chapter21 student
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PHYSICS CHAPTER 21
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CHAPTER 21:
Alternating current
(5 Hours)
www.kms
.matrik.e
du
.my/physics
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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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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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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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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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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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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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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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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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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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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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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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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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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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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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At the end of this chapter, students should be able to:
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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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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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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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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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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Impedance in a pure resistor
From the definition of the impedance, hence
t0
0I
0V
0
I0V
TT2
1 T2T
2
3
Figure 21.8aFigure 21.8b: phasor diagram
VI
(21.11)
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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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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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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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Impedance in a pure capacitor
From the definition of the impedance, hence
Figure 21.10aFigure 21.10b: phasor diagram
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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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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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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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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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
Figure 21.13aFigure 21.13b: phasor diagram
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Impedance in a pure inductor
From the definition of the impedance, hence
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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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)
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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
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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
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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
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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
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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
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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)
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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
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RL series circuit
Consider an AC source of rms voltage Vis connected in series
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
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The phasor diagram of the RL series circuit is shown in Figure8.16b.
Based on the phasor diagram, the rms supply voltage V(or totalvoltage) of the circuit is given by
LVV
I
Figure 21.15b: phasor diagramRV
22
LR VVV
22LIXIR
22
LXRIV (21.20)
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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
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RCL series circuit
Consider an AC source of rms voltage Vis connected in series
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
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The phasor diagram of the RL series circuit is shown in Figure21.17b.
Based on the phasor diagram, the rms supply voltage V(or totalvoltage) of the circuit is given by
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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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
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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)
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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
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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.
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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
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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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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
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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
V314;V115V;153 LCR VVV
IRVR
LX88.5314
26153 I
26R
LL IXV
Hz50f
L5024.53 fLXL 2
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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
V314;V115V;153 LCR VVV
CX88.5115 CC IXV 6.19CX
fC
XC2
1
C502
16.19
41062.1170.021
LC
f
2
1r
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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
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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
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At the end of this chapter, students should be able to:
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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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
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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)
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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
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The average (or mean) power Pav
of the pure capacitor is given
by
tPP 2sin
2
10
Power being absorbed
Figure 21.22
avP
tPP 2sin2
10av
t0
P
TT2
1 T2T
2
3
2
0P
Power being returned to supply
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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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The average (or mean) power Pav
of the pure inductor is given
by
tPP 2sin
2
10
Power being absorbed
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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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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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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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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