Title: UNIT 21 : ALTERNATING CURRENT
1UNIT 21 ALTERNATING CURRENT (5 Hours)
21.1 Alternating current 21.2 Root mean square
(rms) 21.3 Resistance, reactance and
impedance 21.4 Power and power factor
2SUBTOPIC
21.1 Alternating Current (1 hour)
LEARNING OUTCOMES
At the end of this lesson, students should be
able to
- Define alternating current (AC).
- Sketch and interpret sinusoidal AC waveform.
- Write and use sinusoidal voltage and current
- equations.
2
321.1 Alternating current
- An alternating current (ac) is the electrical
current which varies periodically with time in
direction and magnitude.
- An ac circuit and ac generator, provide an
alternating current.
- The usual circuit-diagram symbol for an ac
- source is .
4 The output of an ac generator is sinusoidal
and varies with time.
where
I instantaneous current _at_ current at time t (in
Ampere)
5where
V instantaneous voltage _at_ voltage at time t (in
Volt)
6Vo
Io
T/2
T
- The output of an ac generator is sinusoidal
- and varies with time.
Equation for the current ( I )
Equation for the voltage ( V )
7- Terminology in a.c.
- 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)
- Equation
- Peak (maximum) current ( Io )
- Definition Magnitude of the maximum current.
- Peak (maximum) voltage ( Vo )
- Definition Magnitude of the maximum voltage.
- Angular frequency ( ? )
- Equation
- Unit radian per second
(rads-1)
8SUBTOPIC
21.2 Root Mean Square (rms) (1 hour)
LEARNING OUTCOMES
At the end of this lesson, students should be
able to
- Define root mean square (rms), current and
voltage for AC source. - Use ,
8
921.2 Root mean square (rms)
Root mean square current (Irms) is defined as the
effective value of a.c. which produces the same
power (mean/average power) as the steady d.c.
when the current passes through the same resistor.
the average or mean value of current in a
half-cycle flows of current in a certain direction
10- The r.m.s (root mean square) current means the
- square root of the average value of the current.
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.
V
VVo sin ?t
11- Most household electricity is 240 V AC which
- means that Vrms is 240 V.
12Example 21.2.1
A sinusoidal, 60.0 Hz, ac voltage is read to be
120 V by an ordinary voltmeter.
- What is the maximum value the voltage takes on
- during a cycle?
b) What is the equation for the voltage ?
a)
b)
13Example 21.2.2
A voltage V 60 sin 120pt is applied across a 20
O resistor.
- What will an ac ammeter in series with the
- resistor read ?
b) Calculate the peak current and mean power.
14Example 21.2.3
The alternating potential difference shown above
is connected across a resistor of 10 k?.
Calculate a. the r.m.s. current, b. the
frequency, c. the mean power dissipated in the
resistor.
15Solution 21.2.3
16Exercise 21.2
An ac current is given as I 5 sin (200t) where
the clockwise direction of the current is
positive. Find
- The peak current
- The current when t 1/100 s
- The frequency and period of the oscillation.
5 A , 4.55 A, 31.88 Hz, 0.0314 s
17SUBTOPIC
21.3 Resistance, reactance and impedance (2 hours)
LEARNING OUTCOMES
At the end of this lesson, students should be
able to
- Sketch and use phasor diagram and sinusoidal
waveform to show the phase relationship between
current and voltage for a single component
circuit consisting of - i) Pure resistor
- ii) Pure capacitor
- iii) Pure inductor
17
18- b) Define and use
- i) capacitive reactance,
- ii) inductive reactance,
- iii) impedance,
, and - phase angle,
- c) Use phasor diagram to analyse voltage,
current, and impedance of series circuit of - i) RC
- ii) RL
- iii) RLC
18
1921.3 Resistance, reactance and impedance
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 sinusoidal alternating
- quantity such as current and voltage.
- It also being used to determine the phase
- difference between current and voltage
in ac circuit.
20Phasor diagram
y
y
Ao
N
P
O
- The projection of OP on the vertical axis (Oy)
is ON, - represents the instantaneous value.
- Ao is the peak value of the quantity.
21Resistance, reactance and impedance
Key Term/O 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)
22i) Pure Resistor in the AC Circuit
VR
Phasor diagram
23i) 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).
24i) Pure Resistor in the AC Circuit
- The resistance in a pure resistor is
A resistor in ac circuit dissipates energy in the
form of heat
25ii) Pure Capacitor in the AC Circuit
- Pure capacitor means that no resistance and
- self-inductance effect in the a.c. circuit.
VR
Phasor diagram
26ii) Pure Capacitor in the AC Circuit
- When an alternating voltage is applied across a
- capacitor, the voltage reaches its maximum
value - one quarter of a cycle after the current
reaches its - maximum value,( )
- The voltage across the capacitor VC at any
instant - is equal to the supply voltage V and is given
by
- The charge accumulates on the plates of the
- capacitor is
- The current flows in the ac circuit is
27ii) Pure Capacitor in the AC Circuit
and
or
- The phase difference between V and I is
28ii) Pure Capacitor in the AC Circuit
- 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.
- The capacitive reactance in a pure capacitor is
- The capacitive reactance is defined as
29ii) Pure Capacitor in the AC Circuit
- For the first half of the cycle where the power
is negative, the power is returned to the
circuit. For the second half cycle where the
power is positive, the capacitor is saving the
power.
30Example 21.3.1
ii) Pure Capacitor in the AC Circuit
An 8.00 µF capacitor is connected to the
terminals of an AC generator with an rms voltage
of 150 V and a frequency of 60.0 Hz. Find the
capacitive reactance rms current and the peak
current in the circuit.
Capacitive reactance,
rms current,
Peak current ?
31iii) Pure Inductor in the AC Circuit
- Pure inductor means that no resistance and
capacitance effect in the a.c. circuit.
VL
Phasor diagram
32iii) Pure Inductor in the AC Circuit
- When a sinusoidal voltage is applied across a
- inductor, the voltage reaches its maximum
value one quarter of a cycle before the current
reaches its maximum value,( )
- 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
33iii) 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 ,
0
or
where
34iii) 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.
- The inductive reactance in a pure
inductor is
35iii) Pure Inductor in the AC Circuit
- The inductive reactance is defined as
- 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.
36Example 21.3.2
iii) Pure Inductor in the AC Circuit
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.
Example 21.3.3
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.
37i) RC in series circuit
- In the circuit diagram
- VR and VC represent the instantaneous voltage
- across the resistor and the capacitor.
- In the phasor diagram
- VR and VC represent the peak voltage across the
- resistor and the capacitor.
38i) RC in series circuit
Note
39i) RC in series 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.
and
40i) RC in series circuit
Impedance diagram
- From the phasor diagrams,
- The impedance in RC
- circuit,
I leads V by F
or
41i) RC in series circuit
Z
R
f
0
Graph of Z against f
42Example 21.3.4
i) RC in series circuit
An alternating current of angular frequency of
1.0 x 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.
43ii) RL in series circuit
- The voltage across the resistor VR and the
capacitor - VL are
44ii) RL in series 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.
and
45ii) RL in series circuit
Impedance diagram
- From the phasor diagrams,
- The impedance in RC
- circuit,
V leads I by F
or
46ii) RL in series circuit
Z
R
f
0
Graph of Z against f
47iii) RLC in series circuit
48iii) RLC in series circuit
49iii) RLC in series circuit
- The voltage across the inductor VL , resistor VR
and - capacitor VC are
50iii) RLC in series circuit
- 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.
51iii) RLC in series circuit
Impedance diagram
- From the phasor diagrams,
- The impedance in RLC
- circuit,
V leads I by F
52Resonance in RLC series 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 series resonance circuit is used for tuning
a radio receiver.
Graph of impedance Z, inductive reactance XL,
capacitive reactance XC and resistance R with
frequency.
53Resonance in RLC circuit
The graph shows that
- at low frequency, impedance Z
- is large because 1/?C is large.
- at high frequency, impedance Z
- is high because ?L is large.
- at resonance, impedance Z is minimum (ZR)
- which is
resonant frequency
and I is maximum
54(No Transcript)
55Example 21.3.5
iii) RLC in series circuit
A series circuit contains a 50 O resistor
adjacent to a 200 mH inductor attached to a
0.050µF capacitor, all connected across an ac
generator with a terminal sinusoidal voltage of
150 V effective.
- What is the resonant frequency ? (1.59 kHz)
- What voltages will be measured by voltmeters
across each element at resonance ? (150V,6kV) - c) What is the voltage across the series
combination of the inductor and capacitor ? - Write the equation for the supply voltage at fr.
56Example 21.3.6
iii) RLC in series circuit
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.
57Solution 21.3.6
iii) RLC in series circuit
58Exercise 21.3
iii) RLC in series circuit
A series RLC circuit has a resistance of 25.0 O,
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
- The total impedance of the circuit
- The rms current in the circuit
- The phase angle between the voltage and the
current.
64.9 O , 1.85 A, 67.3o
59SUBTOPIC
21.4 Power and power factor (1 hour)
LEARNING OUTCOMES
At the end of this lesson, students should be
able to
- Apply
- i) average power,
- ii) instantaneous power,
- iii) power factor,
- in AC circuit
consisting of R, RC, RL and RLC in series
59
6021.4 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 given by the average power (Pave) i.e
61(for RLC circuit)
Impedance diagram
.. (2)
(2) into (1)
Vrms supply voltage
62- The term cos ? is called the power factor.
- The power factor (cos ? ) can vary from a
- maximum of 1 to a minimum of 0.
- 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.
63- The power factor can be expressed either as a
- percentage or a decimal.
- A typical circuit has a power factor of less
than 1 - (less than 100).
- Example
- A motor has a power factor of 80 and the motor
- consumes 800 W to operate. In order to operate
- properly, the motor must be supplied with more
power - than it consumes i.e.1000 W .
800 W (consume)
power factor (80)
1000 W (supply)
64Example 21.4.1
An oscillator set for 500 Hz puts out a
sinusoidal voltage of 100 V effective. A 24.0 O
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?
65Solution 21.4.1
66Exercise 21.4
1. 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.
4.4 A, 61.3o , 0.48, 0.23 kW
67Exercise 21.4
2. 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 t he 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.
2.94 x 10-3 H , 4.84 O , 0.163