basic electricity chapter 5

8/3/2019 Basic Electricity Chapter 5

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Chapter 5: ALTERNATING

CURRENT BASICS

By Engr. Ramon A. Alguidano Jr.


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AC Circuits

Voltages and currents are sinusoidal (subject tohigher frequency harmonics and transients)

In Philippines, all AC has a fixed frequency:frequency (f) = 60Hz, period (T) = 16.67 ms

Angular frequency: = 2f = 120

For example

V(t) = Asin(t)= Asin(2ft)

= Asin(120t) in the Philippines


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AC Circuits: Voltage and current in a resistor

Assume V(t) = Asin(t)I(t) = V(t)/R = (A/R) sin(t)

Current and voltage are in-phase for a resistor


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AC Circuits: Voltage and current in a capacitor

Assume V(t) = Asin(t)i(t) = CdV(t)/dt = CAcos(t) = CAsin(t + /2)

Current leads voltage in a capacitor


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Voltage and current in an inductor

Assume V(t) = Asin(t)i(t) = 1/L. V(t)dt = -1/L.Acos(t) = 1/L.Asin(t - /2)

Current lags voltage in an inductor


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Resistance and ReactanceAssuming V(t) = Asin(t):

For resistor i(t) = A/R. sin(t) |V|/|i| = R the resistance ()

For capacitor i(t) = CAsin(t + /2) |V|/|i| = 1/C = XC the capacitive reactance () XC = for = 0 (DC), XC = 0 for =

For inductor i(t) = 1/L.Asin(t - /2) |V|/|i| = L = XL the inductive reactance ()

XL = 0 for

= 0 (DC), XL = for

=


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Resistance and ReactanceR, L & C components are those from previous example(s)


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AC Circuit Example:Series RLC circuit (balanced) Since components are in series, take current as our

reference:

i(t) = 141sin(

t) where

= 100 (f = 50Hz)Resistance = 2.4

Inductance = 1.0mH inductive reactance = 3.16

Capacitance = 1000F capacitive reactance = 3.16

Calculate and plot voltage waveforms


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AC Circuit Example: Series RLC circuit (balanced)


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AC Circuit Example: Series RLC circuit (unbalanced)

Since components are in series, take current as our

reference:

i(t) = 141sin(t) where = 100 (f = 50Hz)

Resistance = 2.40

Inductance = 11mH inductive reactance = 3.46

reactance = 2.89 Capacitance = 1100F capacitive

Calculate and plot voltage waveforms


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Note that in the unbalanced RLC circuit the voltages dueto inductive

and capacitive reactance DO NOT cancel hence theresulting

voltage waveform is out-of-phase with the current. In thiscase, the

current lags the voltage, since the inductive reactance islarger than

the capacitive reactance. So far, we have calculated individual voltage/current

components

mathematically and added them together. This lastexample

required the use of a spreadsheet to do the calculations. Surely there must be an easier way! Well there

isphasor diagrams!


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AC Circuits: Phasors

Phasors aid the analysis of ACcircuits, and are the result of amathematical trick We startwith the following relation:

ej = cos + j sin This function can be drawn on an

Argand diagram as a point onthe unit circle at the angle .

The horizontal and verticalcomponents are the cosines andsines, or the Real and Imaginaryparts respectively.


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AC Circuits: PhasorsNow consider the function:

F(t) = ej(t) = cos(t) + j sin(t)

On the Argand diagram this function rotates around theunit circle at an angular frequency , with the anglesubtended (t) changing with time.

The Real and Imaginary components of this functionboth represent a time-varying sinusoidal function.

The Real component of this function cos(t) is usedto model real - world AC signals, however the maths stillworks even if we include the Imaginary term (whichdoesnt symbolize anything in the real-world).


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AC Circuits: PhasorsNow consider a voltage waveform, modelled as:

V(t) = Aej(t+) = Aej ej(t)

This function consists of two components:ej(t) which we have just described as the rotating

component.This is common to all voltages and currents in an ACcircuit operating at a frequency , and is thereforeignored (but not forgotten).

Aej which is a stationary vector on the Argand diagram,and contains all information about magnitude (A) andphase angle () of the voltage waveform.

This stationary vector is called aPHASOR

.


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AC Circuits: PhasorsTo summarize:

A Phasor is a mathematical tool (a complexquantity) for describing the magnitude andphase angle of a sinusoidal AC signal.

Aej completely characterises the signal V(t) =Acos(t+ ).

How are Phasors used? To represent sinusoidalwaveforms as vectors on the complex plane,which allows us to analyze AC circuits usingvector manipulation!


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Phasor Diagram Example: Unbalanced series RLC circuit


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Phasor Diagram Example

In the Phasor Diagram, note that the phasorcorresponding to inductor voltage lags theresistor voltage and circuit current by 90.

There is also a mathematical trick for describingthe direction of phasors.

The trick involves the use of the complex operator jto describe the impedance of a component inan AC circuit.


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The j operator

To discuss impedance, we first present the followingrelationships:

sin = - j cos , or alternatively, cos = j sin .

This is simply a mathematical way of saying that the sinefunction lags the cosine function by 90, or that thecosine function leads the sine function by 90.

So although j is commonly known as -1, in this contextit is more conveniently considered to be an operatorwhich, when multiplied, advances a phasor through anangle of 90. Similarly, division by j or multiplication by jretards a phasor by 90.


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ImpedanceConsider a series R-L circuit supplied with a current i(t) =

Acos(t)

V(t) = Ri(t) + Ldi(t)/dt= A [Rcos(t) - Lsin(t)]= A [Rcos(t) + jLcos(t)]

= Acos(t)(R + jL)= (R + jXL)i(t)where Z = R + jXL is called the impedance of the circuit.

Similarly, for a series R-C circuit a similar analysis gives:

Z = R jXC

In general:

Z = R +/- jX where X is either inductive (+XL) orcapacitive (-XC)


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Power in AC Circuits

Recall that the instantaneous power in acircuit is given by:

P(t) = V(t) i(t)


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Power in AC Circuits: Resistive Load


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Power in AC Circuits: Resistive LoadAs the voltage and current are both sinusoidal and in-phase, the

instantaneous power varies as a sine-squared function.

However, as engineers we are primarily concerned with theaverage power.

It can easily be shown that the average power is:Pavg = x Vpeak x Ipeak = Vrms x IrmsTo make things more meaningful, electrical engineers useequivalent root-mean-squared (RMS) quantities to describethe magnitude of AC voltages and currents, rather than thepeak value. RMS AC quantities have the same heating-

potential as DC quantities of the same value.It is simple to convert from RMS to peak values:

Vpeak = 2 VrmsFrom this point onwards, as per the convention, we will referto all voltages and currents in RMS quantities!


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Power in AC Circuits: Inductive Load


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Power in AC Circuits: Capacitive Load


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Power in AC Circuits: Inductive or Capacitive Load

Note that the average values of the powerwaveforms for inductive and capacitive loads arezero!

Pure Ls and Cs do NOT dissipate any power

they store it up and return it to the circuit each cycle.

For L and C, the maximum instantaneous valuesof the

power are called the REACTIVE POWERS, Q.


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Power in AC Circuits: Reactive Power

For inductors: Q = V2/XL = i2XL

For capacitors: Q = V2/XC = i2XC

In general: Q = V2/X = i2X

In circuits which have both resistance and reactance, both

active and reactive powers are present.

Active (real) power (P) is measured in Watts (W)

Reactive power (Q) is measured in volt-amperes reactive(VAr)


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Power in AC Circuits: Apparent Power

Apparent power (S) is thecombination of active(real) and reactive

powers.The power triangle:

S2 = P2 + Q2

Apparent power ismeasured in volt-amperes(VA)


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Power in AC Circuits: Power Factor

Reactive power is present in a circuit whenever the voltageand current are out-of-phase (by an angle ).

It can be easily shown that the average value of activepower in any AC circuit is P = Vi cos(). The magnitude

of reactive power is Q = Vi sin().

Cos() is known as the power factor (pf) of the circuit.

If V, I and P are measured we can find out all we need toknow about the circuit:

Z = V/I; pf = P/S; R = Z cos(); X = Z sin()


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Power in AC Circuits: Power Factor

Power factor is an important circuit parameter whendesigning an electrical system.

The power factor should be as high as possible to minimizethe plant required to deliver active power to a system.

Electrical plant is normally sized according to apparentpower (VA) ratings* therefore if the power factor is lowthen the plant is being under-utilized. * Plant is normally

sized according to internal losses (related to current andvoltage not power) which can still occur even if there isno active power flow.

basic electricity chapter 5

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