iii) RLC in series circuit XL ω
VL ω XL XC Z
VL VC V
I R
VC VR
XC
Impedance diagram
Phasor diagram • From the phasor diagrams,
• The impedance in RLC V leads I by Φ
circuit,
tan VL VC tan I X L X C
Z Vrms I R2 X L X C 2
VR IR
Irms I
tan X L X C ωL 1
Z R2 X L X C 2 ωC
R tan
R 51
Resonance 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.
XC , X L , R, Z Z
The series resonance XL f
circuit is used for
tuning a radio receiver. R
0 fr XC 1
f f
Graph of impedance Z, inductive reactance
XL, capacitive reactance XC and resistance52R
with frequency.
Resonance in RLC circuit XC , X L, R, Z Z
The graph shows that :
• at low frequency, impedance Z XL f
is large because 1/ωC is large.
• at high frequency, impedance Z R
is high because ωL is large. 0 fr XC 1
f f
• at resonance, impedance Z is minimum (Z=R)
which is XL XC Z R2 X L XC 2
resonant 2fr L 1 Zmin R2 0
frequency Zmin R
2frC
fr 2 1
LC
and I is maximum Irms Vrms Vrms 53
Z R
54
iii) RLC in series circuit
EXERCISE 18.3.5
A series circuit contains a 50 Ω 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.
a)What is the resonant frequency ? (1.59 kHz)
b)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 ?
a)Write the equation for the supply voltage at fr.
55
Example 18.3.6 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.
56
Solution 18.3.6 iii) RLC in series circuit
R = 200 , L = 0.75 H ,Vrms = 250 V, f = 600 Hz.
a) X L L 2.83 k
XC 2.83 k
b) 2.83 x103 1 , C 93.9 nF
2fC
c) Z = R = 200 VL
Vrms Vrms e) VR I
Z R VC
d) I rms 1.25 A 57
Exercise 18.3 iii) RLC in series circuit
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.
64.9 Ω , 1.85 A, 67.3o
58
SUBTOPIC :
18.4 Power and power factor (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, students should
be able to :
a) Apply
i) average power, Pav IrmsVrms cos
ii) instantaneous power, P IV
iii) power factor, cos Pr Pav
Pa IrmsVrms
in AC circuit consisting of R, RC, RL and RLC in
series
59
18.4 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 :
Pave I 2 R Pr Pave IrmsVRrms … (1)
rms
rms voltage
across resistor
60
(for RLC circuit)
VL ω
XL ω
VL VC V XL XC Z
I
VC VR
R
XC
Phasor diagram Impedance diagram
• From the diagrams above,
cos VR and cos R ….. (2)
VZ
(2) into (1)
V=rms Pave I 2 Z cos cos Pave
supply rms
voltage Pa
or Papparent Pa
61
Pave IrmsV cos
• 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.
62
• 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 .
power factor cos Pave 800 W (consume)
(80%)
Pa 1000 W (supply)
63
Example 18.4.1
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?
d) Calculate the power supply.
e) Find the power factor.
f) What is the phase angle? 64
Solution 18.4.1
f=500 Hz , V=100 V , R=24.0 Ω , C=10.0μF,
L=50.0 mH.
a) I rms Vrms 784 mA
Z
b) VR IR 18.8 V
VL IXL 123 V
VC IXC 24.9 V
c) Real power ?
Pave IrmsV cos 65
Pave 0.7841000.188
Pave 14.7 W Pr
d) Power supply,Pa I Vrms rms
0.784(100)
78.4 W
e) Power factor,
cos R 0.188
Z
f ) cos 0.188
cos1(0.188)
79.16o
66
Exercise 18.4
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
67
Exercise 18.4
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 Ω , 0.163
68