PowerPoint slides t/a Jenneson and Harper, Electrical Principles for the Electrical Trades 6e Slides prepared by Andrew OConnell 9-1 Chapter 9 Alternating

PowerPoint slides t/a Jenneson and Harper, Electrical Principles for the Electrical Trades 6eSlides prepared by Andrew O’Connell

9-1

Chapter 9Alternating current circuits


9-2

Purpose

• This chapter provides detailed information about alternating current circuits

• It describes the effects of AC on inductors and capacitors including resonance

• It covers series and parallel AC circuits and power in AC circuits


9-3

Introduction to alternating current circuits

• Inductors and capacitors behave differently when they are connected to an AC supply compared to when they are connected to a DC supply

• Ohm’s Law still works for instantaneous values, but the effects of time and frequency mean that phasors are a necessary tool for AC circuit analysis


9-4

• AC power is dependent on the phase angle which adds the complication of power factor and power factor correction

• Alternating current using sine waves is the basis of worldwide electricity distribution

• The convention is to use RMS values when discussing AC, unless stated otherwise

(cont.)

Introduction to alternating current circuits


9-5

Resistance in AC circuits

• Figure 9.1 shows that the voltage and current are in phase for a resistive load connected to an AC supply


9-6

• Ohm’s Law is applicable to the resistive load connected to an AC supply—that is:

IRMS = VRMS / R

• By plotting the instantaneous values of power, a power curve has been drawn

Resistance in AC circuits (cont.)


9-7

• Points to note regarding the power curve include:– the power curve is sinusoidal in shape– there are no negative power values– the power curve completed two cycles for

every cycle of current or voltage– the area under the power curve is equal to

half of the value obtained by multiplying the peak values of voltage and current



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• The result is that, if RMS values of voltage and current are used, the power equations are:

P = VRMS IRMS

P = IRMS2 R

P = VRMS2 / R



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9-10

• At power-line frequencies resistors are assumed to be ‘pure’ or non-inductive resistance. Any inductance that does occur is negligible in most cases

• Examples of resistive loads include incandescent lamps, radiators and electric jug elements



9-11

• At higher frequencies inductance may impact on the circuit, if this is undesirable the resistor can be wound to negate any inductance (Figure 9.2)



9-12

• Capacitive effects must be dealt with by spacing the conductors far enough apart that the capacitance becomes negligible

• Since the current and voltage are in phase for resistive AC circuits, Ohm’s Law is applicable

• Kirchoff’s Current and Voltage Laws also apply

• For purely resistive AC circuits, the rules for DC circuits continue to work



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• A change in current flow in an inductive circuit results in an induced EMF that opposes the change in current flow

• In an inductive AC circuit the current is continually changing, thus an induced EMF is always opposing the change in current flow

• This opposition to current flow is called inductive reactance

Inductance in AC circuits


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• The current flow is always lagging behind the applied voltage by 90oE in a purely inductive circuit

Inductance in AC circuits (cont.)


9-15

• The value of inductive reactance in a circuit depends on the inductance and the rate of change of current flow, which depends on the supply frequency.

XL = 2πfL

Where:

XL = The inductive reactance, in ohms

f = The frequency, in hertz

L = The inductance, in henry



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9-17

• Ohm’s Law applies in an inductive AC circuit but the opposition to current flow is the inductive reactance:

I = V / XL

Where:

I = The current, in amperes

V = The voltage, in volts




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9-19

Exercise:

A 230 V 50 Hz supply is applied to a choke coil of negligible resistance and the current through the coil is 2.5 A. Determine the inductance of the coil.



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Solution:



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• When inductors are connected in series to an AC supply the total inductive reactance is the sum of the inductive reactance of each of the individual inductors.

XLtotal = XL1 + XL2 + XL3 + ….. + XLn

• Similarly, the total inductance can be calculated:

Ltotal = L1 + L2 + L3 + ….. + Ln



9-22

Exercise:

Two inductors, one with an inductive reactance of 11 Ω, and the second with an inductive reactance of 12 Ω are connected in series across a 230 V 50 Hz supply.

(a) What is the total inductive reactance?

(b) What is the total current?



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Solution:



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• When inductors are connected in parallel to an AC supply the reciprocal of the total inductive reactance is the sum of the reciprocal of the inductive reactance of each of the individual inductors:

1/XLtotal = 1/XL1 + 1/XL2 + 1/XL3 + ….. + 1/XLn

• Similarly, the total inductance can be calculated:

1/Ltotal = 1/L1 + 1/L2 + 1/L3 + ….. + 1/Ln



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• The average power consumed by a pure inductor is zero

• As the current begins to flow in the inductor the energy is used to produce the magnetic field

• When the current falls the magnetic field collapses and energy is returned to the supply

• This is shown in Figure 9.3



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• Notice that the power curve is sinusoidal in shape (when the voltage and current are sinusoidal), it has twice the frequency of the applied EMF and that the power returned to the supply is shown below the axis


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Capacitors in AC circuits

• The charging and discharging of a purely capacitive circuit when it is connected to an AC supply causes the current to lead the supply voltage by 90oE


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• A capacitor reacts to the connected supply by either charging or discharging

• The current that flows is a displacement current, the value of which is affected by the supply voltage and frequency and the capacity of the capacitor

• The capacitor has the property of capacitive reactance

Capacitors in AC circuits (cont.)


9-31

XC = 1/(2πfC)

Where:

XC = The capacitive reactance, in ohms

f = The frequency, in hertz

C = The capacitance, in farads



9-32

• Ohm’s Law applies in a capacitive AC circuit but the opposition to current flow is the capacitive reactance:

I = V / XC

Where:I = The current, in amperesV = The voltage, in volts




9-33

Exercise:

Calculate the current drawn by a 16 μF capacitor when connected to a 230 V 50 Hz supply.



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Solution:



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• Connecting capacitors in series results in an effective increase in the plate separation

• This reduces the capacitance and therefore increases the capacitive reactance

XCtotal = XC1 + XC2 + XC3 + ….. + XCn



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• Connecting capacitors in parallel results in an effective increase in the plate area. This increases the capacitance and therefore decreases the capacitive reactance.

1/XCtotal = 1/XC1 + 1/XC2 + 1/XC3 + ….. + 1/XCn



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• The average power consumed by a pure capacitor is zero

• As the current begins to flow in the capacitor the energy is stored in the capacitor

• When the current falls the capacitor returns stored energy to the supply

• This is shown in Figure 9.4



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Quick quiz1. What is the relationship between the voltage

and current for a resistive load connected to an AC supply?

2. What is the opposition to AC current due to inductance called?

3. What determines the inductive reactance in a circuit?

4. What determines the capacitive reactance in a circuit?

5. How much power is consumed by a purely reactive AC circuit?


9-42

Quick quiz answers1. The voltage and current are in phase for a

resistive load connected to an AC supply

2. Inductive reactance

3. Supply frequency and inductance (XL = 2πfL)

4. Supply frequency and capacitance (XC = 1/(2πfC))

5. Average power consumed by a purely reactive AC circuit is zero


9-43

Series R-L-C circuits on AC current

• Practical AC circuits consist of combinations of resistance, inductive reactance and capacitive reactance

• The combined opposition to current flow in AC circuits is called impedance (Z)

• Ohm’s Law is still applicable:

Z = VRMS / IRMS


9-44

• In a series circuit the current is common to all parts of the circuit and the total voltage is the phasor sum of the individual voltage drops

• Figure 9.5 shows the phasor diagram for a series R-L circuit, VZ is the total circuit voltage and ϕ is the phase angle

Series R-L-C circuits (cont.)


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• A practical inductor has some resistance and some inductance, Figure 9.6 shows the phasor diagram for an example practical inductor

• Note that the phase angle is less than 90 degrees

• The total power loss in the inductor is made up of the copper losses and the iron losses



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9-48

• Commercially available capacitors are considered as pure capacitance for all practical purposes

• Figure 9.7 shows the phasor diagram for a series R-C circuit



9-49

• If a circuit has resistance, inductive reactance and capacitive reactance then the total voltage is the phasor sum of the voltages across each of the individual circuit components



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9-51

• Noting that each of the series R-L-C circuits consists of voltage phasors (and each of them is equal to the current times the opposition to current) and that the phasor addition produces a triangle, allows mathematical analysis of the circuit impedance



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Z = √(R2 + (XC – XL)2)

and

ϕ = Tan-1(X/R) = Cos-1(R/Z) = Sin-1(X/Z)

Where:

Z = The impedance, in ohms

R = The resistance, in ohms



X = (XL – XC)



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• If the inductive reactance is greater than the capacitive reactance the total current lags the applied voltage

• If the capacitive reactance is greater than the inductive reactance the total current leads the applied voltage

• If the inductive reactance equals the capacitive reactance the circuit is said to be resonant



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Exercise:

A circuit that has 20 Ω in series with 0.25 H and 80 μF is connected to a 230 V 50 Hz supply.

Determine the impedance of the circuit, the current and its phase angle.



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Solution:



9-57

• In parallel circuits the voltage is common and is therefore used as the reference in phasor diagrams

• A pure inductor in parallel with a resistance will give a phasor diagram like Figure 9.10



9-58

• Since a practical inductor has a phase angle less than 90 degrees the phasor diagram for a practical inductor connected in parallel with a resistance will look like Figure 9.11



9-59

• Figure 9.12 shows the phasor diagram for a capacitor connected in parallel with a resistance



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• The total current in a parallel R-L-C circuit can be determined by phasor addition of the branch currents



9-61

• The impedance triangle method cannot be used for parallel circuits



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Quick quiz

1. What is the combined opposition to current flow in an AC circuit called?

2. Of what is the total power loss in an inductor made up ?

3. How does the amount of inductive and capacitive reactance affect the series circuit current?

1. How can the total impedance of a series R-L-C circuit be determined?

2. How can the total impedance of a parallel R-L-C circuit be determined?


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Quick quiz (cont.)


9-65

Quick quiz answers

1. The combined opposition to current flow in AC circuits is called impedance (Z)

2. The total power loss in the inductor is made up of the copper losses and the iron losses

3. If the inductive reactance is greater than the capacitive reactance the total current lags the applied voltage. If the capacitive reactance is greater than the inductive reactance the total current leads the applied voltage. If the inductive reactance equals the capacitive reactance the circuit is said to be resonant.


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Quick quiz answers (cont.)

4. Using the impedance triangle5. The total current in a parallel R-L-C

circuit can be determined by phasor addition of the branch currents—the total impedance can be calculated using Ohm’s Law


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Power in AC circuits

• A practical inductor consists of resistance and inductance

• The resistance consumes power and the inductance does not consume power


9-69

• The power consumed by the resistance can be calculated from P = VRI

• From the phasor diagram VR = Vcosϕ• Combining these two equations gives the

equation for True Power in AC circuitsP = VIcosϕ

Power in AC circuits (cont.)


9-70

• Since VI is multiplied by cosϕ, cosϕ is known as the power factor (λ)

• If the power factor is not included in the equation the result is the Apparent Power (S)

• That is, S = VI

• Apparent power is measured in volt-amperes



9-71

• Many AC machines are rated in VA to provide information about the current rating of the windings

• The true power and the reactive power can be included as two sides of a triangle known as the power triangle



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9-73

• The third side of the power triangle represents the power associated with the reactive components in the circuit. It is known as reactive power (Q)

Q = VIsinϕ

• Since cosϕ varies between zero and one, the power factor also varies between zero and one



9-74

• Power factor can be calculated from the ratio of true power to apparent power or from the ratio of resistance to reactance in the circuit

Power factor (λ) = cosϕ = R/Z = P/S

• In general the lower the power factor the greater the current required to supply a given amount of power



9-75

• The implications of the extra current include:– larger CSA conductors– larger transformers– higher rated switchgear– fuses of a higher current rating– higher voltage drop across conductors– extra copper losses– decreased efficiency– higher costs to supply electricity



9-76

• The following example demonstrates the effect of power factor on the current required to supply a load with a given amount of power



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9-79

• Low power factor is usually caused by inductive loads such as fluorescent lights, electric motors and transformers

• The supply authority requires power factor to be controlled within an installation


• The power factor of motors and transformers is better when they are operating close to full load

• Fluorescent lights have an improved power factor when a capacitor is connected across the terminals of the light


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9-81

• The power factor of a circuit can be obtained using a voltmeter, ammeter and wattmeter and then dividing the true power by the apparent power



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9-83

• Power factor may be measured directly with a power factor meter

• Power factor can also be determined using a phasor diagram or, if it is a series circuit, by using the impedance triangle


Exercise:An inductor draws a current of 20 A on 230 V DC and 10 A on 230 V AC. Calculate the angle of lag when on AC.


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Solution:



9-86

• Power factor correction is required to minimise the current drawn from the supply

• This is achieved by connecting a capacitor across the terminals of the load



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9-89



9-90

• The connection of the capacitor improves the power factor, reduces the current drawn from the supply and maintains the true power consumed by the load

• Rather than guessing the appropriate capacitor size, it can be calculated



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9-93

• Another method of choosing power factor correction is to consider the reactive power required to improve the power factor to the desired level



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9-95

• Power factor correction can also be achieved using a synchronous motor

• The synchronous motor operates with a leading power factor and provides power factor correction and mechanical effort at the same time



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Resonance

• When the power factor is corrected to unity the circuit resistance becomes the only opposition to current flow

• This situation requires the inductive reactance and the capacitive reactance to be exactly equal (XL = XC)

• This condition is called resonance

• At resonance, energy is transferred between the electromagnetic field of the inductor and the electrostatic field of the capacitor

• Because reactance is frequency dependent there is a specific frequency at which resonance will occur, it is called the resonant frequency


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Resonance (cont.)


9-98

• A series circuit operating at resonance may have voltages higher than the supply voltage across the reactive components

Resonance (cont.)


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Resonance (cont.)


9-100

• Generally series resonant circuits are avoided but they can be used to select specific frequencies

• Frequencies other than the resonant frequency are attenuated

• The resonant frequency can be calculated using fo = 1/(2π√(LC))

Where:fo = The resonant frequency, in hertzL = The inductance, in henryC = The capacitance, in farads

Resonance (cont.)


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Resonance (cont.)


9-102

• Parallel circuit resonance causes a large circulating current to oscillate between the capacitive and inductive branches of the circuit

• Providing an adjustable inductor or capacitor allows the selection of a particular frequency as the resonant frequency. This circuit was traditionally used to tune radio frequencies

Resonance (cont.)


9-103

• Parallel resonance is also used in some test equipment for measuring capacitance

• Frequency dependent transducers also use this phenomenon as a means of monitoring and regulating frequency

• Metal detectors also utilise resonant circuits

Resonance (cont.)


9-104

Quick quiz

1. What is true power?

2. What is apparent power?

3. What is reactive power?

4. What is power factor?

5. What causes resonance?


9-105

Quick quiz answers

1. True power is the power associated with the resistive components of a circuit, P = VIcosϕ

2. Apparent power is the supply voltage multiplied by the supply current, S = VI

3. Reactive power is the power associated with the reactive components in a circuit, Q = VIsinϕ

4. Power factor is the ratio of true power to apparent power, λ = cosϕ = P/S

5. Resonance occurs when the inductive reactance in a circuit equals the capacitive reactance in the circuit


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9-107

Summary

• For resistive AC circuits the current and voltage are in phase

• For purely inductive AC circuits the current lags the voltage by 90 oE

• For purely capacitive AC circuits the current leads the voltage by 90 oE


9-108

• The phase angle (ϕ) of an AC R-L-C circuit depends on the relative proportions of each component in the circuit

• Opposition to current flow in an inductor is called inductive reactance (XL)XL = 2πfL

• Opposition to current flow in a capacitor is called capacitive reactance (XC)XC = 1/(2πfC)

Summary (cont.)


9-109

• The total opposition to current flow in an AC circuit is called impedance (Z)

• In a series circuit, Z = √(R2 + (XL-XC)2)

• In a parallel circuit, IZ = √(IR2 + (IL-IC)2)

• The power factor of a circuit is the ratio of the true power to the apparent power, λ = cosϕ

Summary (cont.)


9-110

• The power triangle shows the relationship between true power, apparent power and reactive power

• The power factor of a circuit can be determined by measurement and calculation

• Low power factor results in an increased current for delivery of a given amount of power

Summary (cont.)


9-111

• Electricity suppliers require control of an installation’s power factor

• Typical installations have a lagging power factor due to the connected inductive loads

• Power factor correction may be achieved using capacitors or synchronous machines both of which have a leading power factor

Summary (cont.)


9-112

• Resonance occurs when the inductive reactance in a circuit is exactly equal to the capacitive reactance in the circuit

• The resonant frequency can be calculated using fo = 1/(2π√(LC))

• Resonance can cause dangerous voltages and currents in circuits

Summary (cont.)

Documents

PowerPoint slides t/a Jenneson and Harper, Electrical Principles for the Electrical Trades 6e Slides prepared by Andrew OConnell 9-1 Chapter 9 Alternating