chapter21 student

8/10/2019 Chapter21 Student

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PHYSICS CHAPTER 21

1

CHAPTER 21:

Alternating current

(5 Hours)

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.matrik.e

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.my/physics


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PHYSICS CHAPTER 21

2

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

a. Define alternating current (AC).

b. Sketch and analyse sinusoidal AC waveform.

c. Write and use sinusoidal voltage and current equations.

Learning Outcome:

24.1 Alternating current (1 hour)


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PHYSICS CHAPTER 21

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is defined as an electric current whose magnitude and

direction change periodically.

Figures 21.1a, 21.1b and 21.1c show three forms of alternating

current.

21.1 Alternating current (AC)

Figure 21.1a: sinusoidal AC

I

t0

TT2

1 T2T2

3

0I

0I


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PHYSICS CHAPTER 21

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I

t0 TT2

1 T2T

2

3

0I

0I

TT2

1 T2T

2

3

Figure 21.1b: saw-tooth AC

Figure 21.1c: square AC

0I

0I

I

t0


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PHYSICS CHAPTER 21

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

difference (voltage) across it and this voltage is alternating as

shown in Figure 21.1d.

V

t0 TT2

1 T2T

2

3

0V

0V

voltagemaximumpeak:0Vwhere

period:T

currentmaximumpeak:0I

Figure 21.1d: sinusoidal alternating voltage


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PHYSICS CHAPTER 21

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Frequency (f)

is defined as a number of complete cycle in one second.

Its unit is hertz (Hz) OR s1.

Period (T)

is defined as a time taken for one complete cycle.

Its unit is second (s). Formulae,

Peak current (I0)

is defined as a magnitude of the maximum current.

Its unit is ampere (A).

21.1.1 Terminology in AC

(21.1)


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PHYSICS CHAPTER 21

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Equation for alternating current (I

),

Equation for alternating voltage (V),

21.1.2 Equations of alternating current and voltage

(21.2)

phase

where locityangular veORfrequencyangular:

currentpeak:0I

gepeak volta:0V

time:t

)2( f

(21.3)


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PHYSICS CHAPTER 21

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

half-cycle flows of current in a certain direction.

Formulae:

21.2 Root mean square (rms)

(21.4)

Note:

Iav for one complete cycle is zero because the currentflows in one direction in one-half of the cycle and in the

opposite direction in the next half of the cycle.


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PHYSICS CHAPTER 21

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

The instantaneous power, Pdelivered to a resistanceR is

The average power, Pav

over one cycle of AC is given by

where is the average value of I2 over one cycle and is

given by

Therefore

21.2.2 Root mean square current (Irms

)

RIP 2

RIP 2av2I

2rms2 II

(21.5)

where ACousinstantane:I


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PHYSICS CHAPTER 21

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Since

and the graph ofI2 against time, tis shown in Figure 21.2.

From Figure 21.2, the shaded region under the curve and

above the dashed line forI0

2/2 have the same are as the

shaded region above the curve and below the dashed line for

I0

2/2.

Thus

thus the square value of current is given bytII sin0

tII 22

0

2sin

2

0I

TT

2

1 T2T

2

3

2

2

0I

t0

2I

Figure 21.2

2

2

02 I

I

(21.6)


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PHYSICS CHAPTER 21

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By equating the eqs. (21.5) and (21.6), the rms current is

Root mean square current (Irms) is defined as the value ofthe steady DC which produces the same power in a resistoras the mean (average) power produced by the AC.

The root mean square (rms) current is the effective value of theAC and can be illustrated as shown in Figure 21.3.

2

2

02

rms

I

I 2

2

0

rms

I

I

I

t0 TT2

1 T2T

2

3

0I

0I

rm sI 0707.0 I

Figure 21.3

(21.7)


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PHYSICS CHAPTER 21

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is defined as the value of the steady direct voltage whichwhen applied across a resistor, produces the same poweras the mean (average) power produced by the alternatingvoltage across the same resistor.

Its formula is

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

21.2.3 Root mean square voltage (Vrms

)

Note:

Equations (21.7) and (21.8) are valid only for a sinusoidalalternating current and voltage.

(21.8)


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PHYSICS CHAPTER 21

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An AC source V=500 sintis connected across a resistor of

250 . Calculatea. the rms current in the resistor,

b. the peak current,

c. the mean power.

Solution :By comparing

Thus the peak voltage is

a. By applying the formulae of rms current, thus

Example 21.1 :

250R tV sin500 to the tVV sin0V5000V

2

0

rms

I

I and R

V

I0

0

2

0rms

R

VI

2250

500


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PHYSICS CHAPTER 21

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Figure 21.4 shows a graph to represent alternating current passes

through a resistor of 10 k. Calculate

a. the rms current,

b. the frequency of the AC,

c. the mean power dissipated from the resistor.

Example 21.2 :

and

40

)A(I

)ms(t0 20 80

02.0

02.0

60

Figure 21.4


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PHYSICS CHAPTER 21

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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 3Rs1040A;02.0

3

0

TI

Tf

1

31040

1

f

2

0rms

II

2

02.0rms I

RIP2

rmsav

322 10101041.1


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PHYSICS CHAPTER 21

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a) Sketch and use phasor diagram and sinusoidalwaveform to show the phase relationship between

current and voltage for a circuit consisting ofi. pure resistor

ii. pure capacitor

iii. pure inductor.

b) Define and use capacitive reactance, inductive

reactance and impedance.c) Use phasor diagram to analyse voltage, current and

impedance of series circuit of :

i. RC

ii. RL

iii. RCL.

Learning Outcome:

21.3 Resistance, reactance and impedance(2 hours)


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21.3.1 Phasor diagram

a. Phasoris defined as a vector that rotate anticlockwiseabout its axis with constant angular velocity.

b. A diagram containing phasor is called phasor diagram.

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

d. It also being used to determine the phase angle (is defined asthe phase difference between current and voltage in ACcircuit).

e. Consider a graph represents sinusoidal AC and sinusoidalalternating voltage waveform as shown in Figure 21.5a.

Meanwhile Figure 21.5b shows the phasor diagram of VandI.

21.3 Resistance, reactance and impedance


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PHYSICS CHAPTER 21

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From the Figure 21.5a:

Thus the phase difference is

Therefore the currentI is in phase with the voltage Vandconstant with time.

t0

0I

0V

0

I0V

TT2

1 T2T

2

3

Figure 21.5aFigure 21.5b: phasor diagram

VI

tII sin0 tVV sin00 tt

and

Note:

valuepositive

radian

valuenegativeLeads

Lags behind

In antiphase


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PHYSICS CHAPTER 21

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The quantity that measures the opposition of a circuit to theAC flows.

It is defined by

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

In a DC circuit, impedance likes the resistance.

21.3.2 Impedance (Z)

(21.9)

2

0V

2

0IOR

(21.10)


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PHYSICS CHAPTER 21

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The symbol of an AC source in the electrical circuit is shown inFigure 21.6.

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

Phase difference between voltage Vand current I

Figure 21.7 shows an AC source connected to a pure resistor R.

21.3.3 Pure resistor in an AC circuit

Figure 21.6

AC source

R

I

RV

VFigure 21.7


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PHYSICS CHAPTER 21

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The alternating current passes through the resistor is given by

The alternating voltage across the resistor VR at any instant isgiven by

Therefore the phase difference between VandIis

In pure resistor, the current I always in phase with the

voltage Vand constant with time.

Figure 21.8a shows the variation of VandIwith time while

Figure 21.8b shows the phasor diagram for VandIin a pureresistor.

tII sin0

IRVR

00 VRI RtI sin0 and

VtVVR sin0where tagesupply vol:V

0 tt


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PHYSICS CHAPTER 21

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Impedance in a pure resistor

From the definition of the impedance, hence

t0

0I

0V

0

I0V

TT2

1 T2T

2

3


VI

(21.11)


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PHYSICS CHAPTER 21

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Pure capacitor means that no resistance and self-inductance

effect in the AC circuit.

Phase difference between voltage Vand current I

Figure 21.9 shows an AC source connected to a pure capacitor

C.

The alternating voltage across the capacitor VCat any instant is

equal to the supply voltage Vand is given by

21.3.4 Pure capacitor in an AC circuit

Figure 21.9

AC source

CV

V

C

I

tVVVC sin0


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PHYSICS CHAPTER 21

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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

tCVdt

dI sin0

tdt

dCV sin0

tCV cos0 00 ICV and

tII cos0OR

2sin

0

tII


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PHYSICS CHAPTER 21

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Therefore the phase difference between VandIis

In the pure capacitor,

the voltage Vlags behind the current I by /2 radians.

OR

the current I leads the voltage V by /2 radians.

Figure 21.10a shows the variation ofV

andI

with time while

Figure 21.10b shows the phasor diagram for VandIin a purecapacitor.

2

tt

rad2


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PHYSICS CHAPTER 21

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Impedance in a pure capacitor



0

0

I

VZ

t0

0I0V

0I0

V

TT

2

1 T2T

2

3

VI

rad2

00 CVI and

0

0

CV

V


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PHYSICS CHAPTER 21

30

From the eq. (21.13), the relationship between capacitive

reactanceXCand frequency fcan be shown by using a graph

in Figure 21.11.

f0

CX

fXC

1

Figure 21.11

Pure inductor means that no resistance and capacitance

effect in the AC circuit.Phase difference between voltage Vand current I

Figure 21.12 shows an AC source connected to a pure inductor

L.

21.3.5 Pure inductor in an AC circuit


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PHYSICS CHAPTER 21

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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

Figure 21.12

tII sin0

dtdILB

tIdt

dL sin0

tLI cos0B (21.14)


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PHYSICS CHAPTER 21

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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. (21.15).Thus

Therefore the phase difference between VandI is

In the pure inductor,

the voltage Vleads the current I by /2 radians.

OR

the current I lags behind the voltage V by /2 radians.

BVtLI cos0 00 VLI and

tVV cos0OR

2sin0 tVV

rad22

tt


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PHYSICS CHAPTER 21

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Figure 21.13a shows the variation of VandIwith time while

Figure 21.13b shows the phasor diagram for VandIin a pure

inductor.

t0

0I

0

V

0I0V

TT2

1 T2T

2

3

V

I

rad2



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PHYSICS CHAPTER 21

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Impedance in a pure inductor


whereXL is known as inductive reactance.

0

0

IVZ 00 LIV and

0

0

I

LI

LXLZ

and

(21.15)

inductortheofinductance-self:LsourceACoffrequency:f


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PHYSICS CHAPTER 21

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Inductive reactance is the opposition of a inductor to the

alternating current flows and is defined by

Inductive reactance is a scalar quantity and its unit is ohm ().

From the eq. (21.16), the relationship between inductive

reactanceXL and the frequencyfcan be shown by using agraph in Figure 21.14.

Figure 21.14

f0

LX

fXL

(21.16)

PHYSICS CHAPTER 21


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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.

Solution :

a. The capacitive reactance of the capacitor is given by

Therefore the capacitance of the capacitor is

Example 21.3 :

V14Hz;60A;1021 rms3

rms

VfI

CXIV rmsrms

fCXC

2

1

CX3102114

C6021

667

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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Solution :

b. The capacitive reactance is inversely proportional to the

frequency, so the capacitive reactance will decrease if the

frequency increases. Since the current in the capacitor is

inversely proportional to the capacitive reactance, therefore

the current will increase when the capacitive reactance

decreases.c. Given

The capacitive reactance is

Hence the new rms current in the capacitor is given by

V14Hz;60A;1021 rms3

rms

VfI

Hz410f

fC

XC

2

1

6

1098.34102

1

CX

CXIV rmsrms 5.9714 rmsI

PHYSICS CHAPTER 21


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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.

Solution :

a. The inductive reactance of the inductor is given by

Thus the rms current in the circuit is

Example 21.4 :

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

fLXL 2 33 10290.01000.12

LXIV rmsrms 82.12.12 rmsI

PHYSICS CHAPTER 21


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Solution :

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 3f

fLXL 2

33 10290.01050.22

LXIV 00

56.422.12 0I

and 2rms0 VV

LXIV 0rms 2

PHYSICS CHAPTER 21


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RCseries circuit

Consider an AC source of rms voltage Vis connected in series

to a resistor R and a capacitor Cas shown in Figure 21.15a.

The rms current I passes through the resistor and the

capacitor is equal because of the series connection between

both components.

21.3.6 RC, RL and RCL series circuit

AC source

R

I

RV

V

CV

C

Figure 21.15a

PHYSICS CHAPTER 21


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I

The rms voltages across the resistorVR

and the capacitor

VC

are given by

The phasor diagram of theRCseries circuit is shown in Figure21.15b.

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

IRVR and

CC IXV

where anglephase:

CV

RV

V

Figure 21.15b: phasor diagram

is an angle between the rmscurrent I and rms supply (or

total) voltage Vof AC circuit.

22

CR VVV 22

CIXIRV

22

C

XRIV (21.17)

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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Rearrange the eq. (21.18), thus the impedance of RCseriescircuit is

From the phasor diagram in Figure 21.15b , the current I leads

the supply voltage Vby radians where

A phasor diagram in terms ofR,XCandZis illustrated in Figure21.15c.

I

VZand

22

CXRI

V

22

CXRZ (21.18)

R

C

VVtan IR

IXCtan

R

XCtan (21.19)

CX Z

R

Figure 21.15c

PHYSICS CHAPTER 21


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RL series circuit


to a resistor R and an inductorL as shown in Figure 21.16a.

The rms voltages across the resistorVR and the inductor VLare given by

AC source

R

I

RV

V

L

LV

Figure 21.16a

IRVR and

LL IXV

PHYSICS CHAPTER 21


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


LVV

I

Figure 21.15b: phasor diagramRV

22

LR VVV

22LIXIR

22

LXRIV (21.20)

PHYSICS CHAPTER 21


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Rearrange the eq. (21.20), thus the impedance of RL seriescircuit is

From the phasor diagram in Figure 21.16b , the supply voltage

Vleads the current I the by radians where

The phasor diagram in terms ofR,XL andZis illustrated inFigure 21.16c.

I

VZand

22

LXRI

V

22

LXRZ (21.22)

R

L

VVtan

IRIXLtan

R

XLtan (21.23)

Figure 21.16c

LXZ

R

PHYSICS CHAPTER 21


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RCL series circuit


to a resistor R, a capacitor Cand an inductor L as shown inFigure 21.17a.

The rms voltages across the resistorVR, the capacitor VCand the inductor V

Lare given by

Figure 21.17a

IRVR

and

CC IXV

AC source

I

V

R

RV CV

C L

LV

LL IXV

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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The phasor diagram of the RL series circuit is shown in Figure21.17b.


I

Figure 21.17b: phasor diagram

22

CLR VVVV 22

CL IXIXIR

(8.24)

LV

R

V

V

CV

CL VV


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PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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is defined as the phenomenon that occurs when the

frequency of the applied voltage is equal to the frequency

of the RCL series circuit.

Figure 21.18 shows the variation of XC,XL,R andZwithfrequencyfof theRCL series circuit.

21.3.7 Resonance in AC circuit

Z

fXL

R

fXC

1

0 f

ZRXX LC ,,,

rfFigure 21.18

PHYSICS CHAPTER 21


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From Figure 21.18, the value of impedance is minimum Zmin

when

where its value is given by

This phenomenon occurs at the frequency fr

known as

resonant frequency.

At resonance in theRCL series circuit, the impedance is

minimumZmin

thus the rms current flows in the circuit is

maximum Imax

and is given by

(8.27)

22 CL XXRZ

02

min RZRZ min

(8.28)

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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maxI

Figure 21.19 shows the rms currentIinRCL series circuitvaries with frequency.

At frequencies above or below the resonant frequency fr,

the rms current I is less than the rms maximum current Imax

as shown in Figure 21.19.

0 f

I

rf

Figure 21.19

PHYSICS CHAPTER 21


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The resonant frequency,frof theRCL series circuit is given by

The series resonance circuit is used for tuning a radio

receiver.

CL XX

CL

1

LC

12 r2 fand

LC

f 12 2r

(21.29)

where frequencangularresonant:

Note:

At resonance, the current I and voltage Vare in phase.

PHYSICS CHAPTER 21


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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.

Solution :

a. The capacitive reactance of the inductor is given by

and the impedance of the circuit is

Example 21.5 :

Hz50V;12;1000F;102 6 fVRC

fCXC

2

1

61025021

CX

22CXRZ

2215921000 Z

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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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

CC IXV

15921038.6 3

Hz50V;12;1000F;102 6 fVRC

IZV

R

XC

tan

1000

1592tan

1

R

XC1

tan

OR


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PHYSICS CHAPTER 21


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Solution :

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

R

CL

V

VV

tan

153

115314tan 1

R

CL

V

VV1

tan

OR

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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Solution :

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


IRVR

LX88.5314

26153 I

26R

LL IXV

Hz50f

L5024.53 fLXL 2

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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Solution :

c. Meanwhile, the capacitive reactance is

thus the capacitance of the capacitor is

d. The resonant frequency is given by


CX88.5115 CC IXV 6.19CX

fC

XC2

1

C502

16.19

41062.1170.021

LC

f

2

1r

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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Exercise 21.1 :

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. : 2.0 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. : 2.83 k

, 2.83 k

; 93.8 nF; 200 ; 1.25 A

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

61

Exercise 21.1 :

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 frequency fis varied

from a low to a high value while the magnitude of Vis keptconstant.

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,7th edition, Nelkon & Parker, Q2, p.423)

ANS. : 400 Hz; 0.001 A; 1 V

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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Apply

a) average power,

b) instantaneous power,

c) power factor,

in AC circuit consisting ofR

,RC

,RL

andRCL

in series.

Learning Outcome:

21.4 Power and power factor (1 hour)

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h.matrik.ed

u.my/physics

PHYSICS CHAPTER 21


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21.4.1 Power of a pure resistor In a pure resistor, the voltage Vand current I are in phase,

thus the instantaneous powerP is given by

Figure 21.21 shows a graph of instantaneous powerPbeingabsorbed by the resistor against time t.

21.4 Power and power factor

tVtI sinsin 00

IVP

tVI 200 sin 000 PVI and

(8.30)

where powerum)peak(maxim:0P

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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The average (or mean) power Pav

being absorbed by the

resistor is given by

tPP 20 sin

Power being absorbed

Figure 21.21

avP

tPP 20av sin

0P

2

0P

t0

P

TT2

1 T2T

2

3

(21.31)

PHYSICS CHAPTER 21


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In a pure capacitor, the current I leads the voltage Vby /2

radians, thus the instantaneous powerP is given by

Figure 21.22 shows a graph of instantaneous powerPof the

pure capacitor against time t.

21.4.2 Power of a pure capacitor

tVtI sincos 00IVP

ttVI cossin00

(21.32)

ttt 2sin2

1

cossin and

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

66


of the pure capacitor is given

by

tPP 2sin

2

10


Figure 21.22

avP

tPP 2sin2

10av

t0

P

TT2

1 T2T

2

3

2

0P

Power being returned to supply

PHYSICS CHAPTER 21


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PHYSICS CHAPTER 21

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In a pure inductor, the voltage V leads the current I by /2

radians, thus the instantaneous powerP is given by

Figure 21.23 shows a graph of instantaneous powerPof thepure inductor against time t.

21.4.3 Power of a pure inductor

tVtI cossin 00IVP

ttVI cossin00

tPP 2sin2

10

ttt 2sin2

1

cossin and

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of the pure inductor is given

by

tPP 2sin

2

10


Figure 21.23

avP

tPP 2sin2

10av 0avP

2

0P

t0

P

TT2

1 T2T

2

3

Power being returned to supply

Note:

The term resistance is not used in pure capacitor and inductor because noheat is dissipated from both devices.

2

0P

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69

In an AC circuit in which there is a resistorR, an inductorL anda capacitor C, the average power Pav is equal to that dissipatedfrom the resistor i.e.

From the phasor diagram of the RCL series circuit as shown inFigure 21.24,

21.4.4 Power and power factor of R, RC, RL and

RCL series circuits

(21.33)

rms values

LV

IRV

V

CV

CL VV

Figure 21.24

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We get

then the eq. (21.33 ) can be written as

where cos is called the power factorof the AC circuit, Pr

is

the average real powerand I2Zis called the apparent power.

Power factor is defined as

cosVVR V

VRcos

cosav IVP IZVand

(21.34)

(21.35)

where IVZIP 2a powerapparent:Note:From the Figure 21.24, the power factor also can be calculated by using theequation below:

IZ

IR

V

VR cos (21.36)

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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 21.7 :

tV 100sin520

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Solution :

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 f2100

tV 100sin520 to the tVV sin010 srad100V;520

V

fCXC

2

1

6101009.1521

CX

fLXL 2 0.49.152


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Solution :

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

R

XX CLtan

35

100400tan

1

RXX CL1tan

OR

cosfactorpower

383cos .

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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 :

a. The capacitive reactance is

and the inductive reactance is

Example 21.8 :

fCXC

2

1

6103.325021

CX

fLXL 2

3105.22502

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


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Solution :

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

Rcos

139

105cos

cosav IVP Z

VIand

cos2

Z

V

755.0139

240 2

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78

Exercise 21.2 :

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, 3rd edition, James S. Walker, Q47, p.834)

ANS. : 0.962; 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, 3rd edition, James S. Walker, Q69, p.835)

ANS. : 3.44 mA, 10.5 nF

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Next ChapterCHAPTER 25 :

Quantization of light

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