Ch 21 Alternating Current
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Transcript of Ch 21 Alternating Current
Topic --- Alternating Current
21 ALTERNATING CURRENT
21.1
Alternating current
21.2
Root mean square (rms)
21.3
Resistance, reactance & impedance
21.4
Power and power factor
Topic --- Alternating Current
(a) Define alternating current
(AC)
(b) Sketch and analyse
sinusoidal AC waveform
(c) Use sinusoidal voltage and
current equations
tII sin0 tVV sin0
21.1 ALTERNATING CURRENT
Topic --- Alternating Current
Alternating current (AC) electricity is the type of electricity commonly used in
homes and businesses throughout the world. While direct current (DC)
electricity flows in one direction through a wire, AC electricity alternates its
direction in a back-and-forth motion.
The direction alternates between 50 and 60 times per second, depending on the
electrical system of the country.
AC created by an AC electric generator, which determines the frequency
the voltage can be readily changed, thus making it more suitable for long-distance transmission than DC electricity
can employ capacitors andinductors in electronic circuitry, allowing for a wide range of applications
21.1 ALTERNATING CURRENT
Topic --- Alternating Current
• is defined as an electric
current which magnitude
& direction change
periodically
• Symbol:
21.1 ALTERNATING CURRENT
Topic --- Alternating Current
CurrentVoltage The output of an ac
generator is sinusoidal and varies with time
locityangular ve ORfrequency angular :
currentpeak : 0I gepeak volta: 0V
time: t
)2( f In general,
21.1 ALTERNATING CURRENT
tII o sintVV o sin
Topic --- Alternating Current
21.2 ROOT MEANC SQUARE (rms)
(a) Define root mean square
(rms) current and voltage
for AC source
(b) Use the following formula,
and2
0rms
II
2
0rms
VV
Topic --- Alternating Current
rmsI
acdcpower averagepower
RIRI ave22
aveII 2
Mean or average current, Iave: the average or mean value of current in a half-cycle flows of current in a certain direction
0
2
0
ave
2III
avermsII 2
The r.m.s (root mean square) current means the square root of the average value of the current
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
• Since ac current
and
• So
and
• The root mean square (rms) current is the effective value of the AC
tII sin0
avermsII 2
avermstsinII
2
0
2
0rms
II
2
1sinsin 22
tt
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
• The equation:
• Most household electricity is 240 V AC which means that Vrms is 240 V
2
0rms
VV
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
A sinusoidal, 60.0 Hz, ac voltage is read to be 120 V by an ordinary voltmeter.
(a) What is the maximum valuethe voltage takes on during a cycle?
(b) b) What is the equation for the voltage?
Solution:
(a)
(b)
2o
rms
VV
)V(Vrmso
2
V 170
tsinVVo
tsin 120170
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
A voltage V= 60 sin 120πt is applied across a 20 Ω resistor.
(a)What will an ac ammeter in series with the resistor read?
(b)Calculate the peak current and mean power.
2o
rms
VV
R
VI rms
rms
)I(Irmso
2
20
II
rms
W90(20)(2.12) 2
2
RIP rmsav
Solution:(a)
(b)
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
1. An AC sourceV = 500 sin t
is connected across a resistor of 250 . Calculate
(a) the rms current in the resistor,(b) the peak current,(c) the mean power.
2. Figure below 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.
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
(a) Sketch and use phasor
diagram and sinusoidal
waveform to show the
phase relationship
between current and
voltage for a circuit
consisting of (i) pure resistor(ii)pure capacitor (iii)pure inductor
(b) Use phasor diagram to
analyse voltage, current,
and impedance of series
circuit of:
(i) RL
(ii)RC
(iii)RLC
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
(c) Define and use:
(i) capacitive reactance,
(ii) inductive reactance,
(ii)impedance,
(iv)phase angle,
(d) Explain
graphically the
dependence of
R, XC, XL and Z
on f and relate it
to resonance
fCXC
2
1
fLX L 2
22
CL XXRZ
R
XX CL tan
Topic --- Alternating Current
A diagram containing phasor is called phasor diagram
Phasor a vector that rotate
anticlockwise about its axis with constant
angular velocity
Used to represent a sinusoidally varying quantitysuch as alternating current (AC) & alternating voltage
Also being used to determine the phase angle the phase difference between current and voltage in AC circuit
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
Phase difference is
22
tt
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Note:
valuepositive
radian valuenegative
Leads
Lags behind
In antiphase
Topic --- Alternating Current
• The projection of OP on the y-axis is ON, represents the instantaneous value
• Ao is the
peak valueof the quantity
t0 TT2
1 T2T2
3
Ao
tAA o siny
ωN
O
P
y
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
• It is defined by
OR
• It is a scalar quantityand its unit is ohm ()
• In a DC circuit, impedance likes the resistance
rms
rms
I
VZ
2
0V
2
0I
0
0
I
VZ
21.3 RESISTANCE, REACTANCE & IMPEDANCE
V = IZ V = IR V = IX
V = IXC
V = IXL
Topic --- Alternating Current
Resistance, R: Opposition to current flow in purely resistive circuit
Reactance, X: Opposition to current flow resulting from inductance or capacitance in ac circuit
Capacitive reactance, XC: Opposition of a capacitor to ac
Inductive reactance, XL: Opposition of an inductor to ac
Impedance, Z: Total opposition to ac (Resistance and reactancecombine to form impedance)
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating CurrentAC is defined as an electric
current which magnitude
& direction change
periodically
In general,
Root mean square current (Irms) is defined as the effective value of a.c. which produces the same power as the steady d.c. when the current passes through the same resistor
2
0rms
II
Root mean square voltage/ p.d (Vrms): the value of the steady direct voltage which when applied across a resistor, produces the same power as the mean (average) powerproduced by the alternating voltage across the same resistor
2
0rms
VV
A diagram containing phasor is called phasor diagram
Phasor a vector that rotate
anticlockwise about its axis with constant
angular velocity
Impedance, Z
Resistance, R
in resistor
Reactance, X
Capacitive reactance, XC
in capacitor, C
Inductive reactance, XL
in inductor, L
V = IZ V = IR V = IX
V = IXC
V = IXL
Topic --- Alternating Current
• The circuit
• The alternating current passes through
the resistor is given by
• The alternating voltage across the
resistor VR at any instant is given by
where, V = supply voltage
tII sin0
IRVR
00 VRI RtI sin0 and
VtVVR sin0
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Pure Capacitor In AC Circuit
Topic --- Alternating Current
• From figure above: I = I0 sin tand V = V0 sin t
• Thus the phase difference is
Impedance in a pure resistor• From the
definition of the impedance, hence
t0
0I
0V
0I0V
TT2
1 T2T
2
3
ω
VI
0 tt
Therefore the current I is in phase with the voltage V and constant with time
RI
V
I
VZ
0
0
rms
rms
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Pure Capacitor In AC Circuit
Topic --- Alternating Current
CIRCUIT
AC source
R
I
RV
V
PHASE DIFFERENCEtII sin0
VtVVR sin0
0
IMPEDANCE, ZPHASORDIAGRAM
VI
RZ
In pure resistor, the current I always in phase with the voltage V and constant with time
RI
V
I
VZ
0
0
rms
rms
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Pure Capacitor In AC Circuit
Topic --- Alternating Current
• The circuit
• The alternating voltage across the capacitor VC at any instant is equal to the supply voltage V and is given by
• The charge accumulates at the plates of the capacitor is
21.3 RESISTANCE, REACTANCE & IMPEDANCE
AC source
CV
V
C
I
tVVVC sin0
CCVQ
tCVQ sin0
Pure Capacitor In AC Circuit
Topic --- Alternating Current
• The charge and current are related by
Hence the equation of AC in the capacitor is
and
OR
dt
dQI
tCVdt
dI sin0
tdt
dCV sin0
tCV cos0 00 ICV
tII cos0
2sin0
tII
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Pure Capacitor In AC Circuit
Topic --- Alternating Current
• From figure above: V = V0 sin t and I = I0 sin (t + /2)
• Thus the phase difference is
the voltage V lags
behind the current Iby /2 radians
ORthe current I leads
the voltage V by /2radians
21.3 RESISTANCE, REACTANCE & IMPEDANCE
t0
0I
0V
0I0V
TT2
1 T2T
2
3
ωVI
rad 2
22
tt
Pure Capacitor In AC Circuit
Topic --- Alternating Current
Impedance in a pure capacitor• From the definition of the
impedance, hence
and
and
where XC:Capacitive
(capacitative)reactance
• Capacitive reactance is the opposition of a capacitor to the alternating current flows and is defined by
• Capacitive reactance is a scalar quantity and its unit is ohm ()
0
0
I
VZ 00 CVI
0
0
CV
V
CXC
Z
1f 2
fCXC
2
1
0
0
rms
rms
I
V
I
VXC
f0
CX
fX C
1
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Pure Capacitor In AC Circuit
Topic --- Alternating Current
CIRCUITPHASEDIFFERENCE
IMPEDANCE, Z PHASOR DIAGRAM
21.3 RESISTANCE, REACTANCE & IMPEDANCE
AC source
CV
V
C
I
tVVVC sin0
2sin0
tII
rad 2
the voltage V lags behindthe current I by /2radians.
ORthe current I leads the voltage V by /2 radians
V
I
rad 2
fCXC
2
1
f0
CX
fX C
1
Pure Capacitor In AC Circuit
Topic --- Alternating Current
An 8.00 μF capacitor is connected to the terminals of an AC generator with an rmsvoltage of 150 V and a frequency of 60.0 Hz. Find the capacitive reactance rmscurrent and the peak currentin the circuit.
Solution:
21.2 ROOT MEAN SQUARE (rms)
fCXC
2
1
0
0
rms
rms
I
V
I
VXC
Pure Capacitor In AC Circuit
Topic --- Alternating Current
• The circuit
• 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
21.3 RESISTANCE, REACTANCE & IMPEDANCE
AC sourceV
I
L
LV
tII sin0
dt
dILB tI
dt
dL sin0
tLI cos0B
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
2sin 0
tLIVLB
Pure Inductor In AC Circuit
Topic --- Alternating Current
• From figure above: I = I0 sin tand V = V0 sin (t + /2)
• Thus the phase difference is
In the pure inductor, the voltage V leads
the current I by /2radians
ORthe current I lags
behind the voltage Vby /2 radians
21.3 RESISTANCE, REACTANCE & IMPEDANCE
22
tt
t0
0I
0V
0I0V
TT2
1 T2T
2
3
V
I
rad 2
ω
Pure Inductor In AC Circuit
Topic --- Alternating Current
Impedance in a pure inductor• From the definition of the
impedance, hence
where XL:inductive reactance
• 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 ()
21.3 RESISTANCE, REACTANCE & IMPEDANCE
0
0
I
VZ 00 LIV and
0
0
I
LI
LXLZ
fLX L 2
f 2and
0
0
rms
rms
I
V
I
VX L
f0
LX
fX L
Pure Inductor In AC Circuit
Topic --- Alternating Current
CIRCUITPHASEDIFFERENCE
IMPEDANCE, Z PHASOR DIAGRAM
21.3 RESISTANCE, REACTANCE & IMPEDANCE
the voltage V leads the current I by /2 radians.
ORthe current I lags behindthe voltage V by /2radians
AC sourceV
I
L
LV
tII sin0tVV cos0
OR
2sin0
tVV
rad 2
V
I
rad 2
f0
LX
fX L
fLX L 2
Pure Inductor In AC Circuit
Topic --- Alternating Current
A coil having an inductance of 0.5 H is connected to a 120 V, 60 Hz power source. If the resistance of the coil is neglected, what is the effective current through the coil.
A 240 V supply with a frequency of 50 Hz causes a current of 3.0 A to flow through an pure inductor. Calculate the inductance of the inductor.
0
0
rms
rms
I
V
I
VX L
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
Since the voltage on a capacitor to the charge on it, the current must lead the voltage in time & phase to conduct charge to the capacitor plates and raise the voltage
V
I
rad 2
When a voltage is applied to an
inductor, it resists the change in
current. The current builds up more slowly than the
voltage, lagging it in time and phase
V
I
rad 2
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating CurrentPure Capacitor in AC Circuit Pure Inductor in AC Circuit
V
I
rad 2
the voltage V leads the current I by /2radians
ORthe current I lags behind the voltage Vby /2 radians
tII sin0
2sin0
tVV
rad 2
IV
tVVVC sin0
2sin0
tII
rad 2
IV
V
I
rad 2
the voltage V lags behind the current Iby /2 radians
ORthe current I leads the voltage V by /2 radians
tII sin0
VtVVR sin0
0
In pure resistor, the current I always in phase with the voltage V and constant with time
VI
Topic --- Alternating Current
Q1 A capacitor has a rms
current of 21 mA at a frequency of 60 Hz when the rms voltage across it is 14 V.(i) What is the capacitance of the capacitor?(ii) If the frequency is increased, will the current in the capacitor increase, decrease or stay the same? Explain.(iii) Calculate the rmscurrent in the capacitor at a frequency of 410 Hz.
Q2 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(i) the current flowing,(ii) the voltage across the capacitor,(iii) the phase angle of the circuit.
Q3 A rms voltage of 12.2
V with a frequency of 1.00 kHz is applied to a 0.290 mH inductor.(i) What is the rmscurrent in the circuit?(ii) Determine the peak current for a frequency of 2.50 kHz.
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
AC source
R
I
RV
V
CV
C
IRVR CC IXV
22
CR VVV
Phasor diagram: the current Ileads the supply voltage V by radians
R
C
V
Vtan
R
XCtan
CXZ
R
22
CXRZ
fCXC
2
1
22
2 1
CRZ
A phasor diagram in terms of R, XC and ZI
CV
RV
V
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
An alternating current of angular frequency of 1.0 x 104
rad s-1 flows through a 10 kresistor and a 0.10 F capacitor which are connected in series. Calculate the rmsvoltage across the capacitor if the rms voltage across the resistor is 20 V.
Solution:
fCXC
2
1
0
0
rms
rms
I
V
I
VXC
RC Series Circuit
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
IRVR
Phasor diagram: the supply voltage V leads the
current I the by radians
22
2 1
CRZ
A phasor diagram in terms of R, XL and Z
AC source
R
I
RV
V
L
LV
LL IXV
LV
V
IRV
22
LR VVV
R
L
V
Vtan
LX
Z
RR
X Ltan
22
LXRZ
fLX L 2
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
IRVR
LL IXV
CC IXV
AC source
I
V
R
RV CV
C L
LV
I
LV
RV
V
CV
CL VV
22
CL XXRIV
R
XX CL tan
LX
Z
CX
CL XX
R
22
CL XXRZ
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Phasor diagram: the supply voltage V
leads the current I the
by radians
Topic --- Alternating Current
A series RLC circuit has a resistance of 25.0 Ω, a capacitance of 50.0 μF, and an inductance of 0.300 H. If the circuit is driven by a 120 V, 60 Hz source, calculate
(a) The total impedance of the circuit
(b) The rms current in the circuit
(c) The phase angle between the voltage and the current.
Suggested Answer:
RCL Series Circuit
21.3 RESISTANCE, REACTANCE & IMPEDANCE
• 64.9 Ω , 1.85 A, 67.3o
Topic --- Alternating Current
• is defined as the phenomenon that occurs when the frequency of the applied voltage is equal to the frequency of the RCL series circuit
• Since resonance in series RLC circuit occurs at particular frequency, so it is used for filtering and tuning purpose as it does not allow unwanted oscillations that would otherwise cause signal distortion, noise and damage to circuit to pass through it
• Figure below shows the variation of XC,
XL, R and Z with frequency f of the RCL
series circuit
Z
fX L
R
fXC
1
0 f
Z
rf
•at low frequency,
impedance Z is large
because 1/ωCis large
• at high frequency,
impedance Z is high
because ωL is large
Topic --- Alternating Current
• From the graph, Zmin at fr
• This will happen when
22
CL XXRZ
02min RZ
RZ min
fr : resonant frequency
R
V
Z
VI
min
max
• The resonant frequency, fr of the RCL series circuit is given by
CL XX
CL
1
LC
12
LC
f1
22
r
LCf
2
1r
• At resonance in the RCL series circuit,
the impedance is minimum Zmin
thus the rms current flows in the circuit is maximum Imax and is given by
21.3 RESISTANCE, REACTANCE & IMPEDANCE
The resistance in the circuit is only came from R
CL XX
Topic --- Alternating Current
A 200 resistor, a 0.75 H inductor and a capacitor of capacitance C are connected in series to an alternating source 250 V, fr = 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(e) Sketch the phasor diagram.
Suggested Answer:(a)
(b)
(c)
(d)
(e)
RCL Series Circuit
21.3 RESISTANCE, REACTANCE & IMPEDANCE
k 2.83
k 2.83
C
L
X
LX
nF 93.9 ,2
1 x102.83 3 C
fC
A1.25R
V
Z
VI rmsrms
rms
VL
VC
VR
I
200RZ
Topic --- Alternating Current
1. Based on the RCL series
circuit in Figure above, 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.
(b) Calculate(i) the current flows in the circuit
if the resistance of the
resistor R is 26 ,
(ii) the inductance and
capacitance if the frequency
of the AC source is 50 Hz,
(iii) the resonant frequency.
21.2 ROOT MEAN SQUARE (rms)
Topic --- Alternating Current
Q2
• 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.
Q3
• 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: 2.0 V
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
Q4 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. 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
Q5 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 Vis 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,7th edition, Nelkon
& Parker, Q2, p.423)• ANS: 400 Hz; 0.001 A; 1 V
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
V
V V
22
LR VVV
22
LXRIV
22
LXRZ
IZV 21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
A circuit is made up of a 3200 pF capacitor connected in series to a 30 H coil of resistance 4 . Calculate(i) impedance at frequency
30 kHz.(ii)resonant frequency.
Solution:C = 3200 10-12 F; L = 3010-6
H; R = 4
(i) Given f = 30103 Hz, The reactance of capacitor and inductor are
PSPM 2009/ 10: Q12(c):
fCXC
2
1
123 10320010302
1
1066.1 3
CX
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
(i) and
Therefore the impedance is givenby
(ii) Apply:
fLX L 2
63 103010302
22
CL XXRZ
2321066.166.54
1654Z
66.5LX
126 10320010302
1
Hz 1014.5 5rf
LCfr
2
1
21.3 RESISTANCE, REACTANCE & IMPEDANCE
Topic --- Alternating Current
Apply
(i) average power,
(ii) instantaneous power,
(iii)power factor,
in AC circuit consisting of R, RC, RL and RCL in series.
21.4 POWER & POWER FACTOR
• Power factor is a way of measuring how efficiently electrical power is being used within a facility's electrical system
cosVIPrmsrms
av
IVP
IV
P
P
P av
a
rcos
Topic --- Alternating Current
• In an ac circuit , the power is only dissipated by a resistance, none is dissipated by inductance or capacitance
• From the phasor diagram of the RCL series circuit
21.4 POWER & POWER FACTOR
RIIVP R2
av
ωLV
IRV
V
CV
CL VV
• We get
• Thenand
cosVVR V
VRcos
cosav IVP IZV
r2
av cos PZIP
where cos is called the
power factor of the AC circuit,
Pr is the average real
power and I2Z
is called the apparent
power
Topic --- Alternating Current
• Power factor is defined as
• From
• the power factor also can be calculated by using the equation below
• When = 0o (cos =+1) ,the circuit is completely resistive or when the circuit is in resonance (RCL)
• When = +90o (cos = 0), the circuit is completely inductive
• When = -90o (cos =0), the circuit is completely capacitive
a
r
2
rcosP
P
ZI
P
IZ
IR
V
VR cos
Z
Rcos
22
CL XXRIV
21.4 POWER & POWER FACTOR
Topic --- Alternating Current
An oscillator set for 500 Hz puts out a sinusoidal voltage of 100 V effective. A 24.0 Ω resistor, a 10.0μF capacitor, and a 50.0 mHinductor in series are wired across the terminals of the oscillator.(a)What will an ammeter in the circuit read?(b)What will a voltmeter read across each element?(c)What is the real power dissipated in the circuit?(d)Calculate the power supply(e)Find the power factor(f) What is the phase angle?
Z
Rcos 1
Z
VI rms
rms
CC
LL
R
IXV
IXV
IRV
cosVIPrmsave
rmsrmsplysupVIP
Z
Rcos
Z
Rcos 1
21.4 POWER & POWER FACTOR
Topic --- Alternating Current
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. V = 520 sin 100t. 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.
21.4 POWER & POWER FACTOR
Topic --- Alternating Current
Suggested Answer:
By comparing
V = 520 sin 100t
to the, V = V0 sin t
Thus
V0 = 520V, = 100 rad s-1
(a) The frequency of AC source is
given by
(b) The capacitive reactance is
and the inductive reactance is
f 2
Hz 9.15ff2100
fCXC
2
1
100CX
6101009.152
1
CX
fLX L 2
400LX
0.49.152
21.4 POWER & POWER FACTOR
Topic --- Alternating Current
(c) The impedance of the circuit is
(d) The peak current in the circuit is
(e) The phase angle between the current and the supply voltage is
ORf. The power factor of the circuit is given by
22CL XXRZ
2210040035
302Z
ZIV 00
302520 0I
A 72.10 I
R
XX CL tan
35
100400tan 1
rad 45.1
R
XX CL1tan
3.83
cosfactorpower 383cos .
117.0factorpower
21.4 POWER & POWER FACTOR
Topic --- Alternating Current
Q1 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.
ANS: 0.755, 313 W
Q2 A coil having inductance 0.14
H and resistance of 12 is connected to an alternating source 110 V, 25 Hz. Calculate(a) the rms current flows in the coil(b) the phase angle between the current and supply voltage(c) the power factor of the circuit(d) the average power loss in the coil.ANS: 4.4 A, 61.3o , 0.48, 0.23
kW
Q3 A series RCL circuit
contains a 5.10 μF capacitor and a generator whose voltage is 11.0 V. At a resonant frequency of 1.30 kHz the power dissipated in the circuit is 25.0 W. Calculate(a) the inductance(b) the resistance(c) the power factor when the generator frequency is 2.31 kHz.
ANS: 2.94 x 10-3 H , 4.84 Ω , 0.163
21.4 POWER & POWER FACTOR
Topic --- Alternating Current
Q4
• An RLC circuit has a resistance of 105 , an inductance of 85.0 mH and a capacitance of 13.2 F.
• What is the power factor of the circuit if it is connected to a 125 Hz AC generator?
• 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
Q5
• A 1.15 k resistor and a 505 mH inductor are connected in series to a 14.2 V,1250 Hz AC generator.
• What is the rms current in the circuit?• What is the capacitance’s 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
21.4 POWER & POWER FACTOR