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In-02-06 A.C. Principles

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					Abu Dhabi National Oil Co. ADNOC Technical Institute

INSTRUMENTATION
INDUSTRIAL ELECTRONICS I

UNIT 6 A.C. PRINCIPLES

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE

UNITS IN THIS COURSE

UNIT 1

THE ELECTRICAL CIRCUIT

UNIT 2

SERIES AND PARALLEL CIRCUITS

UNIT 3

ELECTROMAGNETIC PRINCIPLES

UNIT 4

BASIC ELECTROSTATICS AND THE CAPACITOR

UNIT 5

THE INDUCTOR, CAPACITOR AND D.C.

UNIT 6

A.C. PRINCIPLES

UNIT 7

COMMON ELECTRICAL SYMBOLS

UNIT 8

PRACTICAL TASKS

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE

TABLE OF CONTENTS
Paragraph 6.0 COURSE OBJECTIVES 6.1 INTRODUCTION 6.2 A.C. AND THE INDUCTOR 6.3 A.C. AND THE CAPACITOR 6.3.1 The Inductor-Resistor Combination 6.3.2 The Capacitor-Resistor Combination 6.3.3 RL and C in Combination 6.3.4 Resonance 6.4 WORKED EXAMPLES 6.4.1 Example 1 6.4.2 Example 2 6.4.3 Example 3 6.4.4 Example 4 6.4.5 Example 5 6.4.6 Example 6 6.4.7 Example 7 6.5 SUMMARY OF IMPORTANT FORMULAS Page 4 5 5 6 7 8 9 10 10 10 11 12 13 14 15 16 17

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.0 COURSE OBJECTIVES The student will be able to:        Explain with the aid of a diagram the effects of A.C. on an inductor. Explain with the aid of a diagram the effects of A.C. on a capacitor. State the formula for inductive reactance. State the formula for capacitive reactance. State the formulas for RL and C combination circuits supplied with A.C. Explain the term impedance. Carry out simple calculations using the formulas for RL and C combination circuits supplied with A.C.  Explain resonance.

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.1 INTRODUCTION The aim of this unit is to explain the operation of a capacitor and an inductor when supplied with A.C. both individually and in combination with resistors. 6.2 A.C. AND THE INDUCTOR
VL IL IL

A.C. SUPPLY

L

VL

0 90

TIME

ONE CYCLE

Figure 6-1 A.C. and the Inductor The previous unit showed the effect of applying D.C. to an inductor. The back EMF across the inductor falls from maximum to zero and the current rises from zero to maximum. A.C. is a form of continuously switching D.C. The diagram and the graph (see Figure 6-1) show the effect of supplying A.C. to an inductor. The voltage waveform leads the current waveform by 90. When the voltage is maximum the current is zero. When the current is maximum the voltage is zero and so on. The faster the A.C. changes (the higher the frequency) the greater the back EMF which is produced. The back EMF reduces the current. the formula: XL XL L f


The alternating current resistance

provided by the coil is called INDUCTIVE REACTANCE (X L) and is given by

=

2  f L Ohms.

= Inductive reactance (Ohms) = Coil inductance in Henrys (H) = Frequency in HZ = Mathematical constant (3.142).

This formula must be remembered.
INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09 PAGE 5 OF 17 IN-02-06

ADNOC TECHNICAL INSTITUTE 6.3 A.C. AND THE CAPACITOR
IC IC VC

A.C. SUPPLY

C

VC
0 90

ONE CYCLE

Figure 6-2 A.C. and the Capacitor The previous unit showed the effect of applying D.C. to a capacitor. The charging current starts at a maximum and falls to zero and the voltage across the capacitor starts at zero and rises to a maximum. The effect is the exact opposite to an inductor. The diagram (see Figure 6-2) shows the effect of supplying A.C. to a capacitor. This time the graph shows the current leading the voltage by 90. voltage is zero and so on. The faster the A.C. changes (the frequency) the less time the capacitor has to charge so the current through the device increases. The A.C. resistance (reactance) of a capacitor (XC) goes down as the frequency goes up and is given by the formula XC = 1 2fC XC = Capacitive reactance in Ohms C = Capacitance in Farads (F) f = Frequency in Hz 2
= Mathematical constant (3.142).

When the current is maximum the

Ohms

This formula must be remembered.

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.3.1 The Inductor-Resistor Combination
IS IS R IL

IR
VS L VS R L

RL in series circuit

RL in parallel circuit

Figure 6-3 Inductor - Resistor Combination

Figure 6-3 shows the series and parallel circuits of an RL combination. Calculating the unknown values of V S and IS is difficult because the current and voltage waveforms through the inductor are 90 apart.

A B 90 C

A = Total A.C. Resistance (Z) B = Inductive Reactance (XL) C = Resistance (R)

A simple line diagram is used to illustrate the problem. From the diagram we get the following formulas. These must be remembered. The total A.C. resistance, IMPEDANCE (Z) for a series circuit is given by the formula: Z2 = or Z =
R2  XL2

R2 + XL2

The parallel circuit using the standard parallel formula is

1 1 1  2  2 2 Z R XL
or

1 1 1   2 2 Z R XL

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.3.2 The Capacitor-Resistor Combination
IS R IS

IR
IC

VS

C

VS

C

R

RC in series circuit

RC in parallel circuit

Figure 6-4 Inductor - Capacitor Combination Figure 6-4 shows the series and parallel circuits of an RC combination. The formulas for the total A.C. resistance (IMPEDANCE) of the above circuits follow the same principle as for the resistance and inductor circuits to give:

C

90 B A

A = Total A.C. Resistance (Z) B = Capacitive Reactance (XC) C = Resistance (R) Z2 =R 2 + XC2 or Z =
R
2

Series:

X

2 C

Parallel:

1 1 1  2  2 2 Z R XC
or
1  Z 1 1  2 2 R XC

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.3.3 RL and C in Combination
IS R C L IS IR VS VS R IL L IC

C

RLC in series

RLC in parallel

Figure 6-5 Resistor - Capacitor - Inductor Combination Figure 6-5 shows the series and parallel circuits for an RLC combination. In this case the reactance of the reactive parts oppose each other to give the formulas. Series : Z2 = R2 + (XL ~ XC)2 or Z Parallel : = R2 + (XL ~ XC)2

1 1 1  2  2 Z R (X L ~ Xc)2
or

1  Z

1 1 2  R (XL~ Xc)2

The symbol ~ means take the smaller value from the larger value.

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.3.4 Resonance When the value of XL equals the value of XC then XL  XC = 0. So the formula for impedance changes to: Z2 = R2 Therefore Z = R for both circuits This effect is called RESONANCE. resistive. At resonance; XC=XL At one frequency the circuit is only or

1 1  2 2 Z R

1 2 2 f C
4

2 2



f L

f C L=1 f=

1 2


LC

HZ

The frequency (f) is called the RESONANCE FREQUENCY of the circuit. It produces a circuit that is purely resistive. This circuit is useful because it is used to select one frequency from all others. A range of these circuits is used to select a television channel or radio station. interference. Let's take as an example the radio guide in the Gulf News. The parallel circuit used to get Dubai FM 92 will have a resonance frequency of 92 MHz, Capital Radio 100.5 MHz, Abu Dhabi 810 kHz etc. Each channel transmits at a different frequency to stop

6.4 WORKED EXAMPLES 6.4.1 Example 1 Question :

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE Find the inductive reactance of a 5 mH coil at 50 kHZ. Solution Inductive Reactance (XL) = 2  f L XL = 2  50 x 103 (HZ) 5 x 10 3 (H) XL = 500   XL = 1571  Inductive Reactance = 1571  6.4.2 Example 2 Question Find the capacitive reactance of a 1 F capacitor at 50 kHz. Solution Capacitive Reactance (Xc) =

1 2 f C

Xc =

1 2  50 x10 3 (H ) x1x10 -6 (F) z
10 3 = 3.183  100 

=

Capacitive Reactance = 3.183 

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.4.3 Example 3 Question Find the impedance of a 50 resistor and 100 series to a 50 Hz supply. Solution
CIRCUIT DIAGRAM

F capacitor connected in

50 

50 HZ

100  F

Step 1 Xc =

1 2 fC

Xc =

1 2   50()  100 x 10 -6 (F)

Xc = 31.83  Step 2 Z2 = R2 + XC2 Z2 = 502 + 31.832 Z2 = 2500 + 1013 Z2 = 3513 Z=
3513

Z = 59.27  Circuit Impedance = 59.27 

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.4.4 Example 4 Question Find the impedance of a 50 resistor and inductive reactance of 50 connected in series. Solution
CIRCUIT DIAGRAM

R=50

XL=50 

Z2 = R2 + XL2 Z2 = 2500 + 2500 Z=
5000

Z = 70.71  Circuit Impedance = 70.71 

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.4.5 Example 5 Question

XC=10 

R = 5

Using the diagram above, find the circuit impedance.
1 Z
2



1 X
2
C



1 R2

1 Z
2



1 1 02



1 52

1 Z
2



1 1  100 25

1 Z2

 0.01 + 0.04 = 0.05

Z2 = Z=

1 = 20 0.05
20 = 4.472 

Circuit Impedance = 4.472

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.4.6 Example 6 Question

XC=10 

R=3

XL=8

Using the diagram above, find the circuit impedance. Solution

1 1  2  Z2 R (X ~ X )2
C L

1

1 1  2  Z2 3 (10 - 8)2
1 Z
2

1



1 1  9 4

1 Z
2

 0.1111 + 0.25 = 0.3611

Z=

1 = 1.664  0.3611

Circuit Impedance = 1.664 

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.4.7 Example 7 Question
5 mH 10 

100 pF

Using the circuit diagram above, find (a) (b) Resonant frequency Circuit impedance at resonance

Solution (a) At resonance 0 =

1 HZ 2  LC
1

0 =

2  5 x 10

-3

x 100 x 10 -12

0 =

1012 10 6  HZ 4.443 2  0.5

Resonant frequency = 0.225 MHz

(b) At resonance XL cancels XC and the circuit is resistive only. Circuit impedance = 10 

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 6.5 SUMMARY OF IMPORTANT FORMULAS Ohm’s law and A.C. V=R V =  XC V =  XL V=Z Inductive reactance XL = 2  f L Ohms

Capacitive reactance XC = R and L in series Z2 = R2 + XL2

1 Ohms 2 f C
R and L in parallel

1 1 1 2 = 2 + 2 Z R XL
R and C in parallel

R and C in series Z2 = R2 + XC2

1 1 1 2 = 2 + 2 XC Z R
R L and C in parallel

R L and C in series Z2 = R2 + (XL ~ XC)2

1 1 1 2 = 2 + (X L ~ X C )2 Z R

Resonant Frequency 0 =

1 HZ 2  LC

INDUSTRIAL ELECTRONICS I - A.C PRINCIPLES DATE OF ISSUE 8-DEC-09

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