26771199 matriculation physics alternating current

7/29/2019 26771199 Matriculation Physics Alternating Current

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PHYSICS CHAPTER 8

CHAPTER 8:CHAPTER 8:

Alternating currentAlternating current

(6 Hours)(6 Hours)


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PHYSICS CHAPTER 8

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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:

DefineDefine alternating current (AC).alternating current (AC).

Sketch and useSketch and use sinusoidal AC waveform.sinusoidal AC waveform.

Write and useWrite and use sinusoidal voltage and current equations.sinusoidal voltage and current equations.

Learning Outcome:

8.1 Alternating current (1 hour)

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PHYSICS CHAPTER 8

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is defined as an electric current whose magnitude andan electric current whose magnitude and

direction change periodically.direction change periodically.

Figures 8.1a, 8.1b and 8.1c show three forms of alternating

current.

8.1 Alternating current (AC)

Figure 8.1a: sinusoidal ACFigure 8.1a: sinusoidal AC

I

t0

TT2

1 T2T2

3

0I

0I


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PHYSICS CHAPTER 8

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I

t0 TT2

1 T2T

2

3

0I

0I

TT2

1 T2T

2

3

Figure 8.1b: saw-tooth ACFigure 8.1b: saw-tooth AC

Figure 8.1c: square ACFigure 8.1c: square AC

0I

0I

I

t0


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PHYSICS CHAPTER 8

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When an AC flows through a resistorAC flows through a resistor, there will be a

potential difference (voltage)potential difference (voltage) across it and this voltage is

alternatingalternating as shown in Figure 8.1d.

V

t0 TT2

1 T2T

2

3

0V

0V

( ) voltagemaximumpeak:0Vwhere

period:T

( ) currentmaximumpeak:0I

Figure 8.1d: sinusoidal alternating voltageFigure 8.1d: sinusoidal alternating voltage


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Frequency (Frequency (ff)) is defined as a number of complete cycle in one seconda number of complete cycle in one second.

Its unit is hertz (hertz (HzHz)) OR ss 11.

Period (Period (TT)) is defined as a time taken for one complete cyclea time taken for one complete cycle.

Its unit is second (second (ss)). Formulae,

Peak current (Peak current (II00))

is defined as a magnitude of the maximum currenta magnitude of the maximum current.

Its unit is ampere (ampere (AA)).

8.1.1 Terminology in AC

fT

1= (8.1)(8.1)


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Equation for alternating current (I),

Equation for alternating voltage (V),

8.1.2 Equations of alternating current and voltage

tII sin0= (8.2)(8.2)

tVV sin0= (8.3)(8.3)

phasephase

where locityangular veORfrequencyangular:

currentpeak:0I

gepeak volta:0V

time:t

)2( f=


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DefineDefine root mean square (rms) current and voltage forroot mean square (rms) current and voltage for

AC source.AC source.

UseUse the following formula,the following formula,

Learning Outcome:

8.2 Root mean square (rms) (1 hour)

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2

0rms

II = andand

2

0rms

VV =


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PHYSICS CHAPTER 8

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8.2.1 Mean or Average Current (Iav) is defined as the average or mean value of current in athe average or mean value of current in a

half-cycle flows of current in a certain direction.half-cycle flows of current in a certain direction.

Formulae:

8.2 Root mean square (rms)

( )2

00av

2

III == (8.4)(8.4)

Note:Note:

Iavforone complete cycle is zeroone complete cycle is zero because the currentcurrentflows in one direction in one-half of the cycleflows in one direction in one-half of the cycle and in the

opposite direction in the next half of the cycleopposite direction in the next half of the cycle.


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In calculating average power dissipated by an AC, the mean(average) current is not useful.

The instantaneous power,instantaneous power,PPdelivered to a resistanceRis

The average power,average power,PPavav

over one cycle of ACover one cycle of AC is given by

where is the average value ofthe average value ofII22 over one cycleover one cycle and is

given by

Therefore

8.2.2 Root mean square current (Irms

)

RIP 2=

RIP 2av =2I

( ) 2rms2 II = (8.5)(8.5)

RIP2

rmsav = (8.6)(8.6)

where ACousinstantane:I


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Since

and the graph ofI2 against time, tis shown in Figure 8.2.

From Figure 8.2, the shaded region underunderthe curve and aboveabove

the dashed line forI0

2/2 have the same are as the shaded region

aboveabove the curve and belowbelow the dashed line forI0

2/2.

Thus

thus the square value of current is given bytII sin0=tII 2

2

02 sin=

2

0I

TT

2

1 T2T

2

3

2

2

0I

t0

2I

Figure 8.2Figure 8.2

2

2

02 I

I = (8.7)(8.7)


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By equating the eqs. (8.5) and (8.7), the rms current is

Root mean square current (Root mean square current (IIrmsrms)) is defined as the value of thethe value of thesteady DC which produces the same power in a resistor assteady DC which produces the same power in a resistor as

the mean (average) power produced by the AC.the mean (average) power produced by the AC.

The root mean square (rms) current is the effective valueeffective value of theAC and can be illustrated as shown in Figure 8.3.

( ) 2

2

02

rms

I

I = 2

2

0

rms

I

I =

2

0rms

II = (8.8)(8.8)

I

t0 TT2

1 T2T

2

3

0I

0I

rmsI 0707.0 I



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is defined as the value of the steady direct voltage whichthe value of the steady direct voltage whichwhen applied across a resistor, produces the same powerwhen applied across a resistor, produces the same power

as the mean (average) power produced by the alternatingas the mean (average) power produced by the alternating

voltage across the same resistor.voltage across the same resistor.

Its formula is

The unit of the rms voltage (potential difference) is volt (V)volt (V).

8.2.3 Root mean square voltage (Vrms

)

20

rms VV = (8.9)(8.9)

Note:Note:

Equations (8.8) and (8.9) are valid only for a sinusoidalvalid only for a sinusoidal

alternating current and voltagealternating current and voltage.


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An AC source V=500sin tis connected across a resistor of

250 . Calculate

a. the rms current in the resistor,

b. the peak current,

c. the mean power.

Solution :Solution :By comparing

Thus the peak voltage is

a. By applying the formulae of rms current, thus

Example 1 :

= 250R tV sin500= to the tVV sin0=V5000 =V

2

0

rms

I

I = and RV

I0

0 =

2

0rms

R

VI =

2250

500

= A41.1rms =I


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Solution :Solution :

b. The peak current of AC is given by

c. The mean (average) power of the resistor is

= 250R

2

0rms

II =

241.1 0

I=

A99.10 =I

RIP2

rmsav =

( ) ( )25041.12

=W497av =P


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Figure 8.4 shows a graph to represent alternating current passes

through a resistor of 10 k . Calculatea. the rms current,

b. the frequency of the AC,

c. the mean power dissipated from the resistor.

Example 2 :

and

40

)A(I

)ms(t0 20 80

02.0

02.0

60



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From the graph,

a. By applying the formulae of rms current, thus

b. The frequency of the AC is

c. The mean power dissipated from the resistor is given by

= 1010 3Rs1040A;02.0 30

== TI

Tf

1=

31040

1

=f

Hz25=f

2

0rms

II =

2

02.0rms =I

A1041.12

rms=I

RIP2

rmsav =

( ) ( )322 10101041.1 =

W99.1av =P


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UseUse phasor diagram and sinusoidal waveform to showphasor diagram and sinusoidal waveform to show

the phase relationship between current and voltage for athe phase relationship between current and voltage for a

circuit consisting ofcircuit consisting of pure resistorpure resistor

pure capacitorpure capacitor

pure inductor.pure inductor.

DefineDefine capacitive reactance, inductive reactance andcapacitive reactance, inductive reactance and

impedance.impedance. AnalyseAnalyse voltage, current and phasor diagrams for avoltage, current and phasor diagrams for a

series circuit consisting ofseries circuit consisting of

RCRC

RLRL

RCLRCL..

Learning Outcome:

8.3 Resistance, reactance and impedance (2 hours)

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8.3.1 Phasor diagram PhasorPhasoris defined as a vector that rotate anticlockwise abouta vector that rotate anticlockwise about

its axis with constant angular velocity.its axis with constant angular velocity.

A diagram containing phasor is called phasor diagramphasor diagram.

It is used to represent a sinusoidally varying quantityrepresent a sinusoidally varying quantity suchas alternating current (AC) and alternating voltage.

It also being used to determine the phase anglephase angle (is defined as

the phase difference between current and voltage in ACthe phase difference between current and voltage in AC

circuitcircuit).

Consider a graph represents sinusoidal AC and sinusoidalalternating voltage waveform as shown in Figure 8.5a.

Meanwhile Figure 8.5b shows the phasor diagram ofVandI.

8.3 Resistance, reactance and impedance


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From the Figure 8.5a:

Thus the phase difference is

Therefore the currentIis in phasein phase with the voltage Vandconstant with time.

t0

0I

0V

0

I0V

TT2

1 T2T

2

3

Figure 8.5aFigure 8.5aFigure 8.5b: phasor diagramFigure 8.5b: phasor diagram

VI

tII sin0= tVV sin0=0== tt

and

Note:Note:

valuepositive=

radian =valuenegative=

LeadsLeads

Lags behindLags behind

In antiphaseIn antiphase


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The quantity that measures the opposition of a circuit to themeasures the opposition of a circuit to the

AC flowsAC flows.

It is defined by

It is a scalar quantityscalar quantity and its unit is ohm (ohm ( )).

In a DC circuit, impedance likes the resistanceimpedance likes the resistance.

8.3.2 Impedance (Z)

rms

rms

I

VZ= (8.10)(8.10)

2

0V

2

0I

OR

0

0

I

VZ= (8.11)(8.11)


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The symbol of an AC source in the electrical circuit is shown inFigure 8.6.

Pure resistor means that no capacitance and self-inductanceno capacitance and self-inductanceeffect in the AC circuit.

Phase difference between voltagePhase difference between voltage VVand currentand currentII

Figure 8.7 shows an AC source connected to a pure resistorR.

8.3.3 Pure resistor in an AC circuit


AC source

R

I

RV

VFigure 8.7Figure 8.7


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The alternating current passes through the resistor is given by

The alternating voltage across the resistorVR at any instant isgiven by

Therefore the phase difference between VandIis

In pure resistor, the currentcurrent IIalways in phase with the voltagealways in phase with the voltage

VVand constant with timeand constant with time.

Figure 8.8a shows the variation ofVandIwith time while Figure

8.8b shows the phasor diagram forVandIin a pure resistor.

tII sin0=

IRVR =00 VRI =( )RtI sin0= and

VtVVR == sin0where tagesupply vol:V

0== tt


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Impedance in a pure resistorImpedance in a pure resistor From the definition of the impedance, hence

t0

0I

0V

0

I0V

TT2

1 T2T

2

3


VI

RI

V

I

VZ ===

0

0

rms

rms(8.12)(8.12)


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Pure capacitor means thatno resistance and self-inductanceno resistance and self-inductance

effect in the AC circuit.

Phase difference between voltagePhase difference between voltage VVand currentand currentII

Figure 8.9 shows an AC source connected to a pure capacitorC.

The alternating voltage across the capacitorVCat any instant is

equal to the supply voltage Vand is given by

8.3.4 Pure capacitor in an AC circuit


AC sourceAC source

CV

V

C

I

tVVVC sin0==


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The charge accumulates at the plates of the capacitor is

The charge and current are related by

Hence the equation of AC in the capacitor is

CCVQ =

tCVQ sin0=

dt

dQI=

( )tCVdtdI sin0=

( )tdt

dCV sin0=

tCV cos0= 00 ICV =andtII cos0=

OR

+=2

sin0

tII


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PHYSICS CHAPTER 8

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In the pure capacitor,

the voltage VVlags behindlags behind the currentII by //22 radiansradians.

OR

the currentIIleadsleads the voltage VV by //22 radiansradians.

Figure 8.10a shows the variation ofVandIwith time while

Figure 8.10b shows the phasor diagram forVandIin a purecapacitor.

+= 2

tt

rad2

=


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Impedance in a pure capacitorImpedance in a pure capacitor

From the definition of the impedance, hence


0

0

I

VZ=

t0

0I

0V

0I0

V

TT

2

1 T2T

2

3

VI

rad2

=

00 CVI =and

0

0

CV

V

=


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PHYSICS CHAPTER 8

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whereXC is known as capacitive (capacitative) reactancecapacitive (capacitative) reactance.

Capacitive reactance is the opposition of a capacitor to thethe opposition of a capacitor to the

alternating current flowsalternating current flows and is defined by

Capacitive reactance is a scalar quantityscalar quantity and its unit is ohmohm

(( )) .

CXC

Z ==

1

fCXC

2

1=

f 2=and

(8.13)(8.13)

sourceACoffrequency:fcapacitortheofecapacitanc:C

0

0

rms

rms

IV

IVXC == (8.14)(8.14)


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From the eq. (8.13), the relationship between capacitive

reactanceXCand frequencyfcan be shown by using a graph in

Figure 8.11.

f0

CX

fXC

1


Pure inductor means that no resistance and capacitanceno resistance and capacitance

effect in the AC circuit.Phase difference between voltagePhase difference between voltage VVand currentand currentII Figure 8.12 shows an AC source connected to a pure inductor

L.

8.3.5 Pure inductor in an AC circuit


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The alternating current passes through the inductor is given by

When the AC passes through the inductor, the back emf caused

by the self induction is produced and is given by

AC sourceV

I

L

LV


tII sin0=

dtdIL=B

( )tIdt

dL sin0=

tLI cos0B = (8.15)(8.15)


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At any instant, the supply voltage Vequals to the back emfB

in the inductor but the back emf always oppose the supply

voltage Vrepresents by the negative sign in the eq. (8.15).Thus


In the pure inductor,

the voltage VVleadsleads the currentII by //22 radiansradians.OR

the currentIIlags behindlags behind the voltage VV by //22 radiansradians.

B=VtLI cos0= 00 VLI =and

tVV cos0=OR

+=

2sin0 tVV

rad22

=

+= tt


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Figure 8.13a shows the variation ofVandIwith time whileFigure 8.13b shows the phasor diagram forVandIin a pure

inductor.

t0

0I

0

V

0I0V

TT2

1 T2T

2

3

V

I

rad2

=



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Impedance in a pure inductorImpedance in a pure inductor

From the definition of the impedance, hence

whereXL is known as inductive reactanceinductive reactance.

0

0

IVZ= 00 LIV =and

0

0

I

LI =

LXLZ ==

fLXL 2=

f 2=and

(8.16)(8.16)

inductortheofinductance-self:LsourceACoffrequency:f


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PHYSICS CHAPTER 8

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Inductive reactance is the opposition of a inductor to thethe opposition of a inductor to the

alternating current flowsalternating current flows and is defined by

Inductive reactance is a scalar quantityscalar quantity and its unit is ohmohm

(( )).

From the eq. (8.16), the relationship between inductive

reactanceXL

and the frequencyfcan be shown by using a

graph in Figure 8.14.


f0

LX

fXL

0

0

rms

rms

I

V

I

VXL == (8.17)(8.17)


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A capacitor has a rms current of 21 mA at a frequency of 60 Hz

when the rms voltage across it is 14 V.a. What is the capacitance of the capacitor?

b. If the frequency is increased, will the current in the capacitor

increase, decrease or stay the same? Explain.

c. Calculate the rms current in the capacitor at a frequency of

410 Hz.


a. The capacitive reactance of the capacitor is given by

Therefore the capacitance of the capacitor is

Example 3 :

V14Hz;60A;1021 rms3

rms === VfI

CXIV rmsrms =

fCXC

2

1=

( ) CX3102114 == 667CX

( )C6021

667

=

F1098.36

=C

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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b. The capacitive reactance is inversely proportional to thecapacitive reactance is inversely proportional to the

frequencyfrequency, so the capacitive reactance will decreasecapacitive reactance will decrease if the

frequency increasesfrequency increases. Since the current in the capacitor iscurrent in the capacitor is

inversely proportional to the capacitive reactanceinversely proportional to the capacitive reactance, therefore

the current will increase when the capacitive reactancecurrent will increase when the capacitive reactancedecreasesdecreases.

c. Given

The capacitive reactance is

Hence the new rms current in the capacitor is given by

V14Hz;60A;1021 rms3

rms === VfI

Hz410=f

fC

XC

2

1=

( ) ( )6

1098.34102

1

=

CX

= 5.97CX

CXIV rmsrms = ( )5.9714 rmsI=

A144.0rms =

I

PHYSICS CHAPTER 8


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A rms voltage of 12.2 V with a frequency of 1.00 kHz is applied to a

0.290 mH inductor.a. What is the rms current in the circuit?

b. Determine the peak current for a frequency of 2.50 kHz.


a. The inductive reactance of the inductor is given by

Thus the rms current in the circuit is

Example 4 :

H10290.0Hz;1000.1V;2.12 33rms=== LfV

fLXL 2=( )( )33 10290.01000.12 = = 82.1LX

LXIV rmsrms =( )82.12.12 rmsI=A70.6rms =I

PHYSICS CHAPTER 8


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b. Given

The inductive reactance of the inductor is given by

Thus the peak current in the circuit is

H10290.0Hz;1000.1V;2.12 33rms=== LfV

Hz1050.2 3=f

fLXL 2=( )( )33 10290.01050.22 = = 56.4LX

LXIV 00 =

( )56.422.12 0I=A78.30 =I

and 2rms0 VV =

LXIV 0rms 2 =

PHYSICS CHAPTER 8


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RCRCseries circuitseries circuit Consider an AC source ofrms voltagerms voltage VVis connected in series

to a resistorRand a capacitorCas shown in Figure 8.15a.

The rms currentrms currentIIpasses through the resistor and theresistor and thecapacitor is equalcapacitor is equal because of the series connectionseries connection between

both components.

8.3.5 RC,RL andRCL series circuit

AC sourceAC source

R

I

RV

V

CV

C

Figure 8.15aFigure 8.15a

PHYSICS CHAPTER 8


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I

The rms voltagesrms voltages across the resistorresistorVVRR

and the capacitorcapacitorVVCC

are given by

The phasor diagram of theRCseries circuit is shown in Figure8.15b.

Based on the phasor diagram, the rms supply voltage V(or totalvoltage) of the circuit is given by

IRVR =

andCC IXV =

where anglephase:

CV

RV

V

Figure 8.15b: phasor diagramFigure 8.15b: phasor diagram

is an angle between the rmsan angle between the rmscurrentcurrentIIand rms supply (orand rms supply (or

total) voltagetotal) voltage VVof AC circuit.

22

CR VVV += ( ) ( )22

CIXIRV +=22

C

XRIV += (8.18)(8.18)

PHYSICS CHAPTER 8


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PHYSICS CHAPTER 8

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Rearrange the eq. (8.18), thus the impedance ofRCseriescircuit is

From the phasor diagram in Figure 8.15b , the currentcurrentIIleadsleads

the supply voltagethe supply voltage VVbyby radiansradians where

A phasor diagram in terms ofR,XC andZis illustrated in Figure8.15c.

I

VZ=and

22

CXRI

V+=

22

CXRZ += (8.19)(8.19)

R

C

VV=tan IR

IXC=tan

R

XC=tan (8.20)(8.20)

CX Z

R

Figure 8.15cFigure 8.15c

PHYSICS CHAPTER 8


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RLRL series circuitseries circuit

Consider an AC source ofrms voltagerms voltage VVis connected in series

to a resistorRand an inductorL as shown in Figure 8.16a.

The rms voltagesrms voltages across the resistorresistorVVRR and the inductorinductorVVLLare given by

AC sourceAC source

R

I

RV

V

L

LV


IRVR =and

LL IXV =

PHYSICS CHAPTER 8


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The phasor diagram of theRL series circuit is shown in Figure8.16b.


LVV

I


RV

22

LR VVV +=

( ) ( ) 22 LIXIR +=

22

LXRIV += (8.21)(8.21)

PHYSICS CHAPTER 8


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Rearrange the eq. (8.21), thus the impedance ofRL seriescircuit is

From the phasor diagram in Figure 8.16b , the supply voltagesupply voltage

VVleads the currentleads the currentIIthe bythe by radiansradians where

The phasor diagram in terms ofR,XL andZis illustrated inFigure 8.16c.

I

VZ=and

22

LXR

I

V+=

22

LXRZ += (8.22)(8.22)

R

L

VV=tan

IRIXL=tan

R

XL=tan (8.23)(8.23)


LXZ

R

PHYSICS CHAPTER 8


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RCLRCL series circuitseries circuit

Consider an AC source ofrms voltagerms voltage VVis connected in series

to a resistorR, a capacitorCand an inductorL as shown inFigure 8.17a.

The rms voltagesrms voltages across the resistorresistorVVRR, the capacitorcapacitorVVCC and

the inductorinductorVVLLare given by


IRVR =

and

CC IXV =

AC sourceAC source

I

V

R

RV CV

C L

LV

LL IXV =

PHYSICS CHAPTER 8


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The phasor diagram of theRL series circuit is shown in Figure8.17b.


I


( )22

CLR VVVV +=( ) ( ) 22

CL IXIXIR +=

( )

22

CL XXRIV +=(8.24)(8.24)

LV

RV

V

CV

( )CL VV

PHYSICS CHAPTER 8


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Rearrange the eq. (8.24), thus the impedance ofRL seriescircuit is

From the phasor diagram in Figure 8.17b , the supply voltagesupply voltage

VVleads the currentleads the currentIIthe bythe by radiansradians where

IVZ=and( ) 22 CL XXRI

V

+=

( ) 22 CL XXRZ += (8.25)(8.25)

R

CL

V

VV =tan

( )IR

IXIX CL =

R

XX CL =tan (8.26)(8.26)

PHYSICS CHAPTER 8


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The phasor diagram in terms ofR,XC,X

LandZis illustrated in

Figure 8.17c.


LX

Z

CX

( )CL XX

R

PHYSICS CHAPTER 8


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is defined as the phenomenon that occurs when thephenomenon that occurs when the

frequency of the applied voltage is equal to the frequencyfrequency of the applied voltage is equal to the frequency

of theof theRCLRCL series circuitseries circuit.

Figure 8.18 shows the variation ofXC,X

L,R andZwith

frequencyfof theRCL series circuit.

8.3.6 Resonance in AC circuit

Z

fXL

R

fXC

1

0 f

ZRXX LC ,,,

rfFigure 8.18Figure 8.18

PHYSICS CHAPTER 8


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From Figure 8.18, the value of impedance is minimumis minimumZZminmin

when

where its value is given by

This phenomenon occurs at the frequencyffrrknown asknown as

resonant frequencyresonant frequency.

At resonanceresonance in theRCL series circuit, the impedance isisminimumminimumZZ

minminthus the rms current flowsrms current flows in the circuit is

maximummaximumIImaxmax

and is given by

CL XX = (8.27)(8.27)

( ) 22 CL XXRZ +=

02

min += RZRZ =min

R

V

Z

VI ==

min

max (8.28)(8.28)

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maxI

Figure 8.19 shows the rms currentIinRCL series circuit varies withfrequency.

At frequencies above or below the resonant frequencyfrequencies above or below the resonant frequencyffrr, the rmsrms

currentcurrentIIis less than the rms maximum currentless than the rms maximum currentIImaxmax

as shown in

Figure 8.19.

0 f

I

rf


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The resonant frequency,frof theRCL series circuit is given by

The series resonance circuit is used fortuning a radiotuning a radio

receiverreceiver.

CL XX =

CL

1=

LC

12 = r2 f=and

( )LC

f 12 2r =

LCf

2

1r = (8.29)(8.29)

where frequencyangularresonant:

Note:Note:

At resonance, the currentresonance, the currentIIand voltageand voltage VVare in phaseare in phase.

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A 2 F capacitor and a 1000 resistor are placed in series with

an alternating voltage source of 12 V and frequency of 50 Hz.Calculate

a. the current flowing,

b. the voltage across the capacitor,

c. the phase angle of the circuit.


a. The capacitive reactance of the inductor is given by

and the impedance of the circuit is

Example 5 :

Hz50V;12;1000F;102 6 ==== fVRC

fCXC

2

1=

( ) ( )61025021

=

CX

= 1592CX22

CXRZ += ( ) ( )22

15921000 +=Z= 1880Z

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a. Therefore the current flowing in the circuit is

b. The voltage across the capacitor is given by

c. The phase angle between the current and supply voltage is

( )188012 I=A1038.6 3=I

CC IXV =

( )( )15921038.6 3=V2.10=CV

Hz50V;12;1000F;102 6 ==== fVRC

IZV =

R

XC

=tan

= 1000

1592tan

1

rad01.1=

=

R

XC1

tan

OR

9.57

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Based on theRCL series circuit in Figure 8.20 , the rms voltages across

R,L and Care shown.a. With the aid of the phasor diagram, determine the applied voltage

and the phase angle of the circuit.

Calculate:

b. the current flows in the circuit if the resistance of the resistorR is

26 ,

c. the inductance and capacitance if the frequency of the AC source

is 50 Hz,

d. the resonant frequency.

Example 6 :CR L

V314V153 V115

I


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a. The phasor diagram of the circuit is

and the phase angle is

V314;V115V;153 === LCR VVV

LV

IRV

V

CV

( )CL VV

From the phasor diagram,

the applied voltage Vis

( ) 22 CLR VVVV +=

( ) ( ) 22 115314153 +=V251=V

R

CL

V

VV

=tan

=

153

115314tan 1

rad915.0=

=

R

CL

V

VV1

tan

OR

4.52

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b. Given

SinceR, CandL are connected in series, hence the current

passes through each devices is the same. Therefore

c. Given

The inductive reactance is

thus the inductance of the inductor is

V314;V115V;153 === LCR VVV

IRVR =

( ) LX88.5314 =

A88.5=I( )26153 I=

= 26R

LL IXV =

Hz50=f

= 4.53LX

( )L5024.53 =fLXL 2=

H170.0=L

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PHYSICS CHAPTER 8

59


c. Meanwhile, the capacitive reactance is

thus the capacitance of the capacitor is

d. The resonant frequency is given by

V314;V115V;153 === LCR VVV

( ) CX88.5115 =CC IXV = = 6.19CX

fC

XC2

1=

F1062.1 4=C ( )C502

16.19

=

( ) ( )41062.1170.021

=

LC

f

2

1r =

Hz3.30r =

f

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Exercise 8.1 :

1. An AC current of angular frequency of 1.0 104 rad s1 flows

through a 10 k resistor and a 0.10 F capacitor which areconnected in series. Calculate the rms voltage across the

capacitor if the rms voltage across the resistor is 20 V.

ANS. :ANS. : 2.0 V2.0 V

2. A 200 resistor, a 0.75 H inductor and a capacitor of

capacitance Care connected in series to an alternatingsource 250 V, 600 Hz. Calculate

a. the inductive reactance and capacitive reactance when

resonance is occurred.

b. the capacitance C.c. the impedance of the circuit at resonance.

d. the current flows through the circuit at resonance. Sketch

the phasor diagram of the circuit.

ANS. :ANS. : 2.83 k2.83 k , 2.83 k, 2.83 k ; 93.8 nF; 200; 93.8 nF; 200 ; 1.25 A; 1.25 A

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Exercise 8.1 :

3. A capacitor of capacitance C, a coil of inductanceL, a resistor

of resistanceR and a lamp of negligible resistance are placed

in series with alternating voltage V. Its frequencyfis varied

from a low to a high value while the magnitude ofVis keptconstant.

a. Describe and explain how the brightness of the lampvaries.

b. If V=0.01 V, C=0.4 F,L =0.4 H,R = 10 and the

circuit at resonance, calculate

i. the resonant frequency,

ii. the maximum rms current,

iii. the voltage across the capacitor.

(Advanced Level Physics,7(Advanced Level Physics,7thth edition, Nelkon & Parker, Q2, p.423)edition, Nelkon & Parker, Q2, p.423)

ANS. :ANS. : 400 Hz; 0.001 A; 1 V400 Hz; 0.001 A; 1 V

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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: ApplyApply

average power,average power,

instantaneous power,instantaneous power,

power factor,power factor,

in AC circuit consisting ofin AC circuit consisting ofRR

,,RCRC

,,RLRL

andandRCLRCL

ininseries.series.

Learning Outcome:

8.4 Power and power factor (1 hour)

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cosav IVP =

dt

dWP=

IV

P

P

P av

a

rcos ==

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8.4.1 Power of a pure resistor In a pure resistor, the voltagevoltage VVand currentand currentIIare in phaseare in phase,

thus the instantaneous powerinstantaneous powerPis given by

Figure 8.21 shows a graph of instantaneous powerPbeingabsorbed by the resistor against time t.

8.4 Power and power factor

( ) ( )tVtI sinsin 00=

IVP=

tVI 200 sin= 000 PVI =and

tPP 20 sin= (8.30)(8.30)

where powerum)peak(maxim:0P

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64

The average (or mean) powerPav

being absorbed by the resistor

is given by

tPP 20 sin=Power being absorbedPower being absorbed


avP

tPP 20av sin=

000av2

1

2

1VIPP ==

0P

2

0P

t0

P

TT2

1 T2T

2

3

(8.31)(8.31)

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In a pure capacitor, the currentcurrentII leads the voltageleads the voltage VVbyby //22

radiansradians, thus the instantaneous powerinstantaneous powerPis given by

Figure 8.22 shows a graph of instantaneous powerPof the

pure capacitor against time t.

8.4.2 Power of a pure capacitor

( ) ( )tVtI sincos 00=IVP=

ttVI cossin00=tPP 2sin

2

10

= (8.32)(8.32)ttt 2sin2

1

cossin =and

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66


of the pure capacitor is given

by

tPP 2sin

2

10=

Power being absorbedPower being absorbed


avP

tPP 2sin2

10av =

0av =P

2

0P

t0

P

TT2

1 T2T

2

3

2

0P

Power being returned to supplyPower being returned to supply

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67

In a pure inductor, the voltagevoltage VV leads the currentleads the currentIIbyby //22

radiansradians, thus the instantaneous powerinstantaneous powerPis given by

Figure 8.23 shows a graph of instantaneous powerPof thepure inductor against time t.

8.4.3 Power of a pure inductor

( ) ( )tVtI cossin 00=IVP=

ttVI cossin00=tPP 2sin

2

10

=

ttt 2sin2

1

cossin =and

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PHYSICS CHAPTER 8

68


of the pure inductor is given

by

tPP 2sin

2

10=

Power being absorbedPower being absorbed


avP

tPP 2sin2

10av = 0av =P

2

0P

t0

P

TT2

1 T2T

2

3

Power being returned to supplyPower being returned to supply

Note:Note:

The term resistance is not used in pure capacitor and inductor because

no heat is dissipated from both devicesno heat is dissipated from both devices.

2

0P

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In an AC circuit in which there is a resistorR, an inductorL anda capacitorC, the average powerP

avis equal to that dissipated

from the resistor i.e.

From the phasor diagram of theRCL series circuit as shown inFigure 8.24,

8.4.4 Power and power factor ofR,RC,RL and

RCL series circuits

RIIVP R2

av == (8.33)(8.33)

rms valuesrms values

LV

IRV

V

C

V

( )CL VV


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70

We get

then the eq. (8.33 ) can be written as

where coscos is called the power factorpower factorof the AC circuit,PPrr

is

the average real poweraverage real powerandII22ZZis called the apparent powerapparent power.

Power factor is defined as

cosVVR =V

VR=cos

cosav IVP = IZV =and

r2

av cos PZIP == (8.34)(8.34)

a

r

2

rcos

P

P

ZI

P== (8.35)(8.35)

where IVZIP == 2a powerapparent:Note:Note:From the Figure 8.24, the power factor also can be calculated by using theequation below:

IZ

IR

V

VR ==cosZ

R=cos (8.36)(8.36)

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71

A 100 F capacitor, a 4.0 H inductor and a 35 resistor are

connected in series with an alternating source given by theequation below:

Calculate:

a. the frequency of the source,

b. the capacitive reactance and inductive reactance,

c. the impedance of the circuit,

d. the peak current in the circuit,

e. the phase angle,

f. the power factor of the circuit.

Example 7 :

tV 100sin520=

PHYSICS CHAPTER 8


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By comparing

Thusa. The frequency of AC source is given by

b. The capacitive reactance is

and the inductive reactance is

H0.4F;10100;356 === LCR

f 2=Hz9.15=f

f2100 =

tV 100sin520= to the tVV sin0=1

0 srad100V;520 == V

fCXC

2

1=

= 100CX( ) ( )6101009.152

1

=

CX

fLXL 2=

= 400LX( ) ( )0.49.152=

PHYSICS CHAPTER 8


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c. The impedance of the circuit is

d. The peak current in the circuit is

H0.4F;10100;356 === LCR

( ) 22 CL XXRZ +=

( ) ( ) 22 10040035 +== 302Z

ZIV 00 =( )302520 0I=

A72.10 =I

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e. The phase angle between the current and the supply voltage is

f. The power factor of the circuit is given by

H0.4F;10100;356 === LCR

RXX CL =tan

=

35

100400tan 1

rad45.1=

=

RXX CL1tan

OR3.83

cosfactorpower =

383cos .= 117.0factorpower =

PHYSICS CHAPTER 8


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75

A 22.5 mH inductor, a 105 resistor and a 32.3 F capacitor are

connected in series to the alternating source 240 V, 50 Hz.a. Sketch the phasor diagram for the circuit.

b. Calculate the power factor of the circuit.

c. Determine the average power consumed by the circuit.


a. The capacitive reactance is

and the inductive reactance is

Example 8 :

fCXC

2

1=

= 6.98CX( ) ( )6103.32502

1

=

CX

fLXL 2=

= 07.7LX

( ) ( )3105.22502 =

H105.22F;103.32;105 36 === LCRHz50V;240 == fV

PHYSICS CHAPTER 8


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a. Thus the phasor diagram for the circuit is

b. From the phasor diagram in (a),

the impedance of the circuit is

H105.22F;103.32;105 36 === LCRHz50V;240 == fV

Z

LX

CX

R

( )LC XX

( ) 22 LC XXRZ +=

( ) ( ) 22 07.76.98105 +=

= 139Z PHYSICS CHAPTER 8


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b. and the power factor of the circuit is

c. The average power consumed by the circuit is given by

H105.22F;103.32;105 36 === LCRHz50V;240 == fV

Z

R=cos

755.0cos =139

105cos =

cosav IVP =Z

VI=and

cos2

Z

V=

( ) ( )755.0139

240 2=

W313av =P

PHYSICS CHAPTER 8


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78

Exercise 8.2 :

1. AnRLCcircuit has a resistance of 105 , an inductance of

85.0 mH and a capacitance of 13.2 F.a. What is the power factor of the circuit if it is connected to a

125 Hz AC generator?

b. Will the power factor increase, decrease or stay the same

if the resistance is increased? Explain.

(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q47, p.834)edition, James S. Walker, Q47, p.834)

ANS. :ANS. : 0.962; U think0.962; U think

2. A 1.15 k resistor and a 505 mH inductor are connected in

series to a 14.2 V,1250 Hz AC generator.

a. What is the rms current in the circuit?

b. What is the capacitances value must be inserted in series

with the resistor and inductor to reduce the rms current to half

of the value in part (a)?

(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q69, p.835)edition, James S. Walker, Q69, p.835)

ANS. :ANS. : 3.44 mA, 10.5 nF3.44 mA, 10.5 nF

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ExplainExplain half-wave and full wave rectification by using ahalf-wave and full wave rectification by using a

circuit diagram andcircuit diagram and VV--ttgraph.graph.

ExplainExplain the smoothing of rectified output voltage bythe smoothing of rectified output voltage bycapacitor by using a circuit diagram andcapacitor by using a circuit diagram and VV--ttgraph.graph.

Learning Outcome:

8.5 Rectification (1 hour)

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is defined as the process of converting alternatingthe process of converting alternating current tocurrent to

direct current.direct current. Rectifier:

is a device that allows current to flow in one direction onlya device that allows current to flow in one direction only.

diodesdiodes are usually used as rectifiers.

Diode is said to be forward biasedforward biased when positive terminalpositive terminal of the

diodediode connected to the positive terminalpositive terminal of the batterybattery and viceversa, hence a currentcurrent will be able to flowflow (Figure 8.25a).

8.5 Rectification

++

++ --

--

I I

Diode

Figure 8.25a: forward biasedFigure 8.25a: forward biased

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Diode is said to be reverse biasedreverse biased when positive terminalpositive terminal of

the diodediode connected to the negative terminalnegative terminal of the batterybattery

and vice versa, hence no currentcurrent flows (Figure 8.25b).

There are two types of rectification i.e.

half-wave full-wave

++

++--

--

0=I

Figure 8.25b: reverse biasedFigure 8.25b: reverse biased

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Half-wave rectification means that only one half of an AConly one half of an ACcycle can pass through the rectifier (diode).cycle can pass through the rectifier (diode).

Figure 8.26a shows a half-wave rectification circuit.

8.5.1 Half-wave rectification

t0 T T2

RV

0V

0V

0V

0V

0V

0V

t0 T T2

Vtage,supply vol

RV

DVD

R

A

Bsupply

voltage, V


Figure 8.26bFigure 8.26b


Figure 8.26dFigure 8.26d

t0 T T2

DV

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Explanation:Explanation: First half cycle (Figure 8.26b)

When terminal A is positive, diode is forward biased andoffers low resistance such that a pulse of current flowsthrough the circuit.

There is negligible voltage across the diode, VD

(Figure

8.26c).

Thus the voltage across the resistor, VR

is almost equal to

the supply voltage (Figure 8.26d). Next half cycle (Figure 8.26b)

When terminal B is positive, diode is now reverse biased andhas a very high resistance such that a very small currentflows through it.

The voltage across the diode, VD

is almost equal to the

supply voltage (Figure 8.26c).

The voltage across the resistor, VR is almost zero (Figure8.26d).

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An alternating voltage is thus rectified to give direct current

voltage across the resistor. The current flows through the

resistor in one direction only and only half of each cycle cab

pass through the diode as shown in Figure 8.26e.

Rms value after half-wave rectification:Rms value after half-wave rectification:

In the half-wave rectification, half of the supply voltage issuppressed and therefore the mean square voltage is given by

t0T T2

I

0

I

0I

2

1Mean square valueafter rectification =

Mean square value

before rectification

rect.)wavehalfbefore(

2

rect.)wavehalf(

2

2

1

= VV

Figure 8.26eFigure 8.26e

tV 22

0 sin

2

1=

PHYSICS CHAPTER 8


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Therefore the rms voltage of the half-wave rectificationrms voltage of the half-wave rectification isgiven by

In the similar way as to find the rms voltage of half-waverectification, the rms current of half-wave rectificationrms current of half-wave rectification isgiven by

422

12

0

2

0

rect.)wavehalf(

2 VVV =

=

rect.)wavehalf(

2rms

= VV

4

2

0V

=

2

0rms

VV = (8.37)(8.37)

2

0rms

II = (8.38)(8.38)

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The half-wave rectification only allows half of each AC cycle topass through the diode, but the full-wave rectification allowsallows

both halves of each AC cycle to pass through the diodeboth halves of each AC cycle to pass through the diode. To obtain full-wave rectification, four diode are used and are

arranged in a form known as the diode bridgediode bridge.

Figure 8.27a shows a full-wave rectification circuit.

8.5.2 Full-wave rectification

0V

0VT T2

t0

( )Vtagesupply vol

0Vt0

T T2

R

V


Figure 8.27bFigure 8.27b


A

RVVsupply

voltage, F

B

CD

E

11

22 33

44

R

PHYSICS CHAPTER 8


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ExplanationExplanation First half cycle (Figure 8.27b)

When terminal A is positive, diodes 1 and 2 are forwardbiased and conduct the current.

The current takes the path ABC,R and DEF. Diodes 3 and 4 are reverse biased and hence do not conduct

the current. The voltages across diodes 1 and 2 are negligible, the

voltage across the resistorVR is almost equal to the supplyvoltage (Figure 8.27c)

Next half cycle (Figure 8.27b) When terminal F is positive, diodes 3 and 4 are forward

biased and conduct the current.

The path taken by the current is FEC,R and DBA. Diodes 1 and 2 are reverse biased and hence, do not

conduct the current. The voltage across the resistor is again almost equal to the

supply voltage (Figure 8.27c).

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Both halves of the alternating voltage are rectified. The current

flowing through the resistor is in one direction only i.e. a varying

DC is obtained as shown in Figure 8.27d.

Rms value after full-wave rectificationRms value after full-wave rectification Notice that the negative side of supply voltage is flipped over to

become positive side without being suppressed, thus the rmsrms

voltage and current of full-wave rectificationvoltage and current of full-wave rectification are the samesameas therms voltage and current of supply voltagerms voltage and current of supply voltage and given

by

t00I

T T2

I

Figure 8.27dFigure 8.27d

2

0rms

VV = and

2

0rms

II =

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The output obtained from half-wave and full-wave rectificationsare unidirectional but varying DC.

Usually a steady (constant) DC is required for operating variouselectrical and electronic appliances. To change a varying DCinto a steady (constant) DC, smoothing is necessary.

A simple smoothing circuit consists of a capacitor ( with a large

capacitance >16 F) connected parallel to the resistorR asshown in Figure 8.28.

The capacitor functions as a reservoir to store charges.

8.5.3 Smoothing using Capacitor

++R outputVVR =C

--Rectified unsmoothed

voltage, V

II


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Smoothing of a half-wave rectified voltageSmoothing of a half-wave rectified voltage

Figure 8.29 shows an effects of smoothing a half-wave rectifiedvoltage.

Initially, the half-wave rectified input voltage Vcauses the currentto flow through the resistorR. At the same time, capacitorC

becomes charged to almost the peak value of the input voltage. At A (Figure 8.29), input V(dash line) falls below output VR, the

capacitorCstarts to discharge through the resistorR. Hence thecurrent flow is maintained because of capacitors action.

A B

Rectified unsmooth input

voltage, V

( )outputVVRSmoothed voltage, VR

DischargeCharge

t,timeFigure 8.29Figure 8.29

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Along AB (Figure 8.29), Voutput falls. At B, the rectified currentagain flows to recharge the capacitorCto the peak of the input

voltage V. This process is repeated and hence the output voltage V

R

across the resistorR will look like the variation shown in figure8.29.

Smoothing of a full-wave rectified voltageSmoothing of a full-wave rectified voltage

Figure 8.30 shows an effects of smoothing a full-wave rectifiedvoltage.

A B

Rectified unsmooth input

voltage, V

( )outputVVRSmoothed voltage, V

R

Discharge Charge

t,timeFigure 8.30Figure 8.30

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The explanation of the smoothing process likes for a half-wave

rectified voltage.

The fluctuations of the smoothed output voltagefluctuations of the smoothed output voltage are must

less compare to the half-wave rectifiedless compare to the half-wave rectified.

The smoothing action of the capacitor is due to the large timelarge time

constantconstant , given byRCso the output voltage cannot falloutput voltage cannot fallas rapidly as the rectified unsmoothed input voltageas rapidly as the rectified unsmoothed input voltage.

Therefore a large capacitor performs greater smoothinglarge capacitor performs greater smoothing. However, an initially uncharged capacitoruncharged capacitormay cause a

sudden surge of current through the circuit and damage thedamage the

diodediode.

PHYSICS CHAPTER 8


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Next ChapterCHAPTER 9 :

Quantization of light

26771199 matriculation physics alternating current

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