chapter 6: alternating current - ysl...
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Chapter 6: Alternating
Current
An alternating current is an current that reverses its
direction at regular intervals.
Overview
Alternating
Current
A.C. Through a
Capacitor
A.C. Through a
Inductor
A.C. Through a
Resistor
RCL Circuit in
Series
Phasor Diagram Sinusoidal Waveform
6.1 Alternating Current
Define alternating current (AC)
Sketch and interpret sinusoidal AC waveform
Use sinusoidal voltage and current equations:
Learning Objectives
tVV sin0
tII sin0
Alternating Current
An alternating current (AC) is the electrical
current which varies periodically with time in
direction and magnitude.
The usual circuit-diagram symbol for an AC
source is
Sinusoidal AC Waveform
The output of an AC generator is sinusoidal and
varies with time.
T
VI or
t0
T2
1
00 or VI
00 or VI
Positive half cycle
Negative half cycle
1 Revolution
Peak-to-peak value
Peak value
Sinusoidal AC Waveform
Equation for alternating current (I):
Equation for alternating voltage (V):
where
tII o sin
tVV o sin
Phase
)2(
locityangular ve ORfrequency angular :
f
currentpeak : 0I
gepeak volta: 0V
Phase Angle
Phase denotes the particular point in the cycle
of a waveform, measured as an angle in degrees/
radian
Revision A
(s) t0
0A
0A
A
(s) t0
0A
0A
A
(s) t0
0A
0A
tAA sin0
2sin0
tAA
2sin0
tAA
Terminology in AC
Peak (maximum) current ( I0 )
◦ Definition: Magnitude of the maximum current.
Peak (maximum) voltage ( V0 )
◦ Definition: Magnitude of the maximum voltage.
Angular frequency ( )
Unit: radian per second (rad s-1)
Equation: or
f 2T
2
Terminology in AC
Frequency ( f )
◦ Definition: Number of complete cycle in one
second.
◦ Unit: Hertz (Hz) or s-1
Period ( T )
◦ Definition: Time taken for one complete cycle.
◦ Unit: second (s)
fT
1
Example 1
Figure shows the variation of voltage with time for a sinusoidal AC. Determine
a) the frequency
b) the phase angle at t = 15 s
c) the peak-to-peak voltage and
d) write the expression for the graph
Example 1 – Solution
Example 1 – Solution
Example 2
The current in an AC circuit is given by the
expression:
Sketch a I-t graph for the AC circuit. Determine
the current when t = 50 s.
tI 50sinmA 25
Example 2 – Solution
6.2 Root Mean Square (rms)
Define root mean square (rms) current and
voltage for AC source.
Use the following formula,
**Equations Irms and Vrms are valid only for a
sinusoidal alternating current and voltage**
Learning Objectives
2
0II rms and
2
0VVrms
Average Current (Iave) Average current is defined as the average or mean
value of current in a half-cycle flows of current in a
certain direction.
Iave for one complete cycle is zero because the
current flows in one direction in one-half of the
cycle and in the opposite direction in the next half
of the cycle.
T
I
t0T
2
1 T2
0I
0I
Iav
Iav
0)( aveave II
2
2π
ooave
I
π
II
Root Mean Square Current (Irms)
Root mean square current (Irms) is defined as the
effective value of AC which produces the same power
(mean/average power) as the steady d.c. when the
current passes through the same resistor.
acdc power average power
rms
2
ave
2
ave
2
III
RIRI
2
0rms
II
Root Mean Square Voltage (Vrms)
Root mean square voltage/ p.d (Vrms ) is defined as the
value of the steady direct voltage which when applied
across a resistor, produces the same power as the mean
(average) power produced by the alternating voltage
across the same resistor
2
0
rms
VV
rms
2
ave
2
ave
2
VVV
R
V
R
V
acdc power average power
Example 3
A sinusoidal, 60.0 Hz, ac voltage is read to be 120 V by an
ordinary voltmeter. Determine
a) the maximum value the voltage takes on during a
cycle
b) the equation for the voltage
Example 3 – Solution
Example 4
The alternating potential difference shown above is connected across a resistor of 10 k. Calculate
a) the rms current,
b) the frequency,
c) the mean power dissipated in the resistor.
Example 4 – Solution
Example 4 – Solution
Example 5
An AC source V = 200 sin t is connected across a resistor
of 100 . Calculate
a) the rms current in the resistor.
b) the peak current.
c) the mean power.
Example 5 – Solution
Example 5 – Solution
6.3 Resistance, Reactance and
Impedance Sketch and use phasor diagram and sinusoidal waveform to show
the phase relationship between current and voltage for a single
component circuit consisting of pure resistor, pure capacitor and
pure inductor.
Use phasor diagram to analyze voltage, current, and impedance of
series circuit of RL, RC and RLC.
Define and use capacitive reactance, inductive reactance,
impedance and phase angle.
Explain graphically the dependence of R, XC , XL and Z on f and
relate it to resonance.
Learning Objectives
Phasor Diagram
Phasor is defined as a vector that rotates 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 as alternating current (AC) and alternating voltage.
It also being used to determine the phase angle (is
defined as the phase difference between current and
voltage in AC circuit).
Phasor Diagram
t0 TT2
1 T2
ωN
O
P
Ao tAA o sin
A
Phasor diagram
(not including the circle)
Sinusoidal Waveform
Instantaneous value
P
Resistance, Reactance and
Impedance
Key Term/Ω Meaning
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 reactance combine to
form impedance)
Pure Resistor in the AC Circuit
Pure Resistor in the AC Circuit
The current flows in the resistor is
ωtII sin0
IRV R
ωtRIV sin0R
VωtVV sin0R
00 VRI
tageSupply vol:V
The voltage across the resistor VR at any instant is
and
Pure Resistor in the AC Circuit
ωtII sin0
ωtVV sin0
t0
0I
0V
0I0V
TT2
1 T2T
2
3
ω
VI
Phasor diagram
Pure Resistor in the AC Circuit
The phase difference between V and I is
In pure resistor, the voltage V is in phase with the current I
and constant with time (the current and the voltage reach
their maximum values at the same time).
The resistance in a pure resistor is
IV ωtωt
0
0
0
rms
rms
I
V
I
VR
Pure Resistor in the AC Circuit
The instantaneous power The average power
tVIP
tVtIP
R
VRIIVP
2
00
00
22
sin
sinsin
0
00
2
0
2
rmsave
2
1
2
1
2
1
P
IV
RI
RIP
A resistor in AC circuit dissipates energy in the form of heat
Pure Capacitor in the AC Circuit
Pure Capacitor in the AC Circuit
The voltage on a capacitor depends on the amount of
charge you store on its plates. The charge accumulates on
the plates of the capacitor is
The voltage across the capacitor VC
ωtVV sin0C
CCVQ
tCVQ sin0
Pure Capacitor in the AC Circuit
The current flows in the ac circuit is
dt
dQI
tCVdt
dI sin0
tdt
dCVI sin0
)cos(0 ωtCVI 00 ICV
2sin0
ωtII
and
ωtII cos0
Compare with
general equation
Pure Capacitor in the AC Circuit
t0
0I
0V
0I0V
TT2
1 T2T
2
3
ω
VI
rad 2
Phasor diagram ωtVV sin0
2sin0
ωtII
Pure Capacitor in the AC Circuit
The phase difference between V and I is
In pure capacitor, the voltage V lags behind the current I
by /2 radians or the current I leads the voltage V by /2
radians.
2
Δ
2
ωttΔ
Pure Capacitor in the AC Circuit
The capacitive reactance in a pure capacitor is
0
0
0
0
rms
rmsC
CV
V
I
V
I
VX
Definition
circuitcapacitor purein CXZ
fCCX
2
11C
Pure Capacitor in the AC Circuit
The instantaneous power The average power
tPP
tVIP
tVtIP
R
VRIIVP
2sin2
1
2sin2
1
sincos
0
00
00
22 0ave P
For the first half cycle where the power is positive, the capacitor is
saving the power (in electric field). For the second half of the cycle
where the power is negative, the power is returned to the circuit.
Pure Inductor in the AC Circuit
Pure Inductor in the AC Circuit
When the current flows in the inductor, the back emf
caused by the self-induction is produced and given by
The current flows in the resistor is
ωtII sin0
dt
dILε B
ωtIdt
dL sin0B
ωtωLI cos0B
Pure Inductor in the AC Circuit
At each instant the supply voltage V must be equal to the
back e.m.f B (voltage across the inductor) but the back
e.m.f always oppose the supply voltage V. Hence, the
magnitude of V and B:
ωtωLIV cos0B
2sin0
ωtωLIV
00 LIV
ωtVV cos0
Compare with
general equation
2sin0
ωtVV
Pure Inductor in the AC Circuit
t0
0I
0V
0I0V
TT2
1 T2T
2
3
ω
V
I
rad 2
ωtII sin0
2sin0
ωtVV
Phasor diagram
2
Δ
Pure Inductor in the AC Circuit
The phase difference between V and I is
In pure inductor, the voltage V leads the current I by /2
radians or the current I lags behind the voltage V by /2
radians.
ωtωtΔ
2
Pure Inductor in the AC Circuit
The inductive reactance in a pure inductor is
0
0
0
0
rms
rmsL
I
LI
I
V
I
VX
Definition fLLX 2L
circuitinductor purein LXZ
Pure Inductor in the AC Circuit
The instantaneous power The average power
tPP
tVIP
tVtIP
R
VRIIVP
2sin2
1
2sin2
1
cossin
0
00
00
22 0ave P
For the first half of the cycle where the power is positive, the
inductor is saving the power. For the second half cycle where the
power is negative, the power is returned to the circuit.
Summary 1
Circuit Relationship
between V and I
Impedance, Z
(V = IZ)
Pinstantaneous Pave
Pure
resistor
The voltage V is
in phase with the
current I
Resistor, R
Pure
capacitor
The voltage V
lags behind the
current I by /2
radians
Capacitive
reactance, XC
Pure
inductor
The voltage V
leads the current
I by /2 radians
Inductive
reactance, XL
tVIP 2
00 sin
tPP 2sin
2
10
tPP 2sin
2
10
0ave2
1PP
0ave P
0ave P
fCCX
2
11C
fLLX 2L
Example 6
A capacitor with C = 4700 pF is connected to an AC
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.
Example 6 – Solution
Example 6 – Solution
Example 7
A 240 V rms supply with a frequency of 50 Hz causes an
rms current of 3.0 A to flow through an inductor which can
be taken to have zero resistance. Calculate
a. the reactance of the inductor.
b. the inductance of the inductor.
Example 7 – Solution
Example 7 –Solution
RC Circuit
RC Circuit
IRVR CC IXV
The rms voltage across the
resistor VR and the capacitor VC
are given by:
I
ω
CV
RV
V
Based on the phasor diagram, the
rms supply voltage V (or total
voltage) of the circuit is given by:
22
CR VVV
Phasor diagram
and
I leads V by ϕ
RC Circuit
22
22
22
C
C
CR
XRIV
IXIRV
VVV
2
C
2 XRI
V
ω
CXZ
R
2
C
2 XRZ
I
VZ and
Phasor diagram
ωCX C
1and
RC Circuit
From the phasor diagrams,
Graph of Z against f
R
CtanV
V
R
X Ctan or
RL Circuit
RL Circuit
IRVR
The rms voltage across the
resistor VR and the inductor VL
are given by:
and LL IXV
ω
LVV
IRV
Based on the phasor diagram, the
rms supply voltage V (or total
voltage) of the circuit is given by:
Phasor diagram
22
LR VVV
V leads I by ϕ
RL Circuit
22
22
22
L
L
LR
XRIV
IXIRV
VVV
2
L
2 XRI
V
2
L
2 XRZ
I
VZ and
and ωLX L
Phasor diagram
ω
LXZ
R
RL Circuit
From the phasor diagrams,
Graph of Z against f
R
LtanV
V
R
X Ltan or
RCL Circuit
RCL Circuit
IRVR
The rms voltage across the
resistor VR and the capacitor
VC are given by:
LL IXV CC IXV , ,
I
ωLV
RV
V
CV
CL VV
Phasor diagram
Based on the phasor diagram, the
rms supply voltage V (or total
voltage) of the circuit is given by:
22)( cLR VVVV
V leads I by ϕ
RCL Circuit
22
22
22
)(
)(
CL
CL
cLR
XXRIV
IXIXIRV
VVVV
VL > VC
2
CL
2 XXRZ
ωLX
Z
CX
CL XX
R
Phasor diagram
RCL Circuit
From the phasor diagrams,
Graph of Z against f
R
CLtanV
VV
R
XX CLtan
or
VL < VC ?
VL = VC ?
Example 8
An alternating current of angular frequency of 1.0 × 104
rad s‒1 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.
Example 8 – Solution
Resonance in AC circuit
Resonance is defined as the phenomenon that occurs when
the frequency of the applied voltage is equal to the
frequency of the LRC series circuit.
The graph shows that:
i. at low frequency, impedance Z
is large because 1/ωC is large.
ii. at high frequency, impedance Z
is high because ωL is large.
iii. at resonance frequency,
impedance Z is minimum (Z =
R; XC = XL) and I is
maximum
Resonance in AC circuit
At resonance frequency:
CL
1
LCf
CfLf
2
1
2
12
r
r
r
Resonance
frequency
CL XX
Resonance in AC circuit
At resonance frequency:
2
CL
2 XXRZ
02
min RZ
RZ min
max min, , 1
rmsrms
rms
rms
IZZ
I
Z
VI
Example 9
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.
Calculate:
b. the current flows in the circuit if the resistance of the resistor R is
26 ,
c. the inductance and capacitance if the frequency of the AC source
is 50 Hz,
d. the resonant frequency.
Example 9 – Solution
Example 9 – Solution
Example 9 – Solution
Example 9 – Solution
Example 9 – Solution
6.4 Power and Power Factor
Apply:
i. average power,
ii. instantaneous power,
i. power factor,
in AC circuit consisting of R, RC, RL and RLC in
series.
Learning Objectives
cosrmsrmsave VIP
IVP
rmsrms
av
a
r
VI
P
P
Pcos
Power and Power Factor
In an AC circuit, the power is only dissipated by a
resistance; none is dissipated by inductance or
capacitance.
Therefore, the real power (Pr) that is used or gone is
equal to that dissipated from the resistor and given by
the average power (Pave)
rmsrmsave RVIP
r
2
ave rms PRIP
Power and Power Factor
For RCL circuit:
From the diagrams above
Z
R
V
V cos and cos R
Power and Power Factor
cos
or
cos
rmsrmsave
2
ave rms
VIP
ZIP
aapparent PP
a
avecosP
PRearranging
Example 10
A 10 F capacitor, a 2.0 H inductor and a 20 resistor are
connected in series with an alternating source given by the
equation below :
Calculate :
a. the frequency of the source.
b. the capacitive reactance and inductive reactance.
c. the impedance of the circuit.
d. the maximum (peak) current in the circuit.
e. the phase angle.
f. the mean power of the circuit.
tV 300sin300
Example 10 – Solution
Example 10 – Solution
Example 10 – Solution
Example 10 – Solution
Example 10 – Solution
Example 11
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 mH inductor 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?
Example 11 – Solution
Example 11 – Solution
Example 11 – Solution
Summary
Alternating
Current
A.C. Through a
Capacitor
A.C. Through a
Inductor
A.C. Through a
Resistor
RCL Circuit in
Series
Phasor Diagram Sinusoidal Waveform
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