matriculation physics alternating current (2).pdf

1PHYSICS CHAPTER 8

CHAPTER 8: CHAPTER 8: Alternating currentAlternating current

(6 Hours)(6 Hours)

PHYSICS CHAPTER 8

2

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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ics

PHYSICS CHAPTER 8

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is defined as an electric current whose magnitude and an 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 TT21 T2

T23

0I

0I

PHYSICS CHAPTER 8

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I

t0 TT21 T2

T23

0I

0I

TT21 T2

T23

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

Figure 8.1c: square ACFigure 8.1c: square AC

0I

0I

I

t0

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 TT21 T2

T23

0V

0V

( ) voltagemaximumpeak: 0Vwhereperiod:T

( ) current maximumpeak: 0IFigure 8.1d: sinusoidal alternating voltageFigure 8.1d: sinusoidal alternating voltage

PHYSICS CHAPTER 8

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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 ss11.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)

PHYSICS CHAPTER 8

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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 ve ORfrequency angular :

currentpeak : 0Igepeak volta: 0V

time: t

)2( fpi =

PHYSICS CHAPTER 8

8

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine root mean square (rms) current and voltage for root mean square (rms) current and voltage for

AC source.AC source. Use Use the following formula,the following formula,

Learning Outcome:

8.2 Root mean square (rms) (1 hour)

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20

rmsII = andand

20

rmsVV =

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 a the 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)

( )200

av2

pipi

III == (8.4)(8.4)

Note:Note:

Iav for one complete cycle is zero one complete cycle is zero because the current current flows 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.

PHYSICS CHAPTER 8

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

The instantaneous power, instantaneous power, PP delivered to a resistance R is

The average power, average power, PPavav over one cycle of AC over one cycle of AC is given by

where is the average value of the average value of II22 over one cycle over 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)

RIP 2rmsav = (8.6)(8.6)

where AC ousinstantane:I

PHYSICS CHAPTER 8

11

Since

and the graph of I2 against time, t is shown in Figure 8.2.

From Figure 8.2, the shaded region under under the curve and above above the dashed line for I02/2 have the same are as the shaded region above above the curve and belowbelow the dashed line for I02/2.Thus

thus the square value of current is given bytII sin0=tII 220

2 sin=

20I

TT21 T2T

23

2

20I

t0

2I

Figure 8.2Figure 8.2

2

202 II = (8.7)(8.7)

PHYSICS CHAPTER 8

12

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 the the value of the steady DC which produces the same power in a resistor as steady 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 the AC and can be illustrated as shown in Figure 8.3.

( )2

202

rmsII =

2

20

rmsII =

20

rmsII = (8.8)(8.8)

I

t0 TT21 T2

T23

0I

0I

rmsI 0707.0 I


PHYSICS CHAPTER 8

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is defined as the value of the steady direct voltage which the value of the steady direct voltage which when applied across a resistor, produces the same power when applied across a resistor, produces the same power as the mean (average) power produced by the alternating as 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

rmsVV = (8.9)(8.9)

Note:Note:Equations (8.8) and (8.9) are valid only for a sinusoidal valid only for a sinusoidal alternating current and voltagealternating current and voltage.

PHYSICS CHAPTER 8

14

An AC source V=500 sin t is connected across a resistor of 250 . Calculatea. the rms current in the resistor,b. the peak current,c. the mean power.Solution :Solution :By comparingThus the peak voltage isa. By applying the formulae of rms current, thus

Example 1 :

= 250RtV sin500= to the tVV sin0=

V 5000 =V

20

rmsII = and

RVI 00 =

20

rms RVI =

2250500

= A 41.1rms =I

PHYSICS CHAPTER 8

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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

20

rmsII =

241.1 0I=

A 99.10 =I

RIP 2rmsav =

( ) ( )25041.1 2= W497av =P

PHYSICS CHAPTER 8

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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 :

and40

)A(I

)ms(t0 20 80

02.0

02.0

60


PHYSICS CHAPTER 8

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Solution :Solution :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 3Rs 1040 A; 02.0 30

== TI

Tf 1= 31040

1

=f

Hz 25=f

20

rmsII =

202.0

rms =I

A 1041.1 2rms=I

RIP 2rmsav =

( ) ( )322 10101041.1 = W99.1av =P

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

the phase relationship between current and voltage for a the phase relationship between current and voltage for a circuit consisting of circuit consisting of pure resistorpure resistor pure capacitor pure capacitor pure inductor.pure inductor.

DefineDefine capacitive reactance, inductive reactance and capacitive reactance, inductive reactance and impedance.impedance.

AnalyseAnalyse voltage, current and phasor diagrams for a voltage, current and phasor diagrams for a series circuit consisting ofseries circuit consisting of RCRC RL RL RCLRCL..

Learning Outcome:

8.3 Resistance, reactance and impedance (2 hours)

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PHYSICS CHAPTER 8

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8.3.1 Phasor diagram PhasorPhasor is defined as a vector that rotate anticlockwise about a 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 such

as 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 AC the phase difference between current and voltage in AC circuitcircuit).

Consider a graph represents sinusoidal AC and sinusoidal alternating voltage waveform as shown in Figure 8.5a. Meanwhile Figure 8.5b shows the phasor diagram of V and I.

8.3 Resistance, reactance and impedance

PHYSICS CHAPTER 8

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From the Figure 8.5a:Thus the phase difference is

Therefore the current I is in phasein phase with the voltage V and constant with time.

t0

0I0V

0I0V

TT21 T2

T23

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 pi = valuenegative=

LeadsLeads

Lags behindLags behindIn antiphaseIn antiphase

PHYSICS CHAPTER 8

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

IVZ = (8.10)(8.10)

20V

20I

OR

0

0

IVZ = (8.11)(8.11)

PHYSICS CHAPTER 8

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

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

Phase difference between voltage Phase difference between voltage VV and current and current II Figure 8.7 shows an AC source connected to a pure resistor R.

8.3.3 Pure resistor in an AC circuit


AC source

R

I

RV

VFigure 8.7Figure 8.7

PHYSICS CHAPTER 8

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The alternating current passes through the resistor is given by

The alternating voltage across the resistor VR at any instant is given by

Therefore the phase difference between V and I is

In pure resistor, the current current II always in phase with the voltage always in phase with the voltage VV and constant with time and constant with time.

Figure 8.8a shows the variation of V and I with time while Figure 8.8b shows the phasor diagram for V and I in a pure resistor.

tII sin0=

IRVR =00 VRI =( ) RtI sin0= and

VtVVR == sin0where tagesupply vol:V

0== tt

PHYSICS CHAPTER 8

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

t0

0I0V

0I0V

TT21 T2

T23


VI

RIV

IVZ ===

0

0

rms

rms (8.12)(8.12)

PHYSICS CHAPTER 8

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Pure capacitor means that no resistance and self-inductanceno resistance and self-inductance effect in the AC circuit.

Phase difference between voltage Phase difference between voltage VV and current and current II Figure 8.9 shows an AC source connected to a pure capacitor C.

The alternating voltage across the capacitor VC at any instant is equal to the supply voltage V and is given by

8.3.4 Pure capacitor in an AC circuit


AC sourceAC source

CV

V

C

I

tVVVC sin0==

PHYSICS CHAPTER 8

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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=

dtdQI =

( )tCVdtdI sin0=

( )tdtdCV sin0=

tCV cos0= 00 ICV =andtII cos0=

OR

+=

2sin0

pitII

PHYSICS CHAPTER 8

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In the pure capacitor, the voltage VV lags behindlags behind the current II by pipi //22 radians radians.

ORthe current II leadsleads the voltage VV by pipi //22 radians radians.

Figure 8.10a shows the variation of V and I with time while Figure 8.10b shows the phasor diagram for V and I in a pure capacitor.

+=

2pi

ttrad

2pi =

PHYSICS CHAPTER 8

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Impedance in a pure capacitorImpedance in a pure capacitor From the definition of the impedance, hence


0

0

IVZ =

t0

0I0V

0I0V

TT21 T2

T23

VI

rad 2pi =

00 CVI =and

0

0

CVV

=

PHYSICS CHAPTER 8

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where XC is known as capacitive (capacitative) reactancecapacitive (capacitative) reactance.

Capacitive reactance is the opposition of a capacitor to the the 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 ohm ohm (( )) .

CXCZ ==

1

fCX C

pi21

=

fpi 2=and

(8.13)(8.13)

source AC offrequency : fcapacitor theof ecapacitanc: C

0

0

rms

rms

IV

IVX C == (8.14)(8.14)

PHYSICS CHAPTER 8

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From the eq. (8.13), the relationship between capacitive reactance XC and frequency f can be shown by using a graph in Figure 8.11.

f0

CX

fX C

1


Pure inductor means that no resistance and capacitance no resistance and capacitance effect in the AC circuit.

Phase difference between voltage Phase difference between voltage VV and current and current II Figure 8.12 shows an AC source connected to a pure inductor

L.

8.3.5 Pure inductor in an AC circuit

PHYSICS CHAPTER 8

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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

( )tIdtdL sin0=

tLI cos0B = (8.15)(8.15)

PHYSICS CHAPTER 8

32

At any instant, the supply voltage V equals to the back emf B in the inductor but the back emf always oppose the supply voltage V represents by the negative sign in the eq. (8.15).Thus


In the pure inductor, the voltage VV leadsleads the current II by pipi //22 radians radians.

ORthe current II lags behindlags behind the voltage VV by pipi //22 radians radians.

B=VtLI cos0= 00 VLI =andtVV cos0=

OR

+=

2sin0

pitVV

rad 22pi

pi

=

+= tt

PHYSICS CHAPTER 8

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Figure 8.13a shows the variation of V and I with time while Figure 8.13b shows the phasor diagram for V and I in a pure inductor.

t0

0I0V

0I0V

TT21 T2T

23

V

I

rad 2pi =


PHYSICS CHAPTER 8

34

Impedance in a pure inductorImpedance in a pure inductor From the definition of the impedance, hence

where XL is known as inductive reactanceinductive reactance.

0

0

IVZ = 00 LIV =and

0

0

ILI

=

LXLZ ==

fLX L pi2=

fpi 2=and

(8.16)(8.16)

inductor theof inductance-self: Lsource AC offrequency : f

PHYSICS CHAPTER 8

35

Inductive reactance is the opposition of a inductor to the the 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 ohm ohm (( )).

From the eq. (8.16), the relationship between inductive reactance XL and the frequency f can be shown by using a graph in Figure 8.14.

Figure 8.14Figure 8.14f0

LXfX L

0

0

rms

rms

IV

IVX L == (8.17)(8.17)

PHYSICS CHAPTER 8

36

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.Solution :Solution :a. The capacitive reactance of the capacitor is given by

Therefore the capacitance of the capacitor is

Example 3 :

V 14 Hz; 60 A; 1021 rms3

rms === VfI

CXIV rmsrms =

fCX C

pi21

=

( ) CX3102114 == 667CX

( )C6021667

pi=

F 1098.3 6=C

PHYSICS CHAPTER 8

37

Solution :Solution :b. The capacitive reactance is inversely proportional to the capacitive 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 is current in the capacitor is inversely proportional to the capacitive reactanceinversely proportional to the capacitive reactance, therefore the current will increase when the capacitive reactance current will increase when the capacitive reactance decreasesdecreases.c. Given The capacitive reactance is

Hence the new rms current in the capacitor is given by

V 14 Hz; 60 A; 1021 rms3

rms === VfI

Hz 410 =f

fCX C

pi21

= ( ) ( )61098.34102 1 = piCX= 5.97CX

CXIV rmsrms = ( )5.9714 rmsI=A 144.0rms =I

PHYSICS CHAPTER 8

38

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.Solution :Solution :a. The inductive reactance of the inductor is given by

Thus the rms current in the circuit is

Example 4 :

H 10290.0 Hz; 1000.1 V; 2.12 33rms=== LfV

fLX L pi2= ( )( )33 10290.01000.12 = pi= 82.1LX

LXIV rmsrms =( )82.12.12 rmsI=A 70.6rms =I

PHYSICS CHAPTER 8

39

Solution :Solution :b. Given The inductive reactance of the inductor is given by

Thus the peak current in the circuit is

H 10290.0 Hz; 1000.1 V; 2.12 33rms=== LfV

Hz 1050.2 3=f

fLX L pi2= ( )( )33 10290.01050.22 = pi= 56.4LX

LXIV 00 =

( )56.422.12 0I=A 78.30 =I

and 2rms0 VV =

LXIV 0rms 2 =

PHYSICS CHAPTER 8

40

RCRC series circuit series circuit Consider an AC source of rms voltage rms voltage VV is connected in series

to a resistor R and a capacitor C as shown in Figure 8.15a.

The rms current rms current II passes through the resistor and the resistor and the capacitor is equalcapacitor is equal because of the series connectionseries connection between both components.

8.3.5 RC, RL and RCL series circuit

AC sourceAC source

R

I

RV

V

CV

C

Figure 8.15aFigure 8.15a

PHYSICS CHAPTER 8

41

I

The rms voltagesrms voltages across the resistorresistor VVRR and the capacitor capacitor VVCC are given by

The phasor diagram of the RC series circuit is shown in Figure 8.15b.

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

IRVR = and CC IXV =

where angle phase:

CV

RV

V

Figure 8.15b: phasor diagramFigure 8.15b: phasor diagram

is an angle between the rms an angle between the rms current current II and rms supply (or and rms supply (or total) voltage total) voltage VV of AC circuit.

22CR VVV += ( ) ( ) 22 CIXIRV +=

22CXRIV += (8.18)(8.18)

PHYSICS CHAPTER 8

42

Rearrange the eq. (8.18), thus the impedance of RC series circuit is

From the phasor diagram in Figure 8.15b , the current current II leads leads the supply voltage the supply voltage VV by by radians radians where

A phasor diagram in terms of R, XC and Z is illustrated in Figure 8.15c.

IVZ =and22 CXRI

V+=

22CXRZ += (8.19)(8.19)

R

C

VV

=tanIR

IX C=tan

RX C

=tan (8.20)(8.20)

CX Z

R

Figure 8.15cFigure 8.15c

PHYSICS CHAPTER 8

43

RLRL series circuit series circuit Consider an AC source of rms voltage rms voltage VV is connected in series

to a resistor R and an inductor L as shown in Figure 8.16a.

The rms voltagesrms voltages across the resistorresistor VVRR and the inductor inductor VVLL are given by

AC sourceAC source

R

I

RV

V

L

LV


IRVR =and

LL IXV =

PHYSICS CHAPTER 8

44

The phasor diagram of the RL series circuit is shown in Figure 8.16b.


LVV

I

Figure 8.16b: phasor diagramFigure 8.16b: phasor diagramRV

22LR VVV +=

( ) ( ) 22 LIXIR +=22

LXRIV += (8.21)(8.21)

PHYSICS CHAPTER 8

45

Rearrange the eq. (8.21), thus the impedance of RL series circuit is

From the phasor diagram in Figure 8.16b , the supply voltage supply voltage VV leads the current leads the current II the by the by radians radians where

The phasor diagram in terms of R, XL and Z is illustrated in Figure 8.16c.

IVZ =and22 LXRI

V+=

22LXRZ += (8.22)(8.22)

R

L

VV

=tanIR

IX L=tan

RX L

=tan (8.23)(8.23)


LXZ

R

PHYSICS CHAPTER 8

46

RCLRCL series circuit series circuit Consider an AC source of rms voltage rms voltage VV is connected in series

to a resistor R, a capacitor C and an inductor L as shown in Figure 8.17a.

The rms voltagesrms voltages across the resistorresistor VVRR, the capacitor capacitor VVCC and the inductor inductor VVLL are given by


IRVR =

andCC IXV =

AC sourceAC sourceI

V

R

RV CV

C L

LV

LL IXV =

PHYSICS CHAPTER 8

47

The phasor diagram of the RL series circuit is shown in Figure 8.17b.


I

Figure 8.17b: phasor diagramFigure 8.17b: phasor diagram

( ) 22 CLR VVVV +=( ) ( ) 22 CL IXIXIR +=

( ) 22 CL XXRIV += (8.24)(8.24)

LV

RV

V

CV

( )CL VV

PHYSICS CHAPTER 8

48

Rearrange the eq. (8.24), thus the impedance of RL series circuit is

From the phasor diagram in Figure 8.17b , the supply voltage supply voltage VV leads the current leads the current II the by the by radians radians where

IVZ =and( ) 22 CL XXRI

V+=

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

R

CL

VVV

=tan( )

IRIXIX CL

=

RXX CL

=tan (8.26)(8.26)

PHYSICS CHAPTER 8

49

The phasor diagram in terms of R, XC, XL and Z is illustrated in Figure 8.17c.


LX

Z

CX

( )CL XX

R

PHYSICS CHAPTER 8

50

is defined as the phenomenon that occurs when the phenomenon that occurs when the frequency of the applied voltage is equal to the frequency frequency of the applied voltage is equal to the frequency of the of the RCLRCL series circuit series circuit.

Figure 8.18 shows the variation of XC, XL, R and Z with frequency f of the RCL series circuit.

8.3.6 Resonance in AC circuit

Z

fX L

R

fX C

1

0 f

ZRXX LC ,,,

rfFigure 8.18Figure 8.18

PHYSICS CHAPTER 8

51

From Figure 8.18, the value of impedance is minimumis minimum ZZminmin when

where its value is given by

This phenomenon occurs at the frequency ffrr known as known as resonant frequencyresonant frequency.

At resonance resonance in the RCL series circuit, the impedance is is minimumminimum ZZmin min thus the rms current flows rms current flows in the circuit is maximum maximum IImaxmax and is given by

CL XX = (8.27)(8.27)

( ) 22 CL XXRZ +=02min += RZ

RZ =min

RV

ZVI ==min

max (8.28)(8.28)

PHYSICS CHAPTER 8

52

maxI

Figure 8.19 shows the rms current I in RCL series circuit varies with frequency.

At frequencies above or below the resonant frequency frequencies above or below the resonant frequency ffrr, the rms rms current current II is less than the rms maximum current less than the rms maximum current IImaxmax as shown in Figure 8.19.

0 f

I

rfFigure 8.19Figure 8.19

PHYSICS CHAPTER 8

53

The resonant frequency, fr of the RCL series circuit is given by

The series resonance circuit is used for tuning a radio tuning a radio receiverreceiver.

CL XX =

CL

1=

LC12

= r2 fpi =and

( )LC

f 12 2r =pi

LCf

pi21

r = (8.29)(8.29)

where frequencyangular resonant :

Note:Note:At resonance, the current resonance, the current II and voltage and voltage VV are in phase are in phase.

PHYSICS CHAPTER 8

54

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. Calculatea. the current flowing,b. the voltage across the capacitor,c. the phase angle of the circuit.Solution :Solution :a. The capacitive reactance of the inductor is given by

and the impedance of the circuit is

Example 5 :

Hz 50V; 12 ; 1000 F; 102 6 ==== fVRC

fCX C

pi21

= ( ) ( )6102502 1 = piCX= 1592CX

22CXRZ += ( ) ( ) 22 15921000 +=Z

= 1880Z

PHYSICS CHAPTER 8

55

Solution :Solution :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=A 1038.6 3=I

CC IXV = ( )( )15921038.6 3=V 2.10=CV

Hz 50V; 12 ; 1000 F; 102 6 ==== fVRC

IZV =

RX C

=tan

=

10001592tan 1

rad 01.1=

=

RX C1tan

OR 9.57

PHYSICS CHAPTER 8

56

Based on the RCL series circuit in Figure 8.20 , the rms voltages across R, L and C are 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 resistor R 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

V 314V 153 V 115

I


PHYSICS CHAPTER 8

57

Solution :Solution :a. The phasor diagram of the circuit is

and the phase angle is

V 314 ;V 115 V; 153 === LCR VVV

LV

IRV

V

CV

( )CL VV From the phasor diagram,

the applied voltage V is

( ) 22 CLR VVVV +=( ) ( ) 22 115314153 +=

V 251=V

R

CL

VVV

=tan

=

153115314tan 1

rad 915.0=

=

R

CL

VVV1tan

OR 4.52

PHYSICS CHAPTER 8

58

Solution :Solution :b. Given

Since R, C and L 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

V 314 ;V 115 V; 153 === LCR VVV

IRVR =

( ) LX88.5314 =

A 88.5=I( )26153 I=

= 26R

LL IXV =

Hz 50=f

= 4.53LX

( ) L5024.53 pi=fLX L pi2=

H 170.0=L

PHYSICS CHAPTER 8

59

Solution :Solution :c. Meanwhile, the capacitive reactance is

thus the capacitance of the capacitor is

d. The resonant frequency is given by

V 314 ;V 115 V; 153 === LCR VVV

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

fCX C

pi21

=

F 1062.1 4=C( )C50216.19

pi=

( ) ( )41062.1170.021

=

pi

LCf

pi21

r =

Hz 3.30r =f

PHYSICS CHAPTER 8

60

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 are connected 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 V2. A 200 resistor, a 0.75 H inductor and a capacitor of

capacitance C are connected in series to an alternating source 250 V, 600 Hz. Calculatea. 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

PHYSICS CHAPTER 8

61

Exercise 8.1 :3. A capacitor of capacitance C, a coil of inductance L, a resistor

of resistance R and a lamp of negligible resistance are placed in series with alternating voltage V. Its frequency f is varied from a low to a high value while the magnitude of V is kept constant.a. Describe and explain how the brightness of the lamp varies.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

PHYSICS CHAPTER 8

62

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 of in AC circuit consisting of RR, , RCRC, , RLRL and and RCLRCL in in series.series.

Learning Outcome:

8.4 Power and power factor (1 hour)

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cosav IVP =

dtdWP =

IVP

PP av

a

rcos ==

PHYSICS CHAPTER 8

63

8.4.1 Power of a pure resistor In a pure resistor, the voltage voltage VV and current and current II are in phase are in phase,

thus the instantaneous powerinstantaneous power P is given by

Figure 8.21 shows a graph of instantaneous power P being absorbed 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 power um)peak(maxim: 0P

PHYSICS CHAPTER 8

64

The average (or mean) power Pav being absorbed by the resistor is given by

tPP 20 sin=Power being absorbedPower being absorbed


avP

tPP 20av sin=

000av 21

21 VIPP ==

0P

20P

t0

P

TT21 T2T

23

(8.31)(8.31)

PHYSICS CHAPTER 8

65

In a pure capacitor, the current current II leads the voltage leads the voltage VV by by pipi //22 radiansradians, thus the instantaneous powerinstantaneous power P is given by

Figure 8.22 shows a graph of instantaneous power P of the pure capacitor against time t.

8.4.2 Power of a pure capacitor

( )( )tVtI sincos 00=IVP =

ttVI cossin00=

tPP 2sin21

0= (8.32)(8.32)

ttt 2sin21cossin =and

PHYSICS CHAPTER 8

66

The average (or mean) power Pav of the pure capacitor is given by

tPP 2sin21

0=

Power being absorbedPower being absorbed


avP

tPP 2sin21

0av =

0av =P

20P

t0

P

TT21 T2T

23

20P

Power being returned to supplyPower being returned to supply

PHYSICS CHAPTER 8

67

In a pure inductor, the voltage voltage VV leads the current leads the current II by by pipi //22 radiansradians, thus the instantaneous powerinstantaneous power P is given by

Figure 8.23 shows a graph of instantaneous power P of the pure inductor against time t.

8.4.3 Power of a pure inductor

( )( )tVtI cossin 00=IVP =

ttVI cossin00=

tPP 2sin21

0=

ttt 2sin21cossin =and

PHYSICS CHAPTER 8

68

The average (or mean) power Pav of the pure inductor is given by

tPP 2sin21

0=

Power being absorbedPower being absorbed


avP

tPP 2sin21

0av = 0av =P

20P

t0

P

TT21 T2T

23

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.

20P

PHYSICS CHAPTER 8

69

In an AC circuit in which there is a resistor R, an inductor L and a capacitor C, the average power Pav is equal to that dissipated from the resistor i.e.

From the phasor diagram of the RCL series circuit as shown in Figure 8.24,

8.4.4 Power and power factor of R, RC, RL and RCL series circuits

RIIVP R2

av == (8.33)(8.33)

rms valuesrms values

LV

IRV

V

CV

( )CL VV


PHYSICS CHAPTER 8

70

We get

then the eq. (8.33 ) can be written as

where cos cos is called the power factorpower factor of the AC circuit, PPrr is the average real poweraverage real power and II22ZZ is called the apparent powerapparent power.

Power factor is defined as

cosVVR =VVR

=cos

cosav IVP = IZV =andr

2av cos PZIP == (8.34)(8.34)

a

r2

rcosPP

ZIP

== (8.35)(8.35)where IVZIP == 2a powerapparent : Note:Note:

From the Figure 8.24, the power factor also can be calculated by using the equation below:

IZIR

VVR

==cosZR

=cos (8.36)(8.36)

PHYSICS CHAPTER 8

71

A 100 F capacitor, a 4.0 H inductor and a 35 resistor are connected in series with an alternating source given by the equation 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

72

Solution :Solution :By comparingThus a. The frequency of AC source is given by

b. The capacitive reactance is

and the inductive reactance is

H 0.4 F; 10100 ; 35 6 === LCR

fpi 2=Hz 9.15=f

fpi2100 =

tV 100sin520= to the tVV sin0=1

0 s rad 100 V; 520

== V

fCX C

pi21

=

= 100CX( ) ( )6101009.152 1 = piCX

fLX L pi2=

= 400LX( ) ( )0.49.152pi=

PHYSICS CHAPTER 8

73

Solution :Solution :c. The impedance of the circuit is

d. The peak current in the circuit is

H 0.4 F; 10100 ; 35 6 === LCR

( ) 22 CL XXRZ +=( ) ( ) 22 10040035 +=

= 302Z

ZIV 00 =( )302520 0I=

A 72.10 =I

PHYSICS CHAPTER 8

74

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

f. The power factor of the circuit is given by

H 0.4 F; 10100 ; 35 6 === LCR

RXX CL

=tan

=

35100400tan 1

rad 45.1=

=

RXX CL1tan

OR 3.83

cosfactorpower =383cos .=

117.0factorpower =

PHYSICS CHAPTER 8

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.Solution :Solution :

a. The capacitive reactance is

and the inductive reactance is

Example 8 :

fCX C

pi21

=

= 6.98CX( ) ( )6103.32502

1

=

piCX

fLX L pi2=

= 07.7LX( ) ( )3105.22502 = pi

H 105.22 F; 103.32 ; 105 36 === LCRHz 50 V; 240 == fV

PHYSICS CHAPTER 8

76

Solution :Solution :

a. Thus the phasor diagram for the circuit is

b. From the phasor diagram in (a), the impedance of the circuit is

H 105.22 F; 103.32 ; 105 36 === LCRHz 50 V; 240 == fV

Z

LX

CX

R

( )LC XX

( ) 22 LC XXRZ +=( ) ( ) 22 07.76.98105 +=

= 139Z

PHYSICS CHAPTER 8

77

Solution :Solution :

b. and the power factor of the circuit is

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

H 105.22 F; 103.32 ; 105 36 === LCRHz 50 V; 240 == fV

ZR

=cos755.0cos =

139105cos =

cosav IVP = ZVI =and

cos2

ZV

=

( ) ( )755.0139240 2

=

W313av =P

PHYSICS CHAPTER 8

78

Exercise 8.2 :1. An RLC circuit 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 think2. 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

PHYSICS CHAPTER 8

79

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain half-wave and full wave rectification by using a half-wave and full wave rectification by using a

circuit diagram and circuit diagram and VV--tt graph. graph. ExplainExplain the smoothing of rectified output voltage by the smoothing of rectified output voltage by

capacitor by using a circuit diagram and capacitor by using a circuit diagram and VV--tt graph. graph.

Learning Outcome:

8.5 Rectification (1 hour)

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PHYSICS CHAPTER 8

80

is defined as the process of converting alternating the process of converting alternating current to current 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 vice versa, 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

PHYSICS CHAPTER 8

81

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

PHYSICS CHAPTER 8

82

Half-wave rectification means that only one half of an AC only one half of an AC cycle 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

0Vt0 T T2

V tage,supply vol

RV

DVD

R

A

Bsupply voltage, V


Figure 8.26bFigure 8.26b


Figure 8.26dFigure 8.26d

t0 T T2

DV

PHYSICS CHAPTER 8

83

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

When terminal A is positive, diode is forward biased and offers low resistance such that a pulse of current flows through 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 and

has a very high resistance such that a very small current flows 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 (Figure 8.26d).

PHYSICS CHAPTER 8

84

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 is

suppressed and therefore the mean square voltage is given by

t0 T T2

I

0I

0I

21Mean square value

after rectification = Mean square value

before rectification

rect.) wavehalf before(2

rect.) wavehalf(2

21

= VV

Figure 8.26eFigure 8.26e

tV 220 sin21

=

PHYSICS CHAPTER 8

85

Therefore the rms voltage of the half-wave rectificationrms voltage of the half-wave rectification is given by

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

4221 20

20

rect.) wavehalf(2 VVV =

=

rect.) wavehalf(2

rms

= VV

4

20V

=

20

rmsVV = (8.37)(8.37)

20

rmsII = (8.38)(8.38)

PHYSICS CHAPTER 8

86

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

( )V tagesupply vol

0Vt0 T T2

RV


Figure 8.27bFigure 8.27b


A

RVVsupply voltage, F

B

CD

E

11

22 33

44

R

PHYSICS CHAPTER 8

87

ExplanationExplanation First half cycle (Figure 8.27b)

When terminal A is positive, diodes 1 and 2 are forward biased 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 resistor VR is almost equal to the supply voltage (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).

PHYSICS CHAPTER 8

88

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 rms rms voltage and current of full-wave rectificationvoltage and current of full-wave rectification are the samesame as the rms voltage and current of supply voltagerms voltage and current of supply voltage and given by

t00I

T T2

I

Figure 8.27dFigure 8.27d

20

rmsVV = and

20

rmsII =

PHYSICS CHAPTER 8

89

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

Usually a steady (constant) DC is required for operating various electrical and electronic appliances. To change a varying DC into 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 resistor R as shown 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


PHYSICS CHAPTER 8

90

Smoothing of a half-wave rectified voltageSmoothing of a half-wave rectified voltage Figure 8.29 shows an effects of smoothing a half-wave rectified

voltage.

Initially, the half-wave rectified input voltage V causes the current to flow through the resistor R. At the same time, capacitor C 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 capacitor C starts to discharge through the resistor R. Hence the current flow is maintained because of capacitors action.

A B

Rectified unsmooth input voltage, V

( )outputVVRSmoothed voltage, VR

DischargeCharge

t,timeFigure 8.29Figure 8.29

PHYSICS CHAPTER 8

91

Along AB (Figure 8.29), V output falls. At B, the rectified current again flows to recharge the capacitor C to the peak of the input voltage V.

This process is repeated and hence the output voltage VR across the resistor R will look like the variation shown in figure 8.29.

Smoothing of a full-wave rectified voltageSmoothing of a full-wave rectified voltage Figure 8.30 shows an effects of smoothing a full-wave rectified

voltage.

A B

Rectified unsmooth input voltage, V

( )outputVVRSmoothed voltage, VR

Discharge Charge

t,timeFigure 8.30Figure 8.30

PHYSICS CHAPTER 8

92

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 time large time constant constant , given by RC so the output voltage cannot fall output voltage cannot fall as 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 capacitor may cause a

sudden surge of current through the circuit and damage the damage the diodediode.

93

PHYSICS CHAPTER 8

Next ChapterCHAPTER 9 :

Quantization of light

Slide 1Learning Outcome:8.1 Alternating current (AC)Slide 4Slide 58.1.1 Terminology in AC8.1.2 Equations of alternating current and voltageSlide 88.2 Root mean square (rms)8.2.2 Root mean square current (Irms)Slide 11Slide 128.2.3 Root mean square voltage (Vrms)Slide 14Slide 15Slide 16Slide 17Slide 188.3 Resistance, reactance and impedanceSlide 208.3.2 Impedance (Z)8.3.3 Pure resistor in an AC circuitSlide 23Slide 248.3.4 Pure capacitor in an AC circuitSlide 26Slide 27Slide 28Slide 298.3.5 Pure inductor in an AC circuitSlide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 398.3.5 RC, RL and RCL series circuitSlide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 498.3.6 Resonance in AC circuitSlide 51Slide 52Slide 53Slide 54Slide 55Slide 56Slide 57Slide 58Slide 59Slide 60Slide 61Slide 628.4 Power and power factorSlide 648.4.2 Power of a pure capacitorSlide 668.4.3 Power of a pure inductorSlide 688.4.4 Power and power factor of R, RC, RL and RCL series circuitsSlide 70Slide 71Slide 72Slide 73Slide 74Slide 75Slide 76Slide 77Slide 78Slide 798.5 RectificationSlide 818.5.1 Half-wave rectificationSlide 83Slide 84Slide 858.5.2 Full-wave rectificationSlide 87Slide 888.5.3 Smoothing using CapacitorSlide 90Slide 91Slide 92Slide 93

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