Chapter 6
CP3 1 FYSL
Chapter 6: Alternating Current
6.1 Alternating Current
L.O 6.1.1 Define alternating current (AC)
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 .
Extra Knowledge:
Alternating current can be used to create a changing magnetic field, and changing magnetic
fields can be used to create alternating current. This relationship between alternating current
and magnetic fields makes three important devices possible: alternator, motor and transformer.
L.O 6.1.2 Sketch and interpret sinusoidal AC waveform
L.O 6.1.3 Use sinusoidal voltage and current equations
tII o sin
tVV o sin
Phase (Phase denotes the particular point in the cycle of a
waveform, measured as an angle in degrees/ radian)
Chapter 6
CP3 2 FYSL
Terminology in AC
Peak (maximum) current ( I0 )
Definition: Magnitude of the maximum current.
Peak (maximum) voltage ( V0 )
Definition: Magnitude of the maximum voltage.
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)
Angular frequency ( )
Unit: radian per second (rad s-1)
Equation: f 2 or T
2
Example
Question Solution
Figure shows the variation of voltage with
time for a sinusoidal AC current. Determine
a. the frequency
b. the phase angle
c. the peak-to-peak voltage and
d. write the expression for the graph
The current in an AC circuit is given by the
expression:
tI 50sinmA 25
Sketch a I-t graph for the AC circuit.
Determine the current when t = 50 s.
fT
1
Chapter 6
CP3 3 FYSL
6.2 Root Mean Square (rms)
L.O 6.2.1 Define root mean square (rms) current and voltage for AC source
L.O 6.2.2 Use rms voltage and current equations
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.
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.
**Equations Irms and Vrms are valid only for a sinusoidal alternating current and
voltage**
The average power: Relationship between Pave and P0
2
0
rms
II
2
0
rms
VV
Average value for one complete cycle is zero
R
VRIVIP
2
rms2
rmsrmsrmsave 22
1
22
0
00
00
ave
PVI
VIP
000 VIP
acdc power average power rms
2
ave
2
ave
2
III
RIRI
rms
2
ave
2
ave
2
VVV
R
V
R
V
Chapter 6
CP3 4 FYSL
Example
Question Solution
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
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.
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.
Exercise
Question
A voltage V = 60 sin 120πt is applied across a 20 Ω resistor. Determine
a. the reading of the AC ammeter in series with the resistor
b. the peak current and mean power
Answer:
An AC current is given as I = 5 sin (200t) where the clockwise direction of the current is
positive. Find
a. the peak current
b. the current when t = 1/100 s
c. the frequency and period of the oscillation.
Answer: 5 A; 4.55 A; 31.88 Hz; 0.0314 s
Chapter 6
CP3 5 FYSL
6.3 Resistance, Reactance and Impedance
L.O 6.3.1 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
L.O 6.3.3 Define and use capacitive reactance, inductive reactance, impedance and
phase angle
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).
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)
θ
OP
ONsin
Phasor diagram
(not including the circle) Sinusoidal Waveform
Instantaneous value
Chapter 6
CP3 6 FYSL
Pure resistor in the AC circuit
The current flows in the resistor is
The voltage across the resistor VR at any instant is
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
The instantaneous power
The average power
A resistor in AC circuit dissipates energy in the form of heat
ωtII sin0
IRV R
ωtRIV sin0R
VωtVV sin0R
00 VRI
tageSupply vol:V
and
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rms
I
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sin
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000
2
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2
rmsave2
1
2
1
2
1PIVRIRIP
Chapter 6
CP3 7 FYSL
Pure capacitor in the AC circuit
Pure capacitor means that no resistance and self-inductance effect in the AC circuit.
The voltage across the capacitor VC
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 current flows in the ac circuit is
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.
CCVQ
tCVQ sin0
dt
dQI
tCVdt
dI sin0
tdt
dCVI sin0
)cos(0 ωtCVI 00 ICV
2sin0
ωtII
and
2
2
Δ
ωttΔ
ωtVV sin0C
Chapter 6
CP3 8 FYSL
The capacitive reactance in a pure capacitor is
The instantaneous power
The average power
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.
The relationship between capacitive reactance XC and frequency f
fCCX
CV
V
I
V
I
VX
2
11C
0
0
0
0
rms
rms
C
Definition
tPP
tVIP
tVtIP
R
VRIIVP
2sin2
1
2sin2
1
sincos
0
00
00
22
0ave P
circuitcapacitor purein CXZ
Chapter 6
CP3 9 FYSL
Pure inductor in the AC circuit
Pure inductor means that no resistance and capacitance effect in the AC circuit.
The current flows in the ac circuit is
When the current flows in the inductor, the back emf caused by the self-induction is produced
and given by
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:
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.
ωtII sin0
dt
dILε B
ωtIdt
dL sin0B
ωtωLI cos0B
ωtωLIV cos0B
2sin
2sin
0
0
ωtVV
ωtωLIV 00 LIV and
2
2
Δ
ωtωtΔ
Chapter 6
CP3 10 FYSL
The inductive reactance in a pure inductor is
The instantaneous power
The average power
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.
The relationship between inductive reactance XL and the frequency f
Definition
tPP
tVIP
tVtIP
R
VRIIVP
2sin2
1
2sin2
1
cossin
0
00
00
22
0ave P
circuitinductor purein LXZ
fLLX
I
LI
I
V
I
VX
2L
0
0
0
0
rms
rmsL
Chapter 6
CP3 11 FYSL
L.O 6.3.2 Use phasor diagram to analyze voltage, current, and impedance of series
circuit of RL, RC and RLC
RC Circuit
The total p.d (supply voltage), V across R and C is equal to the vector sum of VR and VC as
shown in the phasor diagram.
From the phasor diagrams, the current I leads the supply voltage V by ϕ radians where
Graph of Z against f
2
C
2
2
C
2
2
C
2
R
XRIV
IXIRV
VVV
2
C
2 XRI
VZ
I leads V by ϕ
R
CtanV
V
R
X Ctan or
The resistance, R
is independent of
frequency
Chapter 6
CP3 12 FYSL
RL Circuit
The total p.d (supply voltage), V across R and L is equal to the vector sum of VR and VL as
shown in the phasor diagram.
From the phasor diagram, the supply voltage V leads the current I by ϕ radians where
Graph of Z against f
V leads I by ϕ
2
L
2
2
L
2
2
L
2
R
XRIV
IXIRV
VVV
2
L
2 XRI
VZ
R
LtanV
V
R
X Ltan or
The resistance, R
is independent of
frequency
Chapter 6
CP3 13 FYSL
RCL Circuit (VL > VC)
The total p.d (supply voltage), V across L, R and C is equal to the vector sum of VL, VR and
VC as shown in the phasor diagram.
From the phasor diagram, the supply voltage V leads the current I by ϕ radians where
Graph of Z against f
V leads I by ϕ
2
CL
2
2
CL
2
2
CL
2
R
)(
)(
XXRIV
IXIXIRV
VVVV
2
CL
2 XXRZ
VL > VC
R
CLtanV
VV
IR
XXI CLtan
or
VL < VC ?
VL = VC ?
Chapter 6
CP3 14 FYSL
L.O 6.3.4 Explain graphically the dependence of R, XC, XL and Z on f and relate it to
resonance
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:
at low frequency, impedance Z is large because 1/ωC is large.
at high frequency, impedance Z is high because ωL is large.
at resonance frequency, impedance Z is minimum (Z = R; XC = XL) and I is maximum
CL
XX
1
CL
LCf
CfLf
2
1
2
12
r
r
r
Resonance
frequency
2
CL
2 XXRZ
02
min RZ
RZ min
max min, , 1
rmsrms
rms
rms
IZZ
I
Z
VI
Chapter 6
CP3 15 FYSL
Example
Question Solution
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.
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.
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.
Chapter 6
CP3 16 FYSL
Question Solution
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.
Exercise
Question
A rms voltage of 12.2 V with a frequency of 1.00 kHz is applied to a 0.29 mH inductor.
a. What is the rms current in the circuit?
b. Determine the peak current for a frequency of 2.50 kHz.
Answer: 6.70A; 3.78 A
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.
Answer: 6.38×10-3 A; 10.2 V; 57.9° or 1.01 rad
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.
Answer: 64.9 Ω , 1.85 A, 67.3o
Chapter 6
CP3 17 FYSL
6.4 Power and Power Factor
L.O 6.4.1 Apply average power, instantaneous power and power factor equations in
AC circuit consisting of RL, RC and RLC in series
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)
For RCL circuit:
From the diagrams above,
Substitute (2) into (1):
The term cos ϕ is called the power factor.
The power factor (cos ϕ ) can vary from a maximum of +1 (or 100%) to a minimum of 0.
When ϕ = 0o (cos ϕ = +1) ,the circuit is completely resistive or when the circuit is in
resonance (RCL).
When ϕ = +90o (cos ϕ = 0) ,the circuit is completely inductive.
When ϕ = -90o (cos ϕ = 0) ,the circuit is completely capacitive.
rmsrmsave RVIP Vrms across resistor
(1)
Z
R
V
V cos and cos R
(2)
cos
or
cos
rmsrmsave
2
ave rms
VIP
ZIP
aapparent PP a
avecosP
PRearranging
r
2
ave rms PRIP
Power used by the load
Power delivered to the load
Chapter 6
CP3 18 FYSL
Example
Question Solution
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.
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?
Exercise
Question
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.
Answer: 4.4 A, 61.3o , 0.48, 0.23 kW
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.
Answer: 2.94 x 10-3 H , 4.84 Ω , 0.163
tV 300sin300