basics of electrical systems

8/7/2019 Basics of Electrical Systems

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Basics of electrical systems



Summary

The Basic Capacitor

Capacitors are one of the f undamental passive

components. In its most basic form, it is composed of

two conductive plates separated by an insulating

dielectric.

The ability to store charge is the definition of

capacitance.

DielectricConductors



Capacitance is the ratio of charge to voltage

QC

V

!

R earranging, the amount of charge on a

capacitor is determined by the size of the

capacitor (C ) and the voltage (V ).

Q CV !

If a 22 QF capacitor is connected to

a 10 V source, the charge is 220 QC

Capacitance



Capacitance

An analogy:

Imagine you store r u bber bands in a

bottle that is nearly f ull.

You could store more r u bber bands

(like charge or Q) in a bigger bottle

(capacitance or C ) or if you push

them in more (voltage or V ). Thus,

Q CV !



A capacitor stores energy in the form of an electric field

that is established by the opposite charges on the two

plates. The energy of a charged capacitor is given by theequation

Capacitance

2

2

1CV W !

where

W = the energy in joules

C = the capacitance in farads

V = the voltage in volts



The capacitance of a capacitor depends on

three physical characteristics.

Summary

128.85 10 F/m r AC d

I ¨ ¸! v © ¹ª º

C is directly proportional to

and the plate area.

the relative dielectric constant

C is inversely proportional to

the distance between the plates

Capacitance



128.85 10 F/m r AC

d

I ¨ ¸! v © ¹ª º

Summary

Find the capacitance of a 4.0 cm diameter

sensor immersed in oil if the plates are

separated by 0.25 mm.

The plate area is

The distance between the plates is

Capacitance

4.0 for oilr I !

3 2

12

3

4.0 1.26 10 m8.85 10 F/m

0.25 10 mC

¨ ¸© ¹! v !© ¹vª º

3

0.25 10 m

v

178 pF

2 2 3 2 0.02 m 1.26 10 m A r T ! ! ! v



Summary

Capacitor types

Mica

Mic

F il

F il

Mic

F il

F ilMic

F il

Mica capacitors are small with high working voltage.

The working voltage is the voltage limit that cannot

be exceeded.



Summary

Capacitor types

Ceramic disk

Solder

Lead wire solderedto silver elec trode

Ceramicdielectric

Dipped phenolic coating

Silv er electrode s deposited ontop and b ottom of ceram ic disk

Ceramic disks are small nonpolarized capacitors They

have relatively high capacitance du

e to highI

r.



Summary

Capacitor types

Plastic Film

Lead ir e

Hi - urityf il electr des

lastic fild ielec tric

Outer r ap f polyester fil

apac itor section(alter na te strips of fil d ielectric andf oil electr odes)

So lder coa ted end

Plastic film capacitors are small and nonpolarized. They

have relatively high capacitance due to larger plate area.



Summary

Capacitor types

Electrolytic (two types)

Symbol for any electrolytic capacitor

Al electrolytic

+

_

Ta electrolytic

Electrolytic capacitors have very high capacitance but

they are not as precise as other types and tend to have

more leakage current. Electrolytic types are polarized.



Summary

Capacitor types

Variable

Variable capacitors typically have small capacitance

values and are usually adjusted manually.

A solid-state device that is used as a variablecapacitor is the varactor diode; it is adjusted with an

electrical signal.



Capacitor labeling

Capacitors use several labeling methods. Small

capacitors values are frequently stamped on them such

as .001 or .01, which have implied units of microfarads.

+

+ +

+

V T T

V T

T

4 7 M F

. 0 2 2

Electrolytic capacitors have larger

values, so are read as QF. The unit is usually

stamped as QF, but some older ones may be

shown as MF or MMF).



A label such as 103 or 104 is read as 10x103

(10,000 pF) or 10x104 (100,000 pF)

respectively. (Third digit is the multiplier.)

Capacitor labeling

When values are marked as 330 or 6800, the

units are picofarads.

What is the value of

each capacitor? Both are 2200 pF.

2 2 2 2 2 0 0



Summary

Series capacitors

When capacitors are connected in series, the total

capacitance is smaller than the smallest one. The

general equation for capacitors in series is

T

1 2 3 T

1

1 1 1 1...

C

C C C C

!

The total capacitance of two capacitors is

T

1 2

11 1

C

C C

!

«or you can use the product-over-sum r ule



Summary

Series capacitors

If a 0.001 QF capacitor is connected

in series with an 800 pF capacitor,the total capacitance is 444 pF

0.001 µF 800 pF

C 1

C 2



Summary

Parallel capacitors

When capacitors are connected in parallel, the total

capacitance is the sum of the individual capacitors.

The general equation for capacitors in parallel is

T 1 2 3...

nC C C C C !

1800 pF

If a 0.001 QF capacitor is

connected in parallel withan 800 pF capacitor, the

total capacitance is

0.001 F 800 pFC 1 C 2



Summary

The RC time constant

When a capacitor is charged

through a series resistor and

dc sou

rce, the charging cu

rveis exponential.

C

R I initia l

t 0

(b) ar ing curr ent

V fina l

t 0

(a) apacitor c ar g ing oltage



Summary

When a capacitor is discharged

through a resistor, the

discharge curve is also an

exponential. (Note that the

current is negative.)

t

t

I i iti l

( ) i r i rr t

V i iti l

( ) it r i r i lt

C

R

The RC time constant



Summary

Universal exponential curves

S pecific values for

current and voltage

can be read from auniversal curve. For

an RC circuit, the

time constant is

RC !

100%

80%

60%

40%

20%

00 1X 2X 3X 4X 5X

99%98%

95%

86%

63%

37%

14%

5%2% 1%

Number of time constants

P e r c e n t o f f i n a l v a l u e R ising exponential

Falling exponential



Summary

The universal curves can be applied to general formulas for

the voltage (or current) curves for RC circuits. The general

voltage formula is

v =V F + (V i V F)et/RC

V F = final value of voltage

V i = initial value of voltage

v = instantaneous value of voltage

The final capacitor voltage is greater than the initial

voltage when the capacitor is charging, or less that the

initial voltage when it is discharging.




Summary

Capacitive reactance

Capacitive reactance is the opposition to ac by a

capacitor. The equation for capacitive reactance is

1

2C X

fC !

The reactance of a 0.047 QF capacitor when a

frequency of 15 kHz is applied is 226 ;



Summary


When capacitors are in series, the total reactance is the sum of the

individual reactances. That is,

Assume three 0.033 QF capacitors are in series with a 2.5 kHz

ac source. What is the total reactance?

5.79 k ;

C( ) C1 C2 C3 Ct ot n X X X X X !

The reactance of each capacitor is

1 1 1.93 k 2 2 2.5 kHz 0.033 F

C X f C ! ! ! ;

C( ) C1 C2 C3

1.93 k 1.93 k 1.93 k

t ot X X X X !

! ; ; ; !



Summary


When capacitors are in parallel, the total reactance is the reciprocal

of the sum of the reciprocals of the individual reactances. That is,

If the three 0.033 QF capacitors from the last example are

placed in parallel with the 2.5 kHz ac source, what is the total

reactance?

643 ;

C( )

C1 C2 C3 C

1

1 1 1 1t ot

n

X

X X X X

!

The reactance of each capacitor is 1.93 k ;

( )

C1 C2 C3

1 1

1 1 1 1 1 1

1.93 k 1.93 k 1.93 k

t ot X

X X X

! ! !

; ; ;



Summary

Capacitive Voltage Divider

Two capacitors in series are commonly used as a capacitive voltage

divider. The capacitors split the output voltage in proportion to their

reactance (and inversely proportional to their capacitance).

What is the output voltage for the capacitive voltage divider?

V out

1

1

1 14.82 k

2 2 33 k z 1000 pFC X

fC ! ! ! ;

2

2

1 1482

2 2 33 k z 0.01 FC X

fC ! ! ! ;

1000 pF

0.01 µF

C 2

C 1

1.0 V f = 33 kHz

2

( )

482 1.0 V =

5.30 k

C out s

C t ot

X V V

X

¨ ¸ ;¨ ¸! !© ¹ © ¹© ¹ ;ª ºª º

( ) 1 2

4.82 k 482 5.30 k C t ot C C

X X X !

! ; ; ! ;

91 mV



Summary

Capacitive Voltage Divider

Instead of using a ratio of reactances in the capacitor voltage divider

equation, you can use a ratio of the total series capacitance to the

output capacitance (multiplied by the input voltage). The result is

the same. For the problem presented in the last slide,

V out

1 2

( )

1 2

1000 pF 0.01 F909 pF

1000 pF 0.01 Ft ot

C C C

C C ! ! !

1000 pF

0.01 µF

C 2

C 1

1.0 V

f = 33 kHz

( )

2

909 pF1.0 V =

0.01 F

t ot

out s

C V V

C

¨ ¸ ¨ ¸! !© ¹ © ¹

ª ºª º91 mV



Summary

Capacitive phase shift

When a sine wave

is applied to a

capacitor, there is a

phase shift between

voltage and current

such that current

always leads thevoltage by 90o.

V C

0

I

0

90o



Summary

Power in a capacitor

Voltage and current are always 90o out of phase.

For this reason, no tr ue power is dissipated by a capacitor,

because stored energy is returned to the circuit.

The rate at which a capacitor stores or returns

energy is called reactive power. The unit for reactive power is the VAR (volt-ampere reactive).

Energy is stored by the capacitor during a portion of the ac

cycle and returned to the source during another portion of

the cycle.



Summary

Power su pply filtering

There are many applications for capacitors. One is in

filters, such as the power su pply filter shown here.

R ectifier

60 Hz acC Load

resistance

The filter smoothes the pulsating dc from the

rectifier.



Summary

Cou pling capacitors

Coupling capacitors are used to pass an ac signal

from one stage to another while blocking dc.

The capacitor isolates dc

between the amplifier stages,

preventing dc in one stage from

affecting the other stage.

0 V 0 V

3 V

A¡

plifier

stage 1

A¡

plifier

stage 2

R2

R1¢

+V

Input Output



Summary

Bypass capacitors

Another application is to bypass an ac signal to ground

but retain a dc value. This is widely done to affect gain

in amplifiers.

The bypass capacitor places

point A at ac ground, keeping

only a dc value at point A.R2

R1

C

0 V

dc pl £

¤

¥

c

Poi¦

§

i¦

cir cui§

wher eonly dc i

¤

r equir ed

0 V

dc only

A



Selected Key Terms

Capacitor

Dielectric

Farad

RC timeconstant

An electrical device consisting of two conductive

plates separated by an insulating material and

possessing the property of capacitance.

The insulating material between the conductive plates of a capacitor.

The unit of capacitance.

A fixed time interval set by the R and C values,that determine the time response of a series RC

circuit. It equals the product of the resistance

and the capacitance.



Capacitivereactance

Instantaneous power p)

True power ( P true)

Reactive power ( Pr )

VAR (volt-ampere

reactive)

The value of power in a circuit at a given

instant of time.

The power that is dissipated in a circuit

usually in the form of heat.

The opposition of a capacitor to sinusoidalcurrent. The unit is the ohm.

Selected Key Terms

The rate at which energy is alternately stored

and returned to the source by a capacitor. The

unit is the VAR .

The unit of reactive power.



Quiz

1. The capacitance of a capacitor will be larger if

a. the spacing between the plates is increased.

b. air replaces oil as the dielectric.

c. the area of the plates is increased.

d. all of the above.



Quiz

2. The major advantage of a mica capacitor over other

types is

a. they have the largest available capacitances. b. their voltage rating is very high

c. they are polarized.




Quiz

3. Electrolytic capacitors are usef ul in applications where

a. a precise value of capacitance is required.

b. low leakage current is required.

c. large capacitance is required.




Quiz

4. If a 0.015 QF capacitor is in series with a 6800 pF

capacitor, the total capacitance is

a. 1568 pF. b. 4678 pF.

c. 6815 pF.

d. 0.022 QF.



Quiz

5. Two capacitors that are initially uncharged are connected

in series with a dc source. Compared to the larger

capacitor, the smaller capacitor will have

a. the same charge.

b. more charge.

c. less voltage.

d. the same voltage.



Quiz

6. When a capacitor is connected through a resistor to a dc

voltage source, the charge on the capacitor will reach 50%

of its final charge in

a. less than one time constant.

b. exactly one time constant.

c. greater than one time constant.

d. answer depends on the amount of voltage.



Quiz

7. When a capacitor is connected through a series resistor

and switch to a dc voltage source, the voltage across the

resistor after the switch is closed has the shape of

a. a straight line.

b. a rising exponential.

c. a falling exponential.

d. none of the above.



Quiz

8. The capacitive reactance of a 100 QF capacitor to 60 Hz

is

a. 6.14 k ; b. 265 ;

c. 37.7 ;

d. 26.5 ;



Quiz

9. If an sine wave from a f unction generator is applied to a

capacitor, the current will

a. lag voltage by 90o

. b. lag voltage by 45o.

c. be in phase with the voltage.

d. none of the above.



Quiz

10. A switched capacitor emulates a

a. smaller capacitor.

b. larger capacitor.

c. battery.

d. resistor.



Quiz

Answers:

1. c

2. b

3. c

4. b

5. a

6. a

7. c

8. d

9. d

10. d



Summary

Sinusoidal response of RC circuits

When both resistance and capacitance are in a series

circuit, the phase angle between the applied voltage and

total current is between 0r and 90r, depending on the

values of resistance and reactance.

R

V R

C

V R lead V S V

C lags V S

I leads V S

I

V S

V C



Summary

Impedance of series RC circuits

In a series RC circuit, the total impedance is the phasor

sum of R and X C .

R is plotted along the positive x-axis.

R

X C is plotted along the negative y-axis.

X C

1tan C X

RU ¨ ¸

! © ¹ª º

Z

It is convenient to reposition the

phasors into the impedance t riangle.

R

X C

Z

Z is the diagonal U U



Summary

Impedance of series RC circuits

R = 1.2 k ;

X C =

960 ;

Sketch the impedance triangle and show the

values for R = 1.2 k ; and X C = 960 ;.

2 21.2 k 0.96 k

1.33 k

Z ! ; ;

! ;

1 0.96 k tan

1.2 k 39

U ;!

;! r

Z = 1.33 k ;

39oU



Summary

Analysis of series RC circuits

Ohm¶s law is applied to series RC circuits using Z ,

V , and I .

V V V I Z I Z Z I

! ! !

Because I is the same everywhere in a series circuit,

you can obtain the voltages across different

components by multiplying the impedance of thatcomponent by the current as shown in the following

example.



Summary

Analysis of series RC circuits

Assume the current in the previous example is 10 mArms.

Sketch the voltage phasor diagram. The impedance

triangle from the previous example is shown for reference.

V R = 12 V

V C =

9.6 V

The voltage phasor diagram can be found from Ohm¶s

law. Multiply each impedance phasor by 10 mA.

x 10 mA

=

R = 1.2 k ;

X C =

960 ; Z = 1.33 k ;

39o

V S = 13.3 V

39o

U U



Summary

Variation of phase angle with frequency

Phasor diagrams that have reactance phasors can only

be drawn for a single frequency because X is a

f unction of frequency.

As frequency changes,the impedance triangle

for an RC circuit changes

as illustrated here

becau

se X C decreaseswith increasing f . This

determines the f r equenc y

r es ponse of RC circuits.

U

Z 3

X C 1

X C 2

X C 3

Z 2

Z 1

12

3

1

2

f

f

f

3

ncreasing f U

U

R



U

(phase lag)

J

J

(phase lag)

Summary

Applications

R

V R

V out

C V out V in

V in

V out

V in

For a given frequency, a series RC circuit can be used to

produce a phase lag by a specific amount between an

input voltage and an output by taking the output across

the capacitor. This circuit is also a basic l ow-pa ss f ilter , a

circuit that passes low frequencies and rejects all others.



U

U

(phase lead)

(phase lead)

V

Summary

Applications

R

V C

V out C

V out V in

V in

V out

V in

R eversing the components in the previous circuit produces

a circuit that is a basic lead network. This circuit is also a

basic hi gh-pa ss f ilter , a circuit that passes high frequencies

and rejects all others. This filter passes high frequencies

down to a frequency called the cut o ff frequency.



Summary

Applications

An application showing how the phase-shift network is

usef ul is the phase-shift oscillator, which uses a

combination of RC networks to produce the required 180o

phase shift for the oscillator.Amplifier

R f

R R R

C C C Phase-shift network



Summary

Sinusoidal response of parallel RC circuits

For parallel circuits, it is usef ul to introduce two new

quantities (susceptance and admittance) and to review

conductance.

Conductance is the reciprocal of resistance. 1G R

!

Admittance is the reciprocal of impedance.

Capacitive susceptance is the reciprocal

of capacitive reactance.

1C

C

B X

!

1Y

Z !



Summary


In a parallel RC circuit, the admittance phasor is the sum

of the conductance and capacitive susceptance phasors.

The magnitude can be expressed as 2 2+C Y G B!

V S G BC

Y

G

BC

U

From the diagram, the phase angle is 1tan C B

GU ¨ ¸

! © ¹ª º



Summary


V S C

0.01 QFY = 1.18 mS

G = 1.0 mS

BC = 0.628 mS

Draw the admittance phasor diagram for the circuit.

f = 10 kHz

R

1.0 k ;

1 1 1.0 mS1.0 k

G R

! ! !; 2 10 kHz 0.01 F 0.628 mSC B T Q! !

The magnitude of the conductance and susceptance are:

2 22 2 1.0 mS 0.628 mS 1.18 mSC Y G B! ! !



Summary

Analysis of parallel RC circuits

Ohm¶s law is applied to parallel RC circuits using

Y , V , and I .

I I

V I V Y Y Y V ! ! !

Because V is the same across all components in a

parallel circuit, you can obtain the current in a given

component by simply multiplying the admittance of the component by the voltage as illustrated in the

following example.



Summary

Analysis of parallel RC circuits

If the voltage in the previous example is 10 V, sketch the

current phasor diagram. The admittance diagram from the

previous example is shown for reference.The current phasor diagram can be found from

Ohm¶s law. Multiply each admittance phasor by 10 V.

x 10 V

=Y = 1.18 mS

G = 1.0 mS

BC = 0.628 mS

I R = 10 mA

I C = 6.28 mA

I S = 11.8 mA



Summary

Phase angle of parallel RC circuits

Notice that the formula for capacitive susceptance is

the reciprocal of capacitive reactance. Thus BC and I C

are directly proportional to f : 2C

B f C T !

I R

I C

I S

U

As frequency increases, BC

and I C must also increase,

so the angle between I R and

I S must increase.



Summary

Equivalent series and parallel RC circuits

For every parallel RC circuit there is an equivalent

series RC circuit at a given frequency.

The equivalent resistance and capacitive

reactance are shown on the impedance triangle:

Z

Req = Z cos U

X C( eq) = Z sin U

U



Summary

Series-Parallel RC circuitsSeries-parallel RC circuits are combinations of both series and

parallel elements. These circuits can be solved by methods from

series and parallel circuits.

For example, the

components in the

green box are in

series:

The components in

the yellow box arein parallel: 2 2

22 2

2 2

C

C

R X Z

R X !

R1 C 1

R2 C 2

Z 1 Z 2

The total impedance can be found by

converting the parallel components to anequivalent series combination, then

adding the result to R1 and X C1 to get the

total reactance.

2 2

1 1 1C Z R X !



Summary

Measuring Phase AngleAn oscilloscope is commonly used to measure phase angle in

reactive circuits. The easiest way to measure phase angle is to

set u p the two signals to have the same apparent amplitude and

measure the period. An example of a Multisim simulation is

shown, but the technique is the same in lab.

Set u p the oscilloscope so that

two waves appear to have the

same amplitude as shown.

Determine the period. For the

wave shown, the period is

20 s8.0 div 160 s

divT

¨ ¸! !© ¹ª º



Summary

Measuring Phase Angle Next, spread the waves out using the SEC/DIV control in order

to make an accurate measurement of the time difference between

the waves. In the case illustrated, the time difference is

5 s4.9 div 24.5 sdivt

¨ ¸( ! !© ¹ª º

The phase shift is calculated from

24.5 s360 360

160 s

t

T

U¨ ¸¨ ¸

! r ! r !© ¹© ¹ª º ª º

55o



Summary

The power triangle

R ecall that in a series RC circuit, you could multiply

the impedance phasors by the current to obtain the

voltage phasors. The earlier example is shown for

review:

V R = 12 V

V C =

9.6 V

x 10 mA

=

R = 1.2 k ;

X C = 960 ; Z = 1.33 k ;

39o

V S = 13.3 V

39o U U



Summary

The power triangle

Multiplying the voltage phasors by I rm s gives the power triangle

(equivalent to multiplying the impedance phasors by I 2). Apparent

power is the product of the magnitude of the current and magnitude of

the voltage and is plotted along the hypotenuse of the power triangle.

The rms current in the earlier example was 10 mA.

Show the power triangle.

P t rue = 120 mW

P r = 96

mVAR

x 10 mA

= 39o

P a = 133 mVA

V R = 12 V

V C =

9.6 VV S = 13.3 V

39oU U



Summary

Power factor

The power factor is the relationship between the

apparent power in volt-amperes and tr ue power in

watts. Volt-amperes multiplied by the power factor

equals tr ue power.

Power factor is defined mathematically as

PF = cos U

The power factor can vary from 0 for a purely reactivecircuit to 1 for a purely resistive circuit.



Summary

Apparent power

Apparent power consists of two components; a tr ue

power component, that does the work, and a

reactive power component, that is simply power

shuttled back and forth between source and load.

P t rue (W)

P r (VAR)

Some components such

as transformers, motors,

and generators are rated

in VA rather than watts. P a (VA)

U



10V

V ou ̈ V in

100 ;

1 FQ

10V

0

10V

0

Summary

Frequency R esponse of RC Circuits

When a signal is applied to an RC circuit, and the output is taken

across the capacitor as shown, the circuit acts as a low-pass filter.

As the frequency increases, the output amplitude decreases.

Plotting the response:

V ou

©

(V)

9.98

8.46

1.570.79

0.1 1 10 20 100f (kHz)

98

7

6

5

4

3

2

1

1 = 1 kHz

8.46V rms10V rms ;100

Q

1.57 V s

10V s

1 = 10 Hz

;100

Q

0.7 V s

10V s

1 = 2 Hz

;100

Q



V

10 V d

0

V

0 V d10 V d 100 ;1 FQ

Summary

Frequency R esponse of RC Circuits

R eversing the components, and taking the output across the resistor as

shown, the circuit acts as a high-pass filter.

As the frequency increases, the output amplitude also increases.


Hz

.6 VV

; Q

kHz

. VV

; Q

kHz

.87VV

; Q

V out (V)

f (kHz)

9 8

5 32

0.6300.01 0.1 1

10

9

8

6

5

3

2

1

10



Impedance

Phase angle

Capacitivesuceptance (BC )

Admittance (Y)

The total opposition to sinusoidal current

expressed in ohms.

Selected Key Terms

The ability of a capacitor to permit current;

the reciprocal of capacitive reactance. The

unit is the siemens (S).

The angle between the source voltage and the

total current in a reactive circuit.

A measure of the ability of a reactive circuit to

permit current; the reciprocal of impedance.

The unit is the siemens (S).



Power factor

Frequencyresponse

Cutoff

frequency

The frequency at which the output voltage of

a filter is 70.7% of the maximum output

voltage.

In electric circuits, the variation of the outputvoltage (or current) over a specified range of

frequencies.

The relationship between volt-amperes and

tr ue power or watts. Volt-amperes multiplied

by the power factor equals tr ue power.

Selected Key Terms



Quiz

1. If you know what the impedance phasor diagram looks

like in a series RC circuit, you can find the voltage phasor

diagram by

a. multiplying each phasor by the current

b. multiplying each phasor by the source voltage

c. dividing each phasor by the source voltage

d. dividing each phasor by the current



Quiz

2. A series RC circuit is driven with a sine wave. If the

output voltage is taken across the resistor, the output

will

a. be in phase with the input.

b. lead the input voltage.

c. lag the input voltage.

d. none of the above



Quiz

3. A series RC circuit is driven with a sine wave. If you

measure 7.07 V across the capacitor and 7.07 V across the

resistor, the voltage across both components is

a. 0 V

b. 5 V

c. 10 V

d. 14.1 V



Quiz

4. If you increase the frequency in a series RC circuit,

a. the total impedance will increase

b. the reactance will not change

c. the phase angle will decrease




Quiz

5. Admittance is the reciprocal of

a. reactance

b. resistance

c. conductance

d. impedance



Quiz

6. Given the admittance phasor diagram of a parallel RC

circuit, you could obtain the current phasor diagram by

a. multiplying each phasor by the voltage

b. multiplying each phasor by the total current

c. dividing each phasor by the voltage

d. dividing each phasor by the total current



Quiz

7. If you increase the frequency in a parallel RC circuit,

a. the total admittance will decrease

b. the total current will not change

c. the phase angle between I R and I S will decrease




Quiz

8. The magnitude of the admittance in a parallel RC circuit

will be larger if

a. the resistance is larger

b. the capacitance is larger

c. both a and b




Quiz

9. The maximum power factor occurs when the phase

angle is

a. 0o

b. 30o

c. 45o

d. 90o



Quiz

10. When power is calculated from voltage and current for an

ac circuit, the voltage and current should be expressed as

a. average values

b. rms values

c. peak values

d. peak-to-peak values



Quiz

Answers:

1. a

2. b

3. c

4. c

5. d

6. a

7. d

8. d

9. a

10. b



When a length of wire is formed into a coil., it becomes a

basic inductor. When there is current in the inductor, a three-

dimensional magnetic field is created.

Summary

The Basic Inductor

A change in current causes

the magnetic field to change.

This in turn induces a

voltage across the inductor

that opposes the original

change in current.

NS



Summary

The Basic Inductor

The effect of inductance is greatly

magnified by adding turns and winding

them on a magnetic material. Large

inductors and transformers are wound

on a core to increase the inductance.

Magnetic core

One henry is the inductance of a coil when a current, changing at a

rate of one ampere per second, induces one volt across the coil. Most

coils are much smaller than 1 H.



Summary

Fara ay¶s la

Fara ay¶s la as intro uce in Chapter 7 an repeate here

because of its importance to in uctors.

The amount of voltage in uce in a coil is irectly proportional to

the rate of change of the magnetic fiel ith respect to the coil.



Summary

e z¶s la

e z¶s la as also i troduced i Chapter 7 a d is a exte sio of

Faraday¶s la , defi i g the directio of the i duced voltage:

Whe the curre t through a coil cha ges a d a i duced voltage

is created as a result of the cha gi g mag etic field, the directio

of the i duced voltage is such that it al ays opposes the cha ge

i the curre t.



Summary

Lenz¶s lawA basic circuit to demonstrate Lenz¶s law is shown.

Initially, the SW is open and there is a small current in

the circuit through L and R1.

R1

SW

R2V S

L

+

+



Summary

SW closes and immediately a voltage appears across L that

tends to oppose any change in current.

R1

SW

R2V S

L

+

Initially, the meter

reads same current

as before the switch

was closed.

Lenz¶s law



Summary

After a time, the current stabilizes at a higher level (due to I 2)

as the voltage decays across the coil.

+

R1

SW

R2V S

L

Lenz¶s law

Later, the meter

reads a higher

current because of

the load change.



Summary

Practical inductors

In addition to inductance, actual inductors have winding

resistance ( RW) due to the resistance of the wire and winding

capacitance (C W) between turns. An equivalent circuit for a

practical inductor incl

uding these effects is shown:

L R

W

C W

Notice that the winding resistance

is in series with the coil and the

winding capacitance is in parallelwith both.



Summary

Types of inductors

Common symbols for inductors (coils) are

Air core Iron core Ferrite core Variable

There are a variety of inductors, depending on the amount of

inductance required and the application. Some, with fine

wires, are encapsulated and may appear like a resistor.



Summary

Factors affecting inductance

Four factors affect the amount of inductance for a coil. The

equation for the inductance of a coil is

2 N A

L l

Q!

where

L = inductance in henries

= number of turns of wire

Q = permeability in H/m (same as Wb/At-m)

l = coil length on meters



Summary

What is the inductance of a 2 cm long, 150 turn coil

wrapped on an low carbon steel core that is 0.5 cm diameter?

The permeability of low carbon steel is 2.5 x104 H/m

(Wb/At-m).

22 5 2 0.0025 m 7.85 10 m A r ! ! ! v

2 N A

Ll

Q!

22 mH

2 4 5 2150 t 2.5 10 Wb/At-m 7.85 10 m

0.02 m

v v!

!



Summary

Practical inductors

Inductors come in a variety of sizes. A few common

ones are shown here.

ncapsulated Torr oid coil Variab le



Summary

Series inductorsWhen inductors are connected in series, the total inductance is

the sum of the individual inductors. The general equation for

inductors in series is

2.18 mH

T 1 2 3 ... n L L L L L!

If a 1.5 mH inductor is

connected in series with an 680

QH inductor, the total

inductance is

L1 L2

1.5 mH 680 HQ



Summary

Parallel inductorsWhen inductors are connected in parallel, the total inductance is

smaller than the smallest one. The general equation for

inductors in parallel is

The total inductance of two inductors is

«or you can use the product-over-sum r ule.

T

1 2 3 T

11 1 1 1

...

L

L L L L

!

T

1 2

1

1 1 L

L L

!



Summary

Parallel inductors

If a 1.5 mH inductor is connected in

parallel with an 680 QH inductor, the total

inductance is468 QH

L1 L2

1.5 mH 680 HQ



Summary

Inductors in dc circuits

When an inductor is connected in

series with a resistor and dc source,

the current change is exponential.

R

L

t0

Current after switch closure

t0

Inductor voltage after switch closure

V initial

I final



Summary

Inductors in dc circuitsThe same shape curves are seen if

a square wave is used for the

source. Pulse response is covered

f urther in Chapter 20.

V S

V L

V R

R

L

V S

S



Summary


S pecific values for

current and voltage can

be read from a universal

curve. For an R L circuit,

the time constant is

L

R!

100%

80%

60%

40%

20%

00 1X 2X 3X 4X 5X

99%98%

95%

86%

63%

37%

14%

5%2% 1%


P e r c e n t o f f i n a l v a l u

eR ising exponential

Falling exponential

S



In a series R L circuit, when

is V R > 2V L?

Summary

100%

80%

60%

40%

20%

00 1X 2X 3X 4X 5X

99%98%

95%

86%

63%

37%

14%

5%2% 1%


P e r c e n t o f f i n a l v a l u

e

R ead the rising

exponential at the 67%level. After 1.1 X

The curves can give

specific information

about an R L circuit.


S



Summary

The universal curves can be applied to general formulas for the current

(or voltage) curves for R L circuits. The general current formula is

i =I F + ( I i I F)e Rt/ L

I F = final value of current I i = initial value of current

i = instantaneous value of current

The final current is greater than the initial current when the

inductive field is building, or less than the initial current when the field

is collapsing.


S



Summary

Inductive reactance

Inductive reactance is the opposition to ac by

an inductor. The equation for inductive

reactance is

The reactance of a 33 QH inductor when a frequency of

550 kHz is applied is 114 ;

2 L

X f L!

S



Summary

Inductive reactanceWhen inductors are in series, the total reactance is the sum of the

individual reactances. That is,

Assume three 220 QH inductors are in series with a 455 kHz

ac source. What is the total reactance?

1.89 k ;

L( ) L1 L2 L3 Lt ot n X X X X X !

The reactance of each inductor is

2 2 455 k z 220 629 X fLT T ! ! ! ;

L( ) L1 L2 L3

629 629 629

t ot X X X X !

! ; ; ; !

S



Summary

Inductive reactanceWhen inductors are in parallel, the total reactance is the reciprocal of

the sum of the reciprocals of the individual reactances. That is,

If the three 220 QH inductors from the last example are placed

in parallel with the 455 kHz ac source, what is the total

reactance?

210 ;

L( )

L1 L2 L3 L

1

1 1 1 1t ot

n

X

X X X X

!

The reactance of each inductor is 629 ;

L( )

L1 L2 L3

1 1

1 1 1 1 1 1

629 629 629

t ot X

X X X

! ! !

; ; ;

S



Summary

Inductive phase shift

When a sine wave is

applied to an inductor,

there is a phase shift

between voltage and

current such that

voltage always leads the

current by 90o.

V L 0

0

90r

I



True Power: Ideally, inductors do not dissipate power. However, a

small amount of power is dissipated in winding resistance given by

the equation:

P tr ue = ( I rms)2

RW

R eactive Power: R eactive power is a measure of the rate at which

the inductor stores and returns energy. One form of the reactive

power equation is:

P r =V rms I rms

The unit for reactive power is the VAR .

Power in an inductor



The quality factor (Q) of a coil is given by the ratio of reactive

power to tr ue power.

Q of a coil

2

2

L

W

I X

Q I R!

For a series circuit, I cancels, leaving

L

W

X Q R

!

K T



Inductor

Winding

Induced

voltage

Inductance

An electrical device formed by a wire wound around a corehaving the property of inductance; also known as a coil.

The loops or turns of wire in an inductor.

Voltage produced as a result of a changing magnetic

field.

The property of an inductor whereby a change in current

causes the inductor to produce a voltage that opposes the

change in current.

Key Terms

Ke Terms



H enry ( H )

RL timeconstant

Inductivereactance

Quality factor

A fixed time interval set by the L and R values, that

determines the time response of a circuit. It equals the

ratio of L/ R.

The opposition of an inductor to sinusoidal current. The

unit is the ohm.

The unit of inductance.

Key Terms

The ratio of reactive power to tr ue power for an inductor.

Quiz



Quiz

1. Assuming all other factors are the same, the inductance of an

inductor will be larger if

a. more turns are added

b. the area is made larger

c. the length is shorter

d. all of the above

Quiz



Quiz

2. The henry is defined as the inductance of a coil when

a. a constant current of one amp develops one volt.

b. one volt is induced due to a change in current of one amp

per second.

c. one amp is induced due to a change in voltage of one volt.

d. the opposition to current is one ohm.

Quiz



Quiz

3. The symbol for a ferrite core inductor is

a.

b.

c.

d.

Quiz



Quiz

4. The symbol for a variable inductor is

a.

b.

c.

d.

Quiz



Quiz

5. The total inductance of a 270 QH inductor connected in series with a

1.2 mH inductor is

a. 220 QH

b. 271 QH

c. 599 QH

d. 1.47 mH

Quiz



Quiz

6. The total inductance of a 270 QH inductor connected in parallel with

a 1.2 mH inductor is

a. 220 QH

b. 271 QH

c. 599 QH

d. 1.47 mH

Quiz



Quiz

7. When an inductor is connected through a series resistor and switch

to a dc voltage source, the voltage across the resistor after the switch

closes has the shape of

a. a straight line

b. a rising exponential

c. a falling exponential


Quiz



Quiz

8. For circuit shown, the time constant is

a. 270 ns

b. 270Qs

c. 270 ms

d. 3.70 s

RV S

270 Q

1.0 ;10

Quiz



Quiz

9. For circuit shown, assume the period of the square wave is 10 times

longer than the time constant. The shape of the voltage across L is

RV S

La.

b.

c.

d.

Quiz



Quiz

10. If a sine wave from a f unction generator is applied to an inductor,

the current will

a. lag voltage by 90o

b. lag voltage by 45o

c. be in phase with the voltage


Quiz



Quiz

Answers:

1. d

2. b

3. d

4. c

5. d

6. a

7. b8. a

9. c

10. a

Summary



Summary

Sinusoidal response of R L circuitsWhen both resistance and inductance are in a series circuit, the

phase angle between the applied voltage and total current is

between 0r and 90r, depending on the values of resistance and

reactance.

I

LR

V R

V R lags V

S V L leads V S

I lags V S

V S

V L

Summary



Summary

Impedance of series R L circuitsIn a series R L circuit, the total impedance is the phasor sum of R

and X L.

R is plotted along the positive x-axis.

R

X L is plotted along the positive y-axis.

X L

1tan L X

RU ¨ ¸

! © ¹ª º

Z

It is convenient to reposition the

phasors into the impedance t riangle.

R

X L

Z Z is the diagonal

U U

Summary



Summary

Impedance of series R L circuits

R = 1.2 k ;

X L =

960 ;

Sketch the impedance triangle and show the values for R

= 1.2 k ; and X L= 960 ;.

2 21.2 k 0.96 k

1.33 k

Z ! ; ;

! ;

1 0.96 k tan

1.2 k

39

U ;!

;! r

Z = 1.33 k ;

39o

Summary



Summary

Analysis of series R L

circu

itsOhm¶s law is applied to series R L circuits using quantities of

Z , V , and I .

V V

V I Z I Z Z I ! ! !

Because I is the same everywhere in a series circuit, you can

obtain the voltage phasors by simply multiplying the

impedance phasors by the current.

Summary



Summary

Analysis of series R L

circu

its

Assume the current in the previous example is 10 mArms. Sketch the

voltage phasors. The impedance triangle from the previous example is

shown for reference.

V R = 12 V

V L=

9.6 V

The voltage phasors can be found from Ohm¶s law.

Multiply each impedance phasor by 10 mA.

x 10 mA

=

R = 1.2 k ;

X L = 960 ;

Z = 1.33 k ;

39o

V S = 13.3 V

39o U U

Summary



Summary

Variation of phase angle with frequ

encyPhasor diagrams that have reactance phasors can only be drawn

for a single frequency because X is a f unction of frequency.

As frequency changes, theimpedance triangle for an R L

circuit changes as illustrated

here because X L

increases with

increasing f . This determines

the f r equenc y r es ponse of R L

circuits.

U

X L3

Increasing f

Z 1

Z

Z 3

32

1

X

2

X 1

R

UU

Summary



U

(p s l )

J

J

Summary

Phase shift

R

V R

V out

L V out V in

V inV out V in

For a given frequency, a series R L circuit can be used to produce a

phase lead by a specific amount between an input voltage and an

output by taking the output across the inductor. This circuit is also a

basic high-pass filter, a circuit that passes high frequencies and rejects

all others.

Summary



J ( l )J

Summary

R

V L

V out

L

V out V in

V inV out

V in

R eversing the components in the previous circuit produces a circuit

that is a basic lag network. This circuit is also a basic low-pass filter,

a circuit that passes low frequencies and rejects all others.

Phase shift

Summary



Summary

Sinu

soidal response of parallel R L

circu

itsFor parallel circuits, it is usef ul to review conductance,

susceptance and admittance, introduced in Chapter 10.

Conductance is the reciprocal of resistance.1

G R!

Admittance is the reciprocal of impedance.

Inductive susceptance is the reciprocal

of inductive reactance.

1 L

L

B X

!

1Y

Z !

Summary



Summary

Sinu


circu

its

V S

G B L

Y

G

B L

In a parallel R L circuit, the admittance phasor is the sum of the

conductance and inductive susceptance phasors.

The magnitude of the susceptance is2 2

+ L

Y G B!

The magnitude of the phase angle is 1tan L B

GU

¨ ¸! © ¹ª º

Summary



Summary

Sinu


circu

its

Y

G

B L

G is plotted along the positive x-axis.

B L

is plotted along the negative y-axis.

1

tanL

B

GU

¨ ¸!

© ¹ª ºY is the diagonal

V S

G B L

Some important points to notice are:

Summary



Su y

Sinu


circu

its

V S L

25.3 mH

Y =

1.18 mS

G = 1.0 mS

B L=

0.629 mS

Draw the admittance phasor diagram for the circuit.

f = 10 kHz

R

1.0 k ;

1 1

1.0 mS1.0 k G R! ! !; 1

0.629 mS2 10 kHz 25.3 mH L B T ! !

The magnitude of the conductance and susceptance are:

2 22 2 1.0 mS 0.629 mS 1.18 mS L

Y G B! ! !

Summary



y

Analysis of parallel R L

circuits

Ohm¶s law is applied to parallel R L circuits using quantities of

Y , V , and I .

I I

Y V I V Y

V Y

! ! !

Because V is the same across all components in a parallel

circuit, you can obtain the current in a given component by

simply multiplying the admittance of the component by the

voltage as illustrated in the following example.

Summary



y

Analysis of parallel R L

circuits

Assume the voltage in the previous example is 10 V. Sketch the

current phasors. The admittance diagram from the previous

example is shown for reference.

The current phasors can be found from Ohm¶s law.

Multiply each admittance phasor by 10 V.

x 10 V

=

I R = 10 mA

I L=

6.29 mAI S =

11.8 mA

Y =

1.18 mS

G = 1.0 mS

B L=

0.629 mS

Summary



y

Phase angle of parallel R

Lcirc

uits

Notice that the formula for inductive susceptance is the

reciprocal of inductive reactance. Thus B L

and I L

are inversely

proportional to f : 1

2 L

B

f LT

!

I R

I L

I S

UAs frequency increases, B L

and I L

decrease, so the angle between I Rand I S must decrease as well.

Summary



y

Series-

Parallel R

Lcirc

uits

Series-parallel R L circuits are combinations of both series and

parallel elements. The solution of these circuits is similar to

resistive combinational circuits but you need to combine reactive

elements using phasors.

The components in theyellow box are in series and

those in the green box are

also in series.

R1

L1

R2

L2

Z 1 Z 2

The two boxes are in parallel. You canfind the branch currents by applying

Ohm¶s law to the source voltage and

the branch impedance.

and

2 2

1 1 1 L Z R X !

2 2

2 2 2 L Z R X !

Summary



y

The power triangle

x 10 mA

=

R ecall that in a series RC or R L circuit, you could multiply the

impedance phasors by the current to obtain the voltage phasors.

The earlier example from this chapter is shown for review:

R = 1.2 k ;

X L =

960 ;

Z = 1.33 k ;

3

9

o

V R = 12 V

V L=

9.6 V

V S = 13.3 V

39o

Summary



y

Multiplying the voltage phasors by I rm s gives the power triangle

(equivalent to multiplying the impedance phasors by I 2). Apparent

power is the product of the magnitude of the current and magnitude of

the voltage and is plotted along the hypotenuse of the power triangle.

The rms current in the earlier example was 10 mA.Show the power triangle.

x 10 mA =

P t rue = 120 mW

P r =

96 mVAR

The power triangle

39o

P a = 133 mVA

V R = 12 V

V L=

9.6 V

V S = 13.3 V

39o

Summary



y

Power factor The power factor was discussed in Chapter 15 and applies to

R L circuits as well as RC circuits. R ecall that it is the

relationship between the apparent power in volt-amperes and

tr ue power in watts. Volt-amperes multiplied by the power

factor equals tr ue power.

Power factor is defined as

PF = cos U

Summary



y

Apparent power Apparent power consists of two components; a tr ue power

component, that does the work, and a reactive power

component, that is simply power shuttled back and forth

between source and load.

P t rue (W)

P r (VAR)

P a (VA)

Power factor corrections for

an inductive load (motors,

generators, etc.) are done by

adding a parallel capacitor,

which has a canceling effect.

Summary



y

Frequency R esponse of R L Circuits

Series R L circuits have a frequency response similar to series RC

circuits. In the case of the low-pass response shown here, the output is

taken across the resistor.

100 ;10

10 dc

V ou

! V i n

10 dc

0

10 V dc

0


100 ; = 1 kHz

8.46 rms10 rms10mH

100 ;

10mH

= 10 kHz

1.57 rms

10 rms

100 ;

10mH

= 0 kHz

0.79 rms

10 rms

V ou

"

(V)

9.98

8.46

1.570.79

0.1 1 10 20 100f (kHz)

9

87

6

5

4

3

2

1

Summary



Frequency R esponse of R L Circuits

R eversing the position of the R and L components, produces the

high-pass response. The output is taken across the inductor.

V i #

10V dc

0

V $ % t

0V dc10V dc

100 ; 10


V out (V)

f (kHz)

9.87

5.32

0.6300.01 0.1 1

10

9

87

6

5

&

3

2

1

10

= 100 Hz0.63 V r s

10 V r s100 10 H = 1 kHz

5.32 V r s10 V r s

100 ; 10 mH = 10 kH

9.87 V r ms10 V r ms

100 ;10mH

Key Terms



Inductive

susceptance (B L )

The ability of an inductor to permit current; the

reciprocal of inductive reactance. The unit is the

siemens (S).

Quiz



1. If the frequency is increased in a series R L circuit, the phase angle

will

a. increase

b. decrease

c. be unchanged

Quiz



2. If you multiply each of the impedance phasors in a series R L circuit

by the current, the result is the

a. voltage phasors

b. power phasors

c. admittance phasors


Quiz



3. For the circuit shown, the output voltage

a. is in phase with the input voltage

b. leads the input voltage

c. lags the input voltage

d. none of the above V in V out

Quiz



4. In a series R L circuit, the phase angle can be found from the equation

a.

b.

c. both of the above are correct

d. none of the above is correct

1tan L X

RU ¨ ¸! © ¹

ª º

1

tan

L

R

V

V U

¨ ¸

!

© ¹ª º

Quiz



5. In a series R L circuit, if the inductive reactance is equal to the

resistance, the source current will lag the source voltage by

a. 0o

b. 30o

c. 45o

d. 90o

Quiz



6. Susceptance is the reciprocal of

a. resistance

b. reactance

c. admittance

d. impedance

Quiz



7. In a parallel R L circuit, the magnitude of the admittance can be

expressed as

a.

b.

c. Y = G + B L

d. 2 2+ L

Y G B!

2 2 L

Y G B!

1

1 1

L

Y

G B

!

Quiz



8. If you increase the frequency in a parallel R L circuit,

a. the total admittance will increase

b. the total current will increase

c. both a and b


Quiz



9. The unit used for measuring tr ue power is the

a. volt-ampere

b. watt

c. volt-ampere-reactive (VAR)

d. kilowatt-hour

Quiz



10. A power factor of zero implies that the

a. circuit is entirely reactive

b. reactive and tr ue power are equal

c. circuit is entirely resistive

d. maximum power is delivered to the load

Quiz



Answers:

1. a

2. a

3. c

4. a

5. c

6. b

7. d

8. d

9. b

10. a

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