Upload
ally-violet
View
253
Download
0
Embed Size (px)
Citation preview
7/30/2019 Note Unit8 Sf027
1/57
SF027 1
UNIT 8: ALTERNATINGCURRENT (A.C.)
7/30/2019 Note Unit8 Sf027
2/57
SF027 2
8.1 Alternating Current (a.c.) Definition is defined as an electric current whose magnitude and
direction change periodically.
Figures 8.1a, 8.1b and 8.1c show three forms of alternating current.
T
I
t0 T2
1 T2
0I
0I
T2
3 Fig. 8.1a : Sinusoidal a.c.
T
I
t0 T2
1 T2
0I
0I
T
2
3Fig. 8.1b : Saw-tooth a.c.
7/30/2019 Note Unit8 Sf027
3/57
SF027 3
When an a.c flows through a resistor, there will be a potentialdifference (voltage) across it and this voltage is alternating voltage as
shown in figure 8.1d.
Fig. 8.1c : Square a.c.T
I
t0 T2
1 T2
0I
0I
T2
3
T
V
t0 T2
1 T2
0V
0V
T2
3
Fig. 8.1d : Sinusoidal alternatingvoltage
gepeak volta:oVwhere
period:T
currentpeak:oI
7/30/2019 Note Unit8 Sf027
4/57
SF027 4
8.1.1 Terminology in a.c.
Frequency (f) Definition: Number of complete cycle in one second.
Unit: hertz (Hz) ors-1
Period (T)
Definition: Time taken for one complete cycle.
Unit: second (s)
Equation :
Peak (maximum) current (Io) Definition: Magnitude of the maximum current.
8.1.2 Equation for alternating current and voltage.
Equation for the current (I) :
Equation for the voltage (V) :
f
1T
tII 0 sin
where frequencangular:
currentpeak:0I
phase
tVV 0 sin
gepeak volta:0V
(8.1a)
(8.1b)
7/30/2019 Note Unit8 Sf027
5/57
SF027 5
8.2 Mean or Average Current (Iav)
Definition is defined as the average or mean value of current in ahalf-cycle flows of current in a certain direction.
Equation :
Note :
Iav
forone complete cycle is zero because the current flows in
one direction in one-half of the cycle and in the opposite direction inthe next half of the cycle.
8.3.1 Root mean square current (Irms) In calculating average power dissipated by an a.c., the mean (average)
current is not useful. The instantaneous power, P delivered to a
resistanceRis
2
ooav
I
I2I
8.3 Root Mean Square Values
RIP 2
(8.2a)
7/30/2019 Note Unit8 Sf027
6/57
SF027 6
The average power, Pav over one cycle is given by
where is the average ofI2 over one cycle and can be written as
therefore the eq. (8.3a) can be written as
Since then eq. (8.3b) becomes
Using a double-angle formula for trigonometry and trigonometryidentity,
thus
2I
RIP 2av
2
rms II
2rms
2 II
RIP 2rmsav
tII 0 sin
(8.3a)
(8.3b)
(8.3c)
tII22
0rms sin
2 22 sincoscos 1 22 sincos and212 2sincos
212
12 cossin and t
t212I
I
2
0
rms cos
7/30/2019 Note Unit8 Sf027
7/57
SF027 7
The mean value ofcos2tfor one cycle is zero, finally eq. (8.3b) canbe written as
Root mean square current (Irms) is defined as the value of thesteady d.c. which produces the same power in a resistor as themean (average) power produced by the a.c.
The root mean square (rms) current is the effective value of the a.c.
8.3.2 Root mean square voltage/p.d. (Vrms
)
Definition is defined as the value of the steady direct voltagewhich when applied across a resistor, produces thesame power as the mean (average) power produced bythe alternating voltage across the same resistor.
Its formula is
Note :
Eq. (8.3d) and eq. (8.3e) are valid only for a sinusoidal alternatingcurrent and voltage (p.d.)
2
II 0rms (8.3d)
2VV 0rms (8.3e)
7/30/2019 Note Unit8 Sf027
8/57
SF027 8
Example 1 :
An a.c. source V=200 sin tis connected across a resistor of 100 .Calculate
a. the r.m.s. current in the resistor.
b. the peak current.c. the mean power.
Solution: R=100
then the peak voltage,V0 = 200 V
a. The r.m.s. current is
b. The peak current is
R
VI rmsrms
t200V sin
A411Irms .
tVV 0 sin
2R
VI 0rms
compare with
and2
VV 0rms
2
II 0rms
A991I0 .
7/30/2019 Note Unit8 Sf027
9/57
SF027 9
c. By applying the equation of mean power, thus
Example 2 :
The alternating potential difference shown above is connected across aresistor of 10 k. Calculate
a. the r.m.s. current,
b. the frequency,
c. the mean power dissipated in the resistor.
RIP 2rmsav
W199Pav
and
040.
)(VoltV
)second(t0 020. 080.
200
200
060.
7/30/2019 Note Unit8 Sf027
10/57
SF027 10
Solution: R=10x103
From the graph, V0 = 200 V and T=0.04 sa. The r.m.s. current is
b. The frequency of the a.c. is
c. By applying the equation of mean power, thus
R
VI rmsrms
A10x411I2
rms
.2R
VI 0rms
and2
VV 0rms
T
1f
Hz025f .
RIP 2rmsav
W991Pav .
7/30/2019 Note Unit8 Sf027
11/57
SF027 11
8.4 Phasor diagram Phasor is defined as a vector that rotate anticlockwise about its
axis with constant angular velocity.
A diagram containing phasor is called phasor diagram.
It is used to represent a sinusoidally varying quantity such asalternating current (a.c.) and alternating voltage (a.v.).
It also being used to determine the phase difference between currentand voltage in a.c. circuit.
8.4.1 In Phase
Consider a graph represents sinusoidal a.c. and sinusoidal a.v. asshown in figure 8.4a.
t0
0V
0V
0I
0I
TT2
1 T2T
2
3
Fig. 8.4a
I V
Fig. 8.4b : Phasor diagram
7/30/2019 Note Unit8 Sf027
12/57
SF027 12
From the figure 8.4a :
Thus the phase difference is
Conclusion : The currentIis in phase with the voltage Vand constantwith time.
8.4.2 Lead
Consider a graph represents sinusoidal a.c. and sinusoidal a.v. asshown in figure 8.4c.
t0
00 IV &
00 IV &
TT2
1 T2T2
3
Fig. 8.4c
tII 0 sin
0
tVV 0 sin
tt
I
V
Fig. 8.4d : Phasor diagram
7/30/2019 Note Unit8 Sf027
13/57
SF027 13
From the figure 8.4c :
Thus the phase difference is
Conclusion : The voltage V leads the currentIby /2 radians andconstant with time.
8.4.3 Lags behind
Consider a graph represents sinusoidal a.c. and sinusoidal a.v. as
shown in figure 8.4e.
t0
00 IV &
00 IV &
TT2
1 T2T
2
3
Fig. 8.4e
tII 0 sin
rad2
tVV 0 cos
t2t
IV
Fig. 8.4f : Phasor diagram
2
tVV 0 sin
7/30/2019 Note Unit8 Sf027
14/57
SF027 14
From the figure 8.4e :
Thus the phase difference is
Conclusion : The voltage V lags behind the currentIby /2 radiansand constant with time.
The quantity that measures the opposition of a circuit to the a.c. flows.
It is defined by
It is a scalar quantity and its unit is ohm ( )
In a d.c. circuit, impedance is the resistanceR.
tII 0 sin
rad2
tVV 0 cos
t2t
2
tVV 0 sin
8.5 Impedance (Z)
rms
rms
I
VZ
2
V0
2
I0
or
0
0
I
VZ (8.5a)
7/30/2019 Note Unit8 Sf027
15/57
SF027 15
8.6 Pure Resistor in the A.C. Circuit The symbol of an a.c. source in the electrical circuit is shown in figure
8.6a.
Pure resistor means that no capacitance and self-inductance effectin the a.c. circuit.
8.6.1 Phase difference between Voltage Vand currentI
Figure 8.6b shows an a.c. source connected to a pure resistorR.
Fig. 8.6a
The current flows in the resistor isgiven by
The voltage across the resistorVR at
any instant is given by
tII 0 sin
IRVR tRIV 0R sin
VtVV 0R sin
and00 VRI
Fig. 8.6b
a.c. source
R
I
RV
V
where tageSupply vol:V
7/30/2019 Note Unit8 Sf027
16/57
SF027 16
Therefore the phase difference between VandIis
In pure resistor, the currentIis in phase with the voltage V.
Figure 8.6c shows the variation ofVandIwith time while Figure 8.6d
shows the phasor diagram forVandIin a pure resistor.
0tt
t0
0V
0V
0I
0I
TT2
1 T2T
2
3
Fig. 8.6c
I V
Fig. 8.6d : Phasor diagram
8 6 2 I d i i t
7/30/2019 Note Unit8 Sf027
17/57
SF027 17
8.6.2 Impedance in a pure resistor
From the definition of the impedance, hence
8.6.3 Power in a pure resistor
Since VandIare in phase, the instantaneous powerP is given by
Therefore the graph of variation of power with time is shown in figure8.6e.
tPP2
0 sin
IVP
RI
V
I
VZ
0
0
rms
rms
tVtIP 00 sinsin
tVIP2
00 sin and 000 PVI
(8.6a)
(8.6b)
tPP 20 sin
where powerum)peak(maxim:0P
t0
0P
TT21 T2T
23
Fig. 8.6e
)(PPower Power being absorbed
The a erage (mean) po er P in a resistor is gi en b
7/30/2019 Note Unit8 Sf027
18/57
SF027 18
The average (mean) powerPav in a resistor is given by
Pure capacitor means that no resistance and self-inductance effectin the a.c. circuit.
8.7.1 Phase difference between Voltage Vand currentI Figure 8.7a shows an a.c. source connected to a pure capacitorC.
tPP 20av sin
2
VIP 00
av
and
2
1t
2 sin
0av
P2
1P or (8.6c)
8.7 Pure Capacitor in the A.C. Circuit
The voltage across the capacitorVCatany instant is equal to the supply
voltage Vand is given by
The charge accumulates at the platesof the capacitor is
tCVQ 0 sin
CCVQ
C0 VtVV sin
Fig. 8.7a
a.c. source
CV
V
C
I
Charge and current are related by
7/30/2019 Note Unit8 Sf027
19/57
SF027 19
Charge and current are related by
Hence the equation of a.c. in capacitor is
Therefore the phase difference between VandIis
In pure capacitor, the currentIleads the voltage V by /2 radians.
dt
dQI
tCVI 0 cos
tCVdtd
I 0 sin
tdt
dCVI 0 sin
and
00
ICV
tII 0 cos or
2tII 0
sin
rad2
t2
t
7/30/2019 Note Unit8 Sf027
20/57
SF027 20
Figure 8.7b shows the variation ofVandIwith time while Figure 8.7c
shows the phasor diagram forVandIin a pure capacitor.
8.7.2 Impedance in a pure capacitor
From the definition of the impedance, hence
t0
0V
0V
0I
0I
TT
2
1 T2T
2
3
Fig. 8.7b
rad2
0
0
CV
VZ
0
0
I
VZ and 00 CVI
I V
Fig. 8.7c : Phasor diagram
1
7/30/2019 Note Unit8 Sf027
21/57
SF027 21
whereXCis known as capacitive (capacitative) reactance.
Capacitive reactance is the opposition of a capacitor to the currentflows and is defined by
The unit of capacitive reactance is ohm ( )
From eq. (8.7a), the relationship between capacitive reactanceXCand
frequencyfcan be shown by using graph in figure 8.7d.
CXC
1Z
(8.7a)fC2
1XC
and f2
0
0
rms
rmsC
I
V
I
VX (8.7b)
sourca.c.offrequency:fcapacitortheofecapacitanc:C
f0
Fig. 8.7d
CX
f
1XC
8 7 3 P i it
7/30/2019 Note Unit8 Sf027
22/57
SF027 22
8.7.3 Power in a pure capacitor
Since the currentIleads the voltage V by /2 radians, the
instantaneous powerP is given by
Therefore the graph of variation of power with time is shown in figure8.7e.
t2P2
1P 0 sin
IVP
tVtIP 00 sincosttVIP 00 cossin and 000 PVI
(8.7c)
t22
1PP 0 sin
t0
2
P0
2
P0
TT2
1 T2T
2
3Fig. 8.7e
)(PPower
t2P2
1P 0 sin
Power being absorbed
Power being returned to supply
Th ( ) P i it i
7/30/2019 Note Unit8 Sf027
23/57
SF027 23
The average (mean) powerPav in a capacitor is
Pure inductor means that no resistance and capacitance effect in thea.c. circuit.
8.8.1 Phase difference between Voltage Vand currentI
Figure 8.8a shows an a.c. source connected to a pure inductorL.
t2P2
1P 0av sin and 0t2 sin
0Pav
8.8 Pure Inductor in the A.C. Circuit
The current flows in the inductor isgiven by
When the current flows in the inductor,the back e.m.f. caused by the selfinduction is produced and given by
tLI0B cos
tIdt
dL 0B sin
Fig. 8.8a
a.c. source
V
I
L
LV
tII 0 sin
dtdILB
(8.8a)
At h i t t th l lt V t b l t th b k f
7/30/2019 Note Unit8 Sf027
24/57
SF027 24
At each instant the supply voltage Vmust be equal to the back e.m.f
B (voltage across the inductor) but the back e.m.f always oppose the
supply voltage Vrepresents by the negative sign in eq. (8.8a).
By comparing the magnitude ofVand B ,thus
Therefore the phase difference between VandIis
In pure inductor,
the voltage Vleads the currentIby /2 radians.
or
the currentIlags behind thevoltage V by /2 radians.
tLIV 0B cos
tVV 0 cos
and00 VLI
rad2
t2
t
or
2
tVV 0 sin
7/30/2019 Note Unit8 Sf027
25/57
SF027 25
Figure 8.8b shows the variation ofVandIwith time while Figure 8.8c
shows the phasor diagram forVandIin a pure inductor.
8.8.2 Impedance in a pure inductor
From the definition of the impedance, hence
0
0
I
LIZ
0
0
I
VZ and 00 LIV
t0
0V
0V
0I
0I
TT
2
1 T2T
2
3
Fig. 8.8b
rad2
I
V
Fig. 8.8c : Phasor diagram
7/30/2019 Note Unit8 Sf027
26/57
SF027 26
whereXL is known as inductive reactance.
Inductive reactance is the opposition of a inductor to the current flowsand is defined by
The unit of inductive reactance is ohm ( )
From eq. (8.8b), the relationship between inductive reactanceXL and
frequencyfcan be shown by using graph in figure 8.8d.
LXLZ
(8.8b)fL2XL
and f2
0
0
rms
rms
L I
V
I
V
X (8.8c)
sourca.c.offrequency:finductortheofinductanceself:L
f0
Fig. 8.8d
LX
fXL
8 8 3 Power in a pure inductor
7/30/2019 Note Unit8 Sf027
27/57
SF027 27
8.8.3 Power in a pure inductor
Since the voltage Vleads currentI the by /2 radians, the
instantaneous powerP is given by
Therefore the graph of variation of power with time is shown in figure8.8e.
t2P2
1P 0 sin
IVP
tVtIP 00 cossinttVIP 00 cossin and 000 PVI
(8.8d)
t22
1PP 0 sin
t0
2
P0
2
P0
TT2
1 T2T
2
3Fig. 8.8e
)(PPower
t2P2
1P 0 sin
Power being absorbed
Power being returned to supply
The average (mean) power P in a inductor is
7/30/2019 Note Unit8 Sf027
28/57
SF027 28
The average (mean) powerPav in a inductor is
Note :
For both pure capacitor and inductor, the average (mean) power iszero because the power is positive for quarter of a cycle andnegative for the next quarter of a cycle in each half cycle.
The term resistance is not used in pure capacitor and inductorbecause no heat is dissipated in both devices.
Example 3 :A capacitor with C=4700 pF is connected to an a.c. supply with r.m.s.
voltage of 240 V and frequency of 50 Hz. Calculate
a. the capacitive reactance.
b. the peak current in the circuit.
Solution: C=4700x10-12 F, Vrms=240 V, f=50 Hz.a. By applying the equation of capacitive reactance, thus
t2P2
1P 0av sin and 0t2 sin
0Pav
10x677X3
C
fC2
1XC
b From the definition of the capacitive reactance thus
7/30/2019 Note Unit8 Sf027
29/57
SF027 29
b. From the definition of the capacitive reactance, thus
Example 4 :
A 240 V r.m.s. supply with a frequency of 50 Hz causes an r.m.s.
current of 3.0 A to flow through an inductor which can be taken to have
zero resistance. Calculatea. the reactance of the inductor.
b. the inductance of the inductor.
Solution:Irms=3.0 A, Vrms=240 V, f=50 Hz.
a. From the definition of the inductive reactance, thus
b. By using the equation of the inductive reactance, thus
rms
rmsC
I
VX and
2
II 0rms
A10x015I4
0 .
80XL
C
rms
0 X
2V
I
rms
rmsL
I
VX
H260L .
fL2XL
2
XL
L
8 9 RC RL d LRC i S i Ci it
7/30/2019 Note Unit8 Sf027
30/57
SF027 30
8.9 RC, RL and LRC in Series Circuit8.9.1 RC in series circuit
Consider an a.c. source with voltage Vis connected in series with a
resistorR and a capacitorCas shown in figure 8.9a. The voltage across the resistorVR andthe capacitorVC are given by
The phasor diagram of RC circuit isshown by figure 8.9b.
Based on the phasor diagram, the total
voltage V(supply voltage) across bothresistorR and capacitorCis
IRVR
CC IXV
2
C
2
R
2
VVV
where
anglephase:
Fig. 8.9a
a.c. source
R
I
RV
V
CV
C
CV
IRV
V
Fig. 8.9b : Phasor diagram
2C22
IXIRV
2C222 XRIV
22
2
C
1
RIV
and
C
1XC
(8.9a)
V
7/30/2019 Note Unit8 Sf027
31/57
SF027 31
Since , hence eq. (8.9a) can be written as
From the phasor diagram (fig. 8.9b), the currentIleads the supplyvoltage Vby radians where
From the eq. (8.9b) and (8.9c), the new phasor diagram in terms ofR,
XCandZcan be sketched (refer to figure 8.9c)
rms
rms
I
VZ
R
C
V
Vtan
2
C
2XRZ
IR
IXCtan
RXCtan
22
2
C
1RZ or (8.9b)
CR1tanor (8.9c)
CX
R
Z
Fig. 8.9c : Phasor diagram in
terms ofR,XCandZ
8 9 2 RL in series circuit
7/30/2019 Note Unit8 Sf027
32/57
SF027 32
8.9.2 RL in series circuit
Consider an a.c. source with voltage Vis connected in series with a
resistorR and an inductorL as shown in figure 8.9d.
The voltage across the resistorVR and
the inductorVL are given by
The phasor diagram of RL circuit isshown by figure 8.9e.
Based on the phasor diagram, the total
voltage V(supply voltage) across bothresistorR and inductorL is
IRVR
LL IXV
2
L
2
R
2VVV
LV
IRV
V
Fig. 8.9e : Phasor diagram
2
L
22 IXIRV
2L222 XRIV 222
LRIV
and LXL
(8.9d)
Fig. 8.9d
a.c. source
R
I
RV
V
L
LV
V
7/30/2019 Note Unit8 Sf027
33/57
SF027 33
Since , hence eq. (8.9d) can be written as
From the phasor diagram (fig. 8.9e), the supply voltage Vleads thecurrentIby radians where
From the eq. (8.9e) and (8.9f), the new phasor diagram in terms ofR,
XL andZcan be sketched (refer to figure 8.9f)
rms
rms
I
VZ
R
L
V
Vtan
2
L
2 XRZ
IR
IXLtan
RXLtan
222LRZ or (8.9e)
RLtanor (8.9f)
Fig. 8.9f : Phasor diagram in
terms ofR,XL andZ
LXZ
R
8 9 3 LRC in series circuit
7/30/2019 Note Unit8 Sf027
34/57
SF027 34
8.9.3 LRC in series circuit
Consider an a.c. source with voltage Vis connected in series with an
inductorL, a resistorR and a capacitorCas shown in figure 8.9g.
The voltage across the inductorVL ,
resistorVR and resistorVR are givenby
The phasor diagram of LRC circuit is
shown by figure 8.9h.
Based on the phasor diagram, the total
voltage V(supply voltage) acrossinductorL , resistorR and capacitorCis
IRVR
CC IXV
2
CL
2
R
2
VVVV
LV
IRV
V
Fig. 8.9h : Phasor diagram
2CL22
IXIXIRV
2CL222
XXRIV
2
CL
2
XXRIV
Fig. 8.9g
a.c. source
I
V
R
RV
L
LV CV
C
LL IXV
CV
CL VV
(8.9g)
V
7/30/2019 Note Unit8 Sf027
35/57
SF027 35
Since , hence eq. (8.9g) can be written as
From the phasor diagram (fig. 8.9h), the supply voltage Vleads the
currentIby radians where
From the eq. (8.9h) and (8.9i), the new phasor diagram in terms ofXL,
,R,XCandZcan be sketched (refer to figure 8.9i)
rms
rms
I
VZ
R
CL
V
VV tan
2
2
C
1LRZ
IR
XXI CL tan
R
XX CL tan
2CL2 XXRZ or
R
C
1L
tanor
Fig. 8.9i : Phasor diagram in terms of
XL,R,XCandZ
(8.9h)
(8.9i)
LX
R
Z
CX
CL XX
8 10 Resonance and Po er in A C Circ it
7/30/2019 Note Unit8 Sf027
36/57
SF027 36
From the graph, the value ofZis minimum when
where its value is
This phenomenon occurs at
frequency offr known asresonant frequency.
8.10 Resonance and Power in A.C. Circuit8.10.1 Resonance in a.c. circuit
Definition is defined as the phenomenon that occurs when thefrequency of the applied voltage is equal to the
frequency of the LRC series circuit. Figure 8.10a shows the variation ofXC,XL,R andZwith frequencyf
of the LRC series a.c circuit.
Z
fXL
R
f
1XC
Fig. 8.10a
0 f
ZRXX LC ,,,
rf
2CL2
XXRZ
0RZ 2 min
CL XX
RZ min
(8.10a)
When the resonance occurred in the LRC series circuit, the Z is
7/30/2019 Note Unit8 Sf027
37/57
SF027 37
When the resonance occurred in the LRC series circuit, the Z isminimum but the r.m.s. current flows in the circuit is maximum andgiven by
The resonant frequencyfrof the LRC series circuit is given by
At frequencies above or below the resonant frequencyfr, the r.m.s.current is less than the maximum current.
The series resonance circuit is used for tuning a radio receiver.
When resonanceoccurred
CL XX LXL
R
V
Z
VI rmsrmsrms
C
1XC where and
C
1L
LC1 rf2 and
hence
LC2
1fr
where
frequencyresonant:r
ffrequencangularresonant:
8.10.2 Power in a.c. circuit
7/30/2019 Note Unit8 Sf027
38/57
SF027 38
We get , then eq. (8.10b)can be written as
wherecos is called the power factor ofthe circuit andPr is the average realpower.
8.10.2 Power in a.c. circuit
No power is absorbed by either inductors or capacitors over a completecycle and therefore the average (mean) power absorbed by bothinductors and capacitors is zero.
In an a.c. circuit in which there is resistanceR, inductanceL and
capacitance C, the total average powerPav is equal to that dissipatedin the resistance i.e.
From the phasor diagram of the LRC series circuit in figure 8.10b,
Rav IVP orRIP2
av
IRV
V
CL VV
Fig. 8.10b
(8.10b)
V
VRcos
cosIVPav
r
2
av PZIP cos
and IZV
(8.10c)
Note :
7/30/2019 Note Unit8 Sf027
39/57
SF027 39
Power factor can also be calculated by using formula below :
From the figure 8.10b,
From the figure 8.10b, Since VandIare not in phase, then the
average apparent powerPa is given by
By substituting Pa=I2Zinto eq. (8.10c), the power factor can be
related to Pa and Prby expression below :
where
IZ
IRcos
(8.10d)
V
VRcos IRVR
Z
Rcos
and IZV
Z
VZIIVP
22
a (8.10e)
a
r
P
Pcos (8.10f)
Example 5 :
7/30/2019 Note Unit8 Sf027
40/57
SF027 40
p
A 10 F capacitor, a 2.0 H inductor and a 20 resistor are connectedin 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 maximum (peak) current in the circuit.
e. the phase angle.
f. the mean power of the circuit.Solution:C=10x10-6F, L=2.0 H, R=20
Gives the peak voltage, V0 = 300 V and
the angular frequency = 300 rad s-1.
a. By using the relationship between andf, hence the frequency is
t300300V sin
t300300V sin tVV 0 sincompare with
f2
z. H847f
b By applying the formulae of capacitive reactance and inductive
7/30/2019 Note Unit8 Sf027
41/57
SF027 41
b. By applying the formulae of capacitive reactance and inductive
reactance, thus
c. By using the formula of impedance for LRC series circuit, thus
d. From the definition of the impedance, thus
Capacitivereactance : 333XC
C
1XC
LXL 600XL Inductivereactance :
2
CL
2
XXRZ 267Z
0
0
rms
rms
I
V
I
VZ
A121I0 .Z
VI 00
e By using the formula of phase angle for LRC series circuit thus
7/30/2019 Note Unit8 Sf027
42/57
SF027 42
e. By using the formula of phase angle for LRC series circuit, thus
f. The mean (average) power is
rad501785 .@.
R
XX CL tan
R
XX CL1tan
RIP2
rmsav
W512Pav .
and
2
II 0rms
RI2
1P
2
0av
OR cosZIP2
rmsav
W512Pav .
cosZI2
1P
2
0av
and
2
II 0rms
Example 6 :
7/30/2019 Note Unit8 Sf027
43/57
SF027 43
A 0.14 H inductor and a 12 resistor are connected in series to the
alternating source 110 V, 25 Hz. Calculate
a. the r.m.s current flows in the inductor.
b. the phase angle between the current and supply voltage.
c. the power factor of the circuit.
d. the average power loss to the surrounding.
Solution: R=12 , L=0.14 H, Vrms=110 V, f =25 Hza. The inductive reactance is
The impedance of the circuit is
Therefore the r.m.s. current is
LXL
A44Irms .
2
L
2XRZ
and f2
25Z
fL2XL 22XL
Z
VI rmsrms
b. By using the formula of phase angle for RL series circuit, thusX
7/30/2019 Note Unit8 Sf027
44/57
SF027 44
c. The power factor is
d. The average power loss is
Example 7 :
Based on the RCL series circuit above, the r.m.s. voltage across R, L
and C is shown.
a. By using the phasor diagram, calculate the supply voltage and the
phase angle in this circuit.
R
XLtan
rad071461 .@.
W232Pav
R
XL1tan
cosfactorpower
cosrmsrmsav VIP
480factorpower .
CRL
V3 27 V141 V133
I
Calculate :
7/30/2019 Note Unit8 Sf027
45/57
SF027 45
b. the current flows in the circuit if the resistance of the resistor R
is 68 .
c. the inductance and capacitance if the frequency of the a.c. source
is 50 Hz.
d. the resonant frequency.
Solution: VR=141 V, VL=327 V, VC=133 Va. The phasor diagram of the circuit is
LV
I
RV
V
CV
CL VV
From the phasor diagram :
The supply voltage Vis
The phase angle is
2CL2
R VVVV V240V
R
CL
V
VV tan
rad942.0054 @.
R
CL1
V
VVtan
b. GivenR =68
7/30/2019 Note Unit8 Sf027
46/57
SF027 46
Since L, R and C are connected in series, hence the current flows in
each devices is the same. Therefore
c. Givenf =50 HzThe inductive reactance is
therefore the inductance of the inductor is
The capacitive reactance is
therefore the inductance of the inductor is
IRVR
A072I .R
VI R
H5030L .
LL IXV 158XL
LXL
f2
XL L
and f2
CCIXV 364X
C
.
F10x954C5.
C
1XC
CfX2
1C
and f2
d. The resonant frequency is
7/30/2019 Note Unit8 Sf027
47/57
SF027 47
Example 8 : (exercise)An a.c. current of angular frequency of 1.0 x 104 rad s-1 flows through a10 k resistor and a 0.10 F capacitor which are connected in series.Calculate the r.m.s. voltage across the capacitor if the r.m.s. voltageacross the resistor is 20 V.Ans. : 2.0 V
Example 9 : (exercise)A 200 resistor, a 0.75 H inductor and a capacitor of capacitance Care connected in series to an alternating source 250 V, 600 Hz.Calculate
a. the inductive reactance and capacitive reactance when resonance isoccurred.
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.
Ans. : 2.83 k, 93.8 nF, 200 , 1.25 A
CL XX
Hz032fr .
C
1L
LC21fr
and rf2
8 11 A C Rectification
7/30/2019 Note Unit8 Sf027
48/57
SF027 48
8.11 A.C. Rectification Definition is defined as the process of converting alternating
current to direct current.
Rectifier :
is a device that allows current to flow in one direction only. diodes are usually used as rectifiers.
Diode is said to be forward biased when positive terminal of thediode connected to the positive terminal of the battery and viceversa, hence a current will be able to flow (figure 8.11a).
Diode is said to be reverse biased when positive terminal of thediode connected to the negative terminal of the battery and viceversa, hence no current flows (figure 8.11b).
Fig. 8.11a : Forward biased
+
+ -
-
I I
Fig. 8.11b : Reverse biased
+
+-
-
0I
Diode
Two types of rectification are
7/30/2019 Note Unit8 Sf027
49/57
SF027 49
t0 T T2
RV
0V
0V
Fig. 8.11f
t0T
T2
DV
0V
0V
Fig. 8.11e
half-wave
full-wave
8.11.1 Half-wave Rectification
Half-wave rectification means that only one half of a cycle passesthrough the rectifier (diode).
Figure 8.11c shows a half-wave rectification circuit.
Fig. 8.11d
0V
0V
t0T T2
VVoltageSupply
Fig. 8.11c
RV
DV
D
R
A
B
V
Supplyvoltage
Explanation:
7/30/2019 Note Unit8 Sf027
50/57
SF027 50
First half cycle (Fig. 8.11d)
When terminal A is positive, diode is forward biased and offerslow resistance such that a pulse of current flows through thecircuit.
There is negligible voltage across the diode, VD (Fig. 8.11e). Thus the voltage across the resistor, VR is almost equal to the
supply voltage (Fig. 8.11f).
Next half cycle (Fig. 8.11d)
When terminal B is positive, diode is now reverse biased and
has a very high resistance such that a very small current flowsthrough it.
The voltage across the diode, VD is almost equal to the supplyvoltage (Fig. 8.11e).
The voltage across the resistor, VR is almost zero (Fig. 8.11f).
An alternating voltage is thus rectified to give direct current voltageacross the resistor. Current flows through the resistor in onedirection only and only half of each cycle passes through is shownin figure 8.11g.
t0T T2
I
0I
0IFig. 8.11g
R.m.s. value after half-wave rectification:
7/30/2019 Note Unit8 Sf027
51/57
SF027 51
In half-wave rectification, half the supply voltage is suppressed andtherefore the mean of the square of the voltage is given by
Therefore the r.m.s voltage of half-wave rectification is given by
2
1Mean of the squareafter rectification
= XMean of the squarebefore rectification
rect.)wavehalfbefore(rect.)wavehalf( 22 V
2
1V
tV 220 sin and2
1t
2 sin
4
V
2
V
2
1V
2
0
2
02
rect.)wavehalf(
rect.)wavehalf( 2rms VV
4
V
V
2
0
rms 2
VV 0
rms(8.11a)
2
V2
0
In similar way as to find the r.m.s. voltage of half-wave rectification,h f h lf ifi i i i b
7/30/2019 Note Unit8 Sf027
52/57
SF027 52
0V
0V
t0T T2
VVoltageSupply
Fig. 8.11i
the r.m.s. current of half-wave rectification is given by
8.11.2 Full-wave Rectification
While half-wave rectification only allows half of each cycle to passthrough the diode, full-wave rectification allows both halves of eachcycle to pass through the diode.
To obtain full-wave rectification, four diode are used and are arranged
in a form known as the diode bridge. Figure 8.11h shows a full-wave rectification circuit.
2
II 0rms (8.11b)
Fig. 8.11h
A
RRVV
Supplyvoltage
F
B
CD
E
1
2 3
4
Fig. 8.11j
0V
t0T
T2
RV
Explanation:
7/30/2019 Note Unit8 Sf027
53/57
SF027 53
First half cycle (Fig. 8.11i)
When terminal A is positive, diodes 1 and 2 are forward biasedand conduct the current.
The current takes the path ABC, R and DEF.
Diodes 3 and 4 are reverse biased and hence, do not conductthe current.
The voltages across diodes 1 and 2 are negligible, the voltage
across the resistorVR is almost equal to the supply voltage(Fig. 8.11j)
Next half cycle (Fig. 8.11i)
When terminal F is positive, diodes 3 and 4 are forward biasedand 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 conductthe current.
The voltage across the resistor is again almost equal to thesupply voltage (Fig. 8.11j).
Both halves of the alternating voltage are rectified. The currentflowing through the resistor is in one direction only i.e. a varyingd.c. is obtained (Fig. 8.11k).
Fig. 8.11k
0I
t0 T T2
I
R.m.s. value after full-wave rectification:
7/30/2019 Note Unit8 Sf027
54/57
SF027 54
Notice that the negative side of supply voltage is flipped over tobecome positive side without being suppressed, thus the r.m.s.voltage of full-wave rectification is the same as ther.m.s.voltage of supply voltage and given by
The output obtained from half-wave and full-wave rectifications areunidirectional but varying d.c.
Usually a steady (constant) d.c. is required for operating variouselectrical and electronic appliances. To change a varying d.c. into asteady (constant) d.c., smoothing is necessary.
A simple smoothing circuit consists of a capacitor ( with a large
capacitance >16 F) connected parallel to the resistorR shown in figure8.12a. The capacitor functions as a reservoir to store charges.
2VV 0rms (8.11b)
8.12 Smoothing using Capacitor
+R outputVVR C
- Fig. 8.12a
Rectified unsmoothed
voltage V
I
8.12.1 Smoothing of a half-wave rectified Voltage
Fi 8 12b h ff f hi h lf ifi d
7/30/2019 Note Unit8 Sf027
55/57
SF027 55
Figure 8.12b shows an effects of smoothing a half-wave rectifiedvoltage.
Initially, the half-wave rectified input voltage Vcauses the current toflow through the resistorR. At the same time, capacitorCbecomescharged to almost the peak value of the input voltage.
At A (Fig. 8.12b), input V(dash line) falls below output VR, the
capacitorCstarts to discharge through the resistorR. Hence thecurrent flow is maintained because of capacitors action.
Along AB (Fig. 8.12b), Voutput falls. At B, the rectified current again
flows to recharge capacitorCto the peak of the input voltage V.
This process is repeated and hence the output voltage VR across the
resistorR will look like the variation shown in figure 8.12b.
Fig. 8.12b
A B
Rectified unsmooth input
voltage V
outputVVR
Smoothed voltage VR
Discharge
Charge
ttime,
8.12.2 Smoothing of a full-wave rectified Voltage
Fi 8 12 h ff t f thi f ll tifi d
7/30/2019 Note Unit8 Sf027
56/57
SF027 56
Figure 8.12c shows an effects of smoothing a full-wave rectifiedvoltage.
The explanation of the smoothing process likes for a half-wave rectifiedvoltage.
The fluctuations of the smoothed output voltage are must less compareto the half-wave rectified.
The smoothing action of the capacitor is due to the large time constant
, given byRCso the output voltage cannot fall as rapidly as therectified unsmoothed input voltage.
Therefore a large capacitor performs greater smoothing. However, aninitially uncharged capacitor may cause a sudden surge of currentthrough the circuit and damage the diode.
Fig. 8.12c
A B
Rectified unsmooth input
voltage V
outputVVR
Smoothed voltage VR
Discharge Charge
ttime,
7/30/2019 Note Unit8 Sf027
57/57
THE END
UNIT 9 :
ELECTROMAGNETIC WAVES