n}
)) x v
_{s}
In Series
Circuit elements are
in series
when identical
current
flows through each element.
In Parallel
Circuit elements are
in parallel
when identical
voltage
flows through each element.
Max current
i
_{s}
= V
_{s}
/r
_{s}
r
_{s }
: resistance
i(t) = V
_{s}
(t) / R
Short Circuit
Circuit element with resistance approaching zero.
R = 0
Open Circuit
Circuit element with resistance approaching infiinty.
R = infinity
Loop
any closed connection of branches.
Mesh
A loop that does not contain other loops.
Node Voltage Method
i = (v
_{a}
-v
_{b}
)/R
Principle of Superposition
i = (v
_{B1}
+ v
_{B2}
)/R
Thevenin Equivalent Circuit
Represented by voltage source v
_{T}
in series with R
_{T}
(equibalent resistance).
Norton Equivalent Circuits
Represented by voltage current source i
_{N}
in parrallel with R
_{N}
.
Method for solving Thevenin & Norton Req.
1. Remove load
2. Zero all independent voltage and current sources
3. Compute total resistance with load removed
R
_{T}
= R
_{N}
Method to compute Thevenin voltage
1. Remove the load
2. Define v
_{OC}
across the open load termnials
3. Apply any circuit analysis to solve v
_{OC}
4. The Thevenin voltage is vT = v
_{OC}
Thevenin voltage is v
_{T}
= v
_{OC}
Method to solve Norton Current
1. Replace the load with a short-circuit
2. Define the short-circuit current i
_{SC}
= i
_{N}
3. Apply any method to solve i
_{SC}
4. Therefore i
_{N}
= i
_{SC}
Norton current = short-circuit current
Ideal Capacitors
Q = CV
units : Farad -> C/V
Capacitors in parallel
C
_{eq}
= C
_{1}
+ C
_{2}
+ C
_{3}
Capacitors in series
1/(1/C
_{1}
+ 1/C
_{2}
+ 1/C
_{3}
)
Periodic Signal Waveform
x(t) = x(t + nT) , n = 1,2,3
Sinusoidal Waveforms
x(t) = Acos(ωt) & Acos(ωt + Φ)
Φ
2π(Δt/T)
phase shift
Asin(ωt) = Acos(ωt - π/2)
ω
2πf
Impedance of a resistor
ZR(jω) = VS(jω) / I(jω) =
R
Impedance of an inductor
ZL(jω) = VS(jω) / I(jω)= ωL∠π/2 =
jωL
Impedance of a capacitor
ZC(jω) = VS(jω) / I(jω)= 1ωC∠−π/2=−j / ωC
=
1 / jωC
the impedance of a circuitelement
Z(jω) = R(jω) + jX(jω)
Circuit law for a capacitor.
i(t) = C*(dv(t) / dt)
Energy stored in a capacitor (J)
W
_{C}
(t) = 1/2*Cv
^{2}
_{C}
(t)
Voltage in an inductor.
v
_{L}
(t) = L*(di
_{L}
/ dt) units : 1 H = 1 V-s/A
Inductors in series add. Inductors in parallel combine according to thesame rules used for resistors connected in parallel.
Energy stored in an inductor (J)
W
_{L}
(t) = 1/2*(Li
^{2}
_{L}
(t))
e
^{jθ}
=
cos θ + j sin θ
Ae
^{jθ}
=
Acos θ + jAsin θ = A∠θ
Energy stored in steady state Capacitor
Energy stored in steady state Inductor
(1/2)*C*V
^{2}
(1/2)*L*V
^{2}
^{ }
Author
mechtech2081
ID
107260
Card Set
Intro to Electrical Engineering
Description
Intro to Electrical Engineering
Updated
2011-11-13T15:46:20Z
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