Download ELECTRICAL QUANTITIES

1 ELECTRICAL QUANTITIES
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The important electrical quantities, their definitions, and their SI units are
summarized below :Table 1-2 Important Derived Quantities
Quantity
Symbol
Definition
Force
Energy
Power
Charge
ƒ
w
p
q
Current
i
Voltage
Electrical
field strength
Magnetic flux
density
Magnetic flux
Unit
Abbreviation
(Alternate)
Push or pull
Ability to do work
Energy/unit charge
Quantity of
electricity
Rate of flow of
charge
Newton
Joule
Watt
Coulomb
N
J
W
C
(kg · m/s²)
(N · m)
(J / s)
(A · s)
Ampere
A
(C / s)
υ
ε
Energy/unit charge
Force/unit charge
Volt
Volt/meter
V
V/m
(W / A)
(N / C)
B
Force/unit charge
momentum
Integral of magnetic
flux density
Tesla
T
(Wb / m²)
Weber
Wb
(T · m²)

In electrical terms, power and energy is expressed by :
p
=
dw
dt
=
vi
and
w
=
∫ p dt
=
∫ vi dt
~
Derived expressions can be checked dimensionally since an analytical
equation must be dimensionally homogenous.
~
The approximate behaviour of real circuit components is used to define five
passive and two active circuit elements displayed below :-
Table 1-3 Electrical Circuit Elements
Element
Resistance
(Conductance)
Unit
Ohm
(siemen)
Inductance
Henry
Symbol
Characteristic
υ = Ri
(i = G υ)
vL
di
dt
Capacitance
Farad
iC
Short circuit
Volt
υ = 0 for any i
dv
dt
Open circuit
i = 0 for any
Voltage source
υ = υs for any i
Current source
~
Ampere
i
1 t
vdt
L  
v
1 t
i dt
C  
I = Is for any υ
Voltage and current sources represent reversible energy transformation.
Resistance is the measure of the ability of a device to dissipate energy
irreversibly ( pR = Ri² )
Inductance is a measure of the ability of a device to store energy in a magnetic
field ( wL = ½ Li² )
Capacitance is the measure of the ability of a device to store energy in a magnetic
field (wc = ½Cv² )
~
Energy stored in an element must be a continuous function of time.
The current in an inductance cannot change instantaneously.
The voltage on a capacitance cannot change instantaneously.
2 CIRCUIT PRINCIPLES
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The algebraic sum of currents into the node is zero (Σi = 0).
The algebraic sum of the voltages around a loop is zero (Σv = 0).
~
The general procedure of formulating equations for circuits is :
1.
Arbitrarily assume a consistent set of currents and voltages.
2.
Write the element equations by applying the element definitions and write
the connection equations by applying Kirchoff's laws.
3. Combine the element and connection equations to obtain the governing
circuit equations in terms of the unknown.
The resulting simultaneous equations can be solved by :
Successive substitution to eliminate all but one unknown, or
Use of determinants and Cramer's rule.
~
Use of loop currents or node voltages may greatly reduce the number of
unknowns and simplify the solution.
Checking the essentials because of the many opportunities for mistakes in sign
and value.
~
Two one-ports are equivalent if they present the same v-I characteristics.
Two passive one - ports are equivalent if they have the same input resistance.
For resistance in series, REQ + Rı +R2 +….. + R n
For conductances in parallel, GEQ = Gı + G2 + ….. + Gn
~
For the voltage divider and current divider,
V2 =
R2
Rı + R2
V
I2 =
G2
Gı + G2
I
=
Rı
Rı + R2
~
Insofar as a load is concerned, any one-port network of resistance elements and energy
sources can be replaced by a series combination (Thévenin) of an ideal voltage source VT
and a resistance RT , or by a parallel combination (Norton) of an ideal current source IN
and a conductance GN,
Where VT = VOC, RT = VOC/SC,IN = ISC / VOC = I/ RT
~
A series combination of voltage V and resistance R can be replaced by a parallel
combination of current I = V/R and conductance G = I/R.
~
If cause and effect are linearly related, the principle of superposition applies and the total
effect of several causes acting simultaneously is equal to the sum of the effects of the
individual causes acting one at a time.
~
Networks involving nonlinear elements can be solved in various ways depending on the
nature of the problem and the form of the data.
If there is a single nonlinear element in an otherwise linear network, construction of a
load line permits a simple graphical solution.
3. SIGNAL WAVEFORM
~
The simplicity of the mathematical relations between exponentials and their derivatives
and integrals makes them valuable in the circuit analysis.
The general decaying exponential is α = A e-t/T
The time constant T is a measure of the rate of decay .
When t = T,
When t = 5T,
~
α/A = 1/e
α/A = 1/e5
= 0.368;
= 0.0067.
Sinusoids are important because; they occur naturally, they are common in power and
communication, they are easy to handle mathematically and they can be used to represent
other periodic waves.
The general sinusoid is a = A cos (wt = α ).
Frequency w = 2πf rad/s; f = 1/T hertz, where period T is in seconds.
~
In the imaginary number jb, j is an operator indicating a 90˚ counterclockwise rotation; in
algebraic manipulation, j²1 = -1 or j = √ -1
The combination of a real number jb defines a complex number W = a + jb = Mθ. For
conversion,
~
a = M cos θ
M = √ a² = b²
b = M sin θ
θ = arctan b/a
A sinusoidal function of time a = A cos (wt + α ) can be interpreted as the real part of the
complex function
A e j(wt + α) = A e jα e jwt
The complex constant A e jα is defined as a phasor A, the transform of a (t).
Phasor calculations follow the rules of complex algebra.
~
Phasor diagrams follow the rules for vector diagrams and can provide an approximate
answer, a quick check, or a clear visualization. The procedure is :1. Transform the functions of time into constant phasors.
2. Perform the indicated operations on phasors, using phasor diagrams.
3. Transform the resulting phasors into function of time.
~
In a periodic function of time, f(t + T) = f(t).
The average value of a periodic current is
Iаv =
1
T

T
0
idt
For a sinusoid, the half-cycle average is (2/π)‫׀‬m =0.636 ‫׀‬m
The effective or rums value of a periodic current is
I=
1
T
t
 I
2
0
dt
For a sinusoid, the rms value I = l m/ √2 = 0.707 l m
On phasor diagrams, use of effective values as read on ac meters.
~
4.
Rms values are significant whenever the square of a variable is the important quantity.
For function curves of simple geometry, the rms value is the square root of the average
ordinate of the 'squared' curve.
NATURAL RESPONSE
~
Forced behaviour is the response to external energy sources. Natural behaviour is the
response to internal stored energy.
~
Many physical systems with one energy-storage element can be described adequately by
first-order integro-differential equations. The general procedure for determining the
natural behaviour of a linear system is:1. Write the governing integro-differential equation.
2. Reduce this to a homogeneous differential equation.
3. Assume an exponential solution with undetermined constants.
4. Determine the exponents from the homogeneous equation.
5. Evaluate the coefficients from the given conditions.
~
In a second-order system with two energy-storage elements, the character of the natural
response is determined by the discriminant.
If the discriminant is positive, the response is overdamped and is represented by the sum
of two decaying exponentials:
st
st
1
 A e A e 2
1
2
If the discriminant is negative, the response is oscillatory and is represented by a damped
sinusoid : A = A ℮ ˉαt sin (wt + θ)
If the discriminant is zero the response is critically damped and is represented by the
sum of two different terms:
  A1 e st  A2 te st
~
Impedance Z (ohms) and admittance Y (mhos) are defined for exponentials.
Where
~
~
v = Zi,
Rz = R,
ZL = sL, and ZC = 1/ sC
I = Yv,
YR = G,
YL = 1/sL and Yc = sC
Impedances and admittances in complicated networks are combined in the same way as
resistances and conductances, respectively. The impedance function Z(s) and Y(s) contain
the same information as the characteristic equation.
The pole-zero diagram contains the essential information of the impedance function or
the admittance function.
A zero of the impedance function indicates the possibility of a current without an applied
voltage, therefore a natural current. A pole of the impedance function indicates function
indicates the possibility of a voltage without an applied current, therefore a natural
voltage.
~
Using the pole-zero concept, the general procedure is :1. Write the impedance or admittance function for the terminals of interest.
2. Determine the poles and zeros of impedance or admittance.
3. Use of the poles and zeros to identify possible components of natural voltage or
current.
4. Evaluate the coefficients for the given conditions.
5.
FORCED RESPONSE
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The waveform of the forcing function determines the waveform of the forced response.
~
Impedance is defined for exponentials of the form i = l0 est .
Impedances are combined in series and parallel, just as resistances are:
In general, the forced response is governed by v = Z(s)i where
ZR (s) = R
ZL (0) = 0
ZC (s) = 1/sC
~
For s = 0, i = l0 = l , a direct current. In this case,
ZR (0) = R
ZL (0) = 0
ZC (0)
=
∞.
An inductance looks like a short circuit to a direct current.
A capacitance looks like an open circuit to a direct voltage.
~
Phasor voltage and current are related by the complex impedance Z = Z(jw) or the
complex admittance Y = Y = Y (jw).
V = ZI
Where
ZR (jw) = R
ZL (jw) = jwl
ZC (jw) = 1/jwc
I = YV
Where
YR (jw) = G
YL (jw) = 1/jwl
YC (jw) = jwC
For phasors, Kirchoff's laws are written V = 0 and I = 0.
~
The steady-state sinusoidal response of a two-terminal network is completely defined by
Z or Y at the terminals where.
Z = Zz = R + jX and Y = Y = G + jB
Where
 =
tanˉ1 (X/R)
and
R = ac resistance in ohms
G =
ac conductance in mhos
X =
=
resistance in ohms
L or -1/C
B =
=
susceptance in mhos
C or -1/L
 =
tanˉ1(B/G)
~
In ac circuits,
The voltage VR is in phase with the current IR.
The voltage VL leads the current IL by 90.
The voltage VC lags the current IC by 90.
~
To determine the forced response to sinusoids :1.
2.
3.
4.
5.
~
Transform time functions to phasors and evaluate complex admittances.
Combine admittances in series or parallel to simplify the circuit.
Determine the desired response in phasor form.
Draw a phasor diagram to check values and display results.
Transform phasors to time functions if required.
Analogous systems are described by similar integro-differential equations.
Corresponding terms in the equations are analog quantities.
The solution of a problem is applicable to all analogous problem.
~
Two networks are dual if the set of transforms that converts the first into the second also
converts the second into the first.
The loop equations of planar network have the form as the node equations of its dual.
The solution of any network automatically provides the solution to its dual.
6.
COMPLETE RESPONSE
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The complete response is the sum or the forced response and the natural response :i = if + in
or
v = vf+ vn
~
The form of the forced response is the same as the form of the forcing function; the
amplitude is determined by the magnitude of the forcing function and the impedance or
admittance.
~
The form of the natural response is determined by the poles or zeros of the impedance or
admittance function; the amplitude is determined by the difference between the initial
energy storage and that indicated by the forcing function.
~
The general procedure for finding complete response is :1.
Write the appropriate impedance or admittance function.
2.
3.
4.
~
Determine the forced response from the forcing function and proper immitance.
Identify the natural components from poles and zeros of Z(s) or Y(s).
Add the forced and natural responses and evaluate the undermined constants
from initial conditions.
In determining constants from initial conditions, make use of continuity of current in an
indutance and voltage across a capacitance. In second-order circuits, also make use
of the expressions:di/dt = v/L and
dv/dt = i/C
~
A rectangular pulse can be generated by the combination of a positive step and, a short
time later, a negative step.
~
An impulse is the limiting case of a pulse of area Mp, amplitude Mp/Tp, and duration Tp
approaches zero.
For a current impulse, the magnitude Mp = ∫ i dt = CV
For a voltage impulse, the magnitude Mp = ∫ v dt = LI.
~
6.
The effect of an impulse is to store energy in an energy-storage element; the impulse
response of a circuit is the natural response of the circuit with the initial stored energy.
STEADY-STATE AC CIRCUITS
~
For sinusoids :Power (average)
P = VI cos θ = IR² R = VR²/R in W
Reactive power
Q = VI sin θ = IX²X = VX²/X in VAR
Apparent power
S = VI = I²Z = V²/Z in VA
Complex power
S = P + jQ
Power factor
= cos θ = P/VI
Reactive factor
= sin θ = Q/VI
~
To determine total apparent power, add power and relative power components separately
and combine vectorially; display on power triangles.
~
A circuit containing inductance and capacitance is in response if the terminal voltage and
current are in phase. At the resonant frequency the power is unity, and the impedance and
admittance are purely real.
Resonant frequency w0 = 1/ LC
~
or
f0 = 1/2 LC
The frequency selectivity of a resonant circuit is determined by bandwidth.
BW = w2 - w1 = w0 /Q
Where
maximum energy stored
Q = 2
Energy dissipated per cycle
For Q  10,
~
w1 = w0 - w0/ 2Q
w2 = w0 + w0/2Q
For resonant circuits :Series:
Q
wo L
1

R
wo CR
VL = VC = QV at w0
Q
wo C
1

G
wo LG
IL = IC QI at w0
Parallel:
Series Y/Y0 = ‫ װ‬Z/Z0 =
1
1
~
A balanced three-phase system consists of three equal single-phase sources connected in
∆ or Y. For balanced three-phase systems in
∆ : Vline = Vphase
Y = Iline = Iphase
~
+ JQ (w/w0 - w0/w)
and
and
In analysing balanced three phase circuits :-
Iline = √3 I phase  -30°
Vline = √3 Vphase  -30°
Draw a carefully oriented and clearly labeled diagram.
Sketch the phasor diagram as a guide and as a check . Analyse one phase and use
symmetry for the other phases.
Total power = 3 Vphase Iphase cos θ = √3 Vline Iline cos θ.