Pearson BTEC Level 4 HNC & Level 5 HND in
Electrical & Electronic Engineering (RQF)
Unit 19
Electrical & Electronic Principles
Unit Code: M/615/1493
Unit Learning Outcome 2
Sinusoidal Circuits
Series Resonance
Learning Outcomes
By the end of the session all learners should be
able to:
• state the conditions for series resonance
• define the terms resonant frequency, Q factor,
selectivity, bandwidth & cut off frequencies
• describe how frequency affects the impedance of
a series resonant circuit
• describe how Q factor affects the selectivity of
a series resonant circuit
• carry out calculations relating to series resonance,
including resonant frequency, Q factor, bandwidth
& other related quantities
Learning Outcomes
By the end of the session most learners should be
able to:
• use a given frequency response for an RLC series
circuit to determine the circuit’s quantities at
resonance
Learning Outcomes
By the end of the session some learners should
be able to:
• begin to carry out calculations involving parallel
resonant circuits
Introduction
Series resonance circuits are one of the most
important circuits used in the electrical &
electronics industry.
They can be found in various forms such as AC
mains filters, noise filters & also in television &
radio tuning circuits.
These television & radio tuning circuits produce a
very selective tuning circuit for the receiving of
the different frequency channels.
Series Resonance
For an AC series circuit we recall that
XL = 2fL &
XC = 1
2fC
From this we see that the reactance of the
inductor increases as frequency increases & the
reactance of the capacitor decreases as
frequency increases:
Series Resonance
Overlaying these two frequency responses on top
of each other reveals the point where XL = XC:
The frequency at which XL = XC is a special
frequency called the resonant frequency fr.
At frequencies less than fr, XL < XC & the circuit
is capacitive; at frequencies greater than fr, XL >
XC & the circuit is inductive.
Series Resonance
At resonance, XL = XC 
2frL = 1  rearranging
2frC
(2fr)2 = 1 
LC
2fr =
1  rearranging
LC
fr =
1
2 LC
where fr is the resonant frequency
Series Resonance
The series resonant circuit comprises of a coil of
inductance L & resistance R connected in series
with a capacitor C. A resistive element is always
present due to the internal resistance of the
source, the internal resistance of the inductor, &
any added resistance used to control the shape of
the frequency response curve:
Series Resonance
The series resonant circuit has a frequency
response characteristic similar to the one shown in
the following diagram:
We note that frequency varies on the horizontal
axis while amplitude varies on the vertical axis.
Series Resonance
This circuit is at resonance when (XL – XC) = 0
i.e. when XL = XC or L = 1/C. These two
reactance's are now equal & opposite &, hence,
they cancel each other out, meaning the only
opposition to current flow is the resistance.
When the circuit is at resonance, the applied
voltage & current are in phase & the current is at
a maximum:
VL = IXL
VR = V
VC = IXC
I
Series Resonance
The key points are as follows:
At resonance:
(1)
XL = XC  VL = VC
(2)
Z = R i.e. the minimum circuit impedance possible
(3)
I = V/R i.e. the maximum current possible
(4)
Applied voltage & current are in phase
Series Resonance
(5)
From earlier, at resonance, XL = XC 
fr =
1
2 LC
(6)
The series resonant circuit is often described as
an acceptor circuit since it has minimum
impedance, & thus maximum current, at the
resonant frequency.
Example 1
An RLC series circuit consists of a coil of
resistance 10, an inductance of 125mH & a
capacitance of 60F. If the circuit is connected
to a 120V AC supply, determine:
a) the resonant frequency
b) the current flowing at resonance
a)
fr =
1 =
2 LC
1
=
2    125  10-3  60  10-6
Example 1
58.12Hz
b)
At resonance, XL = XC & impedance Z = R 
I = V/R = 120/10 = 12A
Example 2
The current at resonance in an RLC series circuit
is 100A. The applied voltage is 2mV at a
frequency of 200kHz & the circuit inductance is
50H. Determine:
a) the value of the circuit resistor
b) the value of the circuit capacitor
a)
At resonance Z = R  R = V/I =
2  10-3 =
100  10-6
20
Example 2
b)
At resonance XL = XC 
2frL = 1  rearranging
2frC
C =
1
=
(2fr)2L
1
(2  200  103)2(50  10-6)
0.0127F or 12.7F
=
Q Factor
At resonance, if R is small compared with XL &
XC, it is possible for VL & VC to have voltages
many times greater than the supply voltage.
Voltage magnification at resonance =
voltage across L or C
supply voltage V
This ratio is a measure of the quality of a circuit
(as a resonator or tuning circuit) & is called the Q
factor.
More formally, Q factor is the ratio of power
stored (reactive power) to the power dissipated
(resistive power) i.e.
Q Factor
Q factor = Power stored = Q =
Power dissipated P
I2(XL or XC) =
I2R
XL or XC
R
We can also write
Q factor = VL = IXL = XL = 2frL
V
IR
R
It can also be proved that
R
Q Factor
Q factor =
1
2frCR
or 1 L
R C
Example 3
A coil having an inductance of 80mH & a
negligible resistance is connected in series with a
capacitance of 0.25F & a resistor of resistance
12.5 across a 100VAC, variable frequency
supply. Determine:
a) the resonant frequency
b) the circuit current at resonance
c) the voltage magnification factor
a)
Resonant frequency
fr = 1 =
2 LC
Example 3
1
=
2 80  10-3  0.25  10-6
1125.4Hz or 1.125kHz
b)
Current at resonance I = V/R = 100/12.5 =
8A
c)
XL = 2frL = 2    1125.4  80  10-3 =
565.6877
VL = IXL = 8  565.6877 =
4525.5V
Example 3
XC = 1
=
1
=
2frC 2    1125.4  0.25  10-6
565.68
Voltage magnification at resonance =
VL/V or VC/V =
4525.5/100 = 45.255
What does this result tell us?
At resonance, the voltage across the reactances
are 45.255 times greater than the supply voltage.
Hence the Q factor of the circuit is 45.255.
Example 4
A series circuit comprises a coil having a
resistance of 2 & an inductance of 60mH, & a
30F capacitor. Determine the Q factor of the
circuit.
At resonance
Q factor =
1
L = 1 60  10-3 =
R
C
22.36
2 30  10-6
Example 5
A coil of negligible resistance & inductance
100mH is connected in series with a capacitance
of 2F & a resistance of 10 across a 50VAC,
variable frequency supply. Determine:
a) the resonant frequency
b) the current at resonance
c) the voltages across the coil & capacitor at
resonance
d) the Q factor of the circuit
a)
Resonant frequency fr = 1
2 LC
=
Example 5
1
=
2 100  10-3  2  10-6
355.88Hz
b)
Current at resonance I = V/R = 50/10 =
5A
c)
XL = 2frL = 2    355.88  100  10-3 =
223.62
VL = IXL = 5  223.62 =
1118.8V
Example 5
XC = 1
=
1
2frC 2    355.88  2  10-6
223.61
VC = IXC = 5  223.62 = 1118.8V
d)
Voltage magnification at resonance =
VL/V or VC/V =
1118.8/50 = 223.37
=
Bandwidth
The bandwidth of a circuit is defined as the
frequency range between the half-power points
when I = Imax or I = 0.7071  Imax.
2
Bandwidth
The bandwidth (BW) equals f1 - f2, where the
frequencies f1 & f2 are referred to as half power
points or cut off frequencies.
The term ‘half power’ can be justified by
consideration of the conditions for maximum &
half power for the series RLC circuit.
At maximum power, when f = fr
Imax = V/R 
Pmax = I2maxR = V2/R
At the half power points
P1 = P2 = I2maxR
2
Bandwidth
Thus, the condition for half power is given when
|I| = Imax = V
2
R2
Selectivity
Selectivity is the ability of a circuit to respond
more readily to signals of a particular frequency
to which it is tuned than to signals of other
frequencies.
The response becomes progressively weaker as
the frequency departs from the resonant
frequency.
Selectivity
The higher the Q-factor, the narrower the
bandwidth & the more selective the circuit.
It may be shown that
Q = fr
or
f2 – f1
f 2 – f 1 = fr
Q
Selectivity
Circuits having high Q factors (i.e. in the order
of 100 to 300) are  useful in communications
engineering.
A high Q-factor in a series power circuit has
disadvantages in that it can lead to dangerously
high voltages across the insulation & may result in
electrical breakdown.
What is the reason for this?
Example 6
A filter in the form of an RLC series circuit is
designed to operate at a resonant frequency of
5kHz. Included within the filter is a 20mH
inductance & 10 resistance. Determine the
bandwidth of the filter.
At resonance
Qr = 2frL = 2  5000  20  10-3 = 628.31
Qr =
R
10
fr
 rearranging
f2 – f1
f2 – f1 = fr = 5  103 =
Qr
628.31
Example 6
7.96Hz
Example 7
An RLC series circuit has a resonant frequency of
2kHz & a Q factor at resonance of 40. If the
impedance of the circuit at resonance is 30,
determine:
a) the inductance
b) the capacitance
c) the bandwidth
d) the lower & upper -3dB frequencies
a)
At resonance R = Z
Q = 2frL  rearranging
R
Example 7
L = QR =
2fr
40  30
=
2    2000
95.5mH
b)
At resonance
XL = XC = 2frL = 2    2000  0.0955 =
1.2k
XC = 1  rearranging
2frC
Example 7
C =
1
2frXC
=
1
=
2    2000  1200
66.3F
c)
Q = fr/BW  rearranging
BW = fr/Q = 2000/40 = 50Hz
d)
The half power frequencies are at 25Hz above &
below the resonant frequency of 2000Hz i.e.
f2 = 2000Hz + 25Hz = 2025Hz
f1 = 2000Hz - 25Hz = 1975Hz
Example 8
Given the following frequency response, find:
a) the Q factor & bandwidth
b) the values of L & R given C = 101.5F
c) the applied voltage
Example 8
a)
From the above frequency response, the resonant
frequency fr = 2800Hz & the bandwidth at the
half power points = 2900Hz – 2700Hz = 200Hz.
Example 8
Qr = fr/BW = 2800/200 =
14
b)
fr = 1
 squaring each side
2 LC
f r2 = 1
 transposing
42LC
L =
1
=
42fr2C
1
42  28002  101.5  10-9
=
Example 8
31.83mH
XL = 2frL = 2    2800  0.03183 =
560
Qr = XL/R  rearranging
R = XL/Qr = 560/14 =
40
c)
Imax = V/R  rearranging
V = Imax  R = 200  10-3  40 =
8V
Exercise Solutions
Question 1
An RLC series circuit consists of a coil of
resistance 15, an inductance of 60mH & a
capacitance of 15F. If the circuit is connected
to a 300mV AC supply, determine:
a) the resonant frequency
b) the pd across each component
a)
fr =
1 =
2 LC
1
2    60  10-3  15  10-9
=
1
0.00006

Question 1
5.3kHz
b)
At resonance, the reactive components (XL & XC)
will be equal but opposite. The impedance at
resonance will, thus, be R alone i.e. Z = R 
I = V/Z = V/R =
300  10-3 =
15
20mA
XL = 2frL =
2    5.3  103  60  10-3 =
Question 1
2    5.3  60 = 636 
2k
VL = IXL =
20  10-3  2  103 =
40V
Note that this voltage leads the supply current
by 90o. Since the reactance of the capacitor will
be the same as that of the inductor, the voltage
developed across the capacitor will be identical
but lagging the supply current by 90o. Thus
VL = VC = 40V
Question 2
An RLC series circuit consists of a coil of
resistance 10, an inductance of 50mH & a
capacitance of 50F. If the circuit is connected
to a 100V AC supply, determine:
a) the resonant frequency
b) the current flowing at resonance
a)
fr =
1 =
2 LC
1
2    50  10-3  50  10-9
=
Question 2
100.66Hz
b)
I = V/R = 100/10 = 10A
Question 3
The current at resonance in an RLC series circuit
is 0.2mA. If the applied voltage is 250mV at a
frequency of 100kHz & the circuit capacitance is
0.04F, find the circuit resistance & inductance.
fr = 1
 squaring each side
2 LC
f r2 = 1
 rearranging
42LC
L =
1
=
42Cfr2
1
=
4  2  0.04  10-6  (100  103)2
Question 3
1/15791 =
63.33H
R = V/I = 250mV/0.2mA = 250/0.2 =
1.25k
Question 4
A coil of resistance 25 & inductance 100mH is
connected in series with a capacitance of 0.12F
across a 200VAC, variable frequency supply.
Determine:
a) the resonant frequency
b) the current at resonance
c) the voltage magnification factor
a)
fr = 1 =
2 LC
1
=
2    100  10-3  0.12  10-6
Question 4
1
=
6.88288  10-4
1.4528kHz
b)
I = V/R = 200/25 = 8A
c)
XL = 2fL = 2    1452.879  100  10-3 =
912.871
XC = 1 =
1
2fC 2    1452.879  0.12  10-3
912.871
=
Question 4
VL = IXL = 8  912.871 = 7302.968V
VC = IXC = 8  912.871 = 7302.968V
Q factor =
Voltage magnification factor at resonance =
VL/V or VC/V =
7302.968/200 =
36.51
Question 5
A coil of 0.5H inductance & 8 resistance is
connected in series with a capacitor across a
200VAC, 50Hz supply. If the current is in phase
with the supply voltage, determine the
capacitance of the capacitor, the PD across its
terminals & the voltage magnification factor
Since the current is in phase with the voltage at
the 50Hz supply frequency, the circuit must  be
at resonance.
XL = 2frL = 2    50  0.5 =
157.08
At resonance, XL = XC = 157.08
Question 5
XC = 1
 rearranging
2frC
C =
1
2frXC
=
1
=
2    50  157.08
20.26F
The current at resonance I = V/R = 200/8 =
25A
VC = IXC = 25  157.08 =
3927V
Question 5
Voltage magnification factor at resonance =
VC/V =
3927/200 =
19.635
Question 6
Calculate the capacitance which must be
connected in series with a 158.31mH inductor to
give a resonant frequency of 400kHz.
fr = 1
 squaring each side
2 LC
f r2 =
1
 rearranging
42LC
C =
1
=
42Lfr2
1
4  2  158.31  10-3  (400  103)2
=
Question 6
1000F
Question 7
For the following RLC series circuit, determine:
a) the Q factor using three different methods
b) the bandwidth
a)
fr = 1 =
1
=
2 LC 2    400  10-3  10  10-6
Question 7
1
=
0.004
79.58
Q factor = 2frL = 2    79.58  400  10-3 =
R
4
200 = 50 or
4
Q factor =
1
=
2frCR
1
= 50 or
2    79.58  10  10-6  4
Question 7
Q factor = 1 L =
R C
1  400  10-3 =
4
10  10-6
¼  200 = 50
b)
Bandwidth = fr/Q =
79.58/50 =
1.59Hz
Question 8
For the series resonant circuit shown, determine:
a) the resonant frequency
b) the current at resonance
c) the voltage across each series component
d) the Q factor at resonance
e) the bandwidth
f) the half power frequencies
g) the power dissipated in the circuit at the two
frequencies
Assume the circuit has a variable frequency
supply.
Question 8
a)
fr = 1 =
2 LC
1
=
2    0.15  100  10-6
41Hz
Question 8
b)
The current at resonance I = V/R = 240/8 =
30A
c)
XL = 2frL = 2    41  0.15 =
38.64
XC = 1
=
1
2frC 2    41  100  10-6
VR = IR = 30  8 = 240V
VL = IXL = 30  38.64 = 1159V
VC = IXC = 30  38.64 = 1159V
= 38.81
Question 8
d)
Qr = 2frL = 2  41  0.15 = 4.83
R
8
Alternatively
Voltage magnification factor at resonance =
VC/V = 1159/240 =
4.83
e)
Q = fr/BW  rearranging
BW = fr/Q = 41/4.83 =
8.5Hz
Question 8
f)
The half power frequencies are at 4.25Hz above
& below the resonant frequency of 41Hz i.e.
f2 = 41Hz + 4.25Hz = 45.25Hz
f1 = 41Hz - 4.25Hz = 36.75Hz
g)
Pmax = Imax2R = 302  8 =
7.2kW
Question 9
A coil of inductance 0.2H & resistance 60 is
connected in parallel with a 20F capacitor across
a 20VAC, variable frequency supply. Determine
the resonant frequency.
fr = 1
1 – R2 =
2 LC
1
L2
1
2 0.2  20  10-6
– 602 =
0.22
1/2  160,000 = 1/2  400 =
63.66Hz
Summary
In an RLC series circuit, resonance occurs when
the supply voltage & current are in phase.
At the resonant frequency of a series resonant
circuit, the impedance is a minimum & equal to the
circuit resistance. The magnitudes of the
inductive & capacitive reactances are equal & 
cancel each other out.
At resonance, the voltages which appear across
the reactive components can be many times
greater than that of the supply voltage. This
magnification, called the voltage magnification
factor in the series resonant circuit, is called the
Q factor.
Summary
An RLC series circuit accepts maximum current
from the source at resonance & for that reason is
called an acceptor circuit.
Below the resonant frequency, the circuit is
entirely capacitive & above the resonant
frequency, the circuit is entirely inductive.
The selectivity of a resonant circuit is its ability
to respond more readily to signals of a particular
frequency to which it is tuned than to signals of
other frequencies.
This means the higher the Q-factor, the
narrower the bandwidth & the more selective the
circuit.
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