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PURE RESISTANCE
Consider the application to a pure resistance R of a voltage of:
v = Vo + V1m sin(t + 1) + V2m sin(2t + 2)+….…+ Vnm sin(nt + n)volts
where n is the number of the harmonic. Since i = v/R and resistance R is independent of
frequency, then the current due to the Vo term is Vo/R, that due to the first harmonic term is
(V1 sin t)/R, that due to the second harmonic term is (V2 sin 2t)/R and so on. Thus the
resulting current waveform is:
I = Vo + V1m sin(t + 1) + V2m sin(2t + 2)+….…+ Vnm sin(nt + n)volts
R
R
R
R
Because the resistance is the same for each harmonic, the amplitude of each voltage harmonic
is reduced by the same factor, i.e. the resistance, The phases of each harmonic are not
changed. The current waveform is thus the same shape as the voltage waveform.
Example
A complex voltage of 5 + 6.4 sin l00t + 1.6 sin 200t V is applied across a resistor having a
resistance of 100 . Determine the current through the resistor.
The complex current will be the sum of the currents due to each of the voltage terms in the
complex voltage. Since the resistance is the same at all frequencies, the complex current will
be:
i = 0.05 + 0.064 sin l00t + 0.016 sin 200t A
Revision
1.
Determine the waveform of the current occurring when a resistor of resistance 1 k
has connected across it the half-wave rectified sinusoidal voltage v = 0.16 + 0.1 cos
l00t + 0.21 cos 200t V.
[i = 0.16 + 0.1 cos l00t + 0.21 cos 200t mA].
PURE INDUCTANCE
Consider the application to a pure inductance L of a voltage of:
v = Vo + V1m sin(t + 1) + V2m sin(2t + 2)+….…+ Vnm sin(nt + n)volts
where n is the number of the harmonic. The impedance of a pure inductance depends on the
frequency, i.e. its reactance XL = 2fL . In an inductive circuit the current lags the voltage
by 90o. The graph shows the variation of reactance with frequency.
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Graph to show variation of Xl with frequency
1500
1400
1300
1200
1100
1000
900
800
700
600
500
400
300
200
100
0
1000
Xl (ohms) 900
800
700
600
500
400
300
200
100
0
Frequency (Hz)
The impedance is 0 when the frequency is 0 and thus the current due to the V o term will be 0.
The current due to the first harmonic will be the voltage of that harmonic divided by the
impedance at that frequency and so V1sin(t + 1 - 900)/L. The current due to the second
harmonic will be the voltage of that harmonic divided by the impedance at that frequency and
so V2 sin(2t + 2 – 90o)/2L. Thus the current waveform will be
I = V1m sin(t + 1 – 90o) + V2m sin(2t + 2)+….…+ Vnm sin(nt + n – 90o)volts
L
2L
nL
Each of the voltage terms has its amplitude altered by a different amount; the phase, however,
is changed by the same amount. The result is that the shape of the current waveform is
different to that of the voltage waveform.
In terms of phasors, since the impedance Z = V/I, then for each of the Fourier series terms we
have:
Zn = Vn = Vn /2 = nL 90o
In
In
nL is the reactance XL for the harmonic concerned. We can represent the above relationship
in complex notation as Zn = jXL = jnL.
Example
A complex voltage of v = 5 + 6.4 sinl00t + 1.6sin200t V is applied across a pure inductor
having a inductance of 100 mH. Determine the current through the inductor.
When the frequency is 0 the reactance is zero. 5 V is a dc and therefore has zero frequency
hence the current due to this term is zero. For the next term the reactance X L = L = 100 x
100 x 10-3 = 10 and the current lags the voltage by 90o and so the current due to this is
6.4/10 = 0.64 sin(l00t – 900) For the net term the reactance, XL = is 200 x 0.100 = 20  and
the current lags the voltage by 90o and so the current due to this harmonic is 0.08 sin(200t 900) A. Thus the current waveform is: i = 0.64 sin(l00t – 900) + 0.08 sin(200t - 900)
2
Revision
2.
Determine the waveform of the current when a pure inductor of inductance 10 mH
has connected across it the half-wave rectified sinusoidal voltage V = 0.32 + 0.5 cos l00t +
0.21 cos 200t V.
[0.5cos(l00t – 90o) + 0.21 cos(200t – 90o)]
PURE CAPACITANCE
Consider a pure capacitor capacitance C when the voltage applied across it is:
v = Vo + V1m sin(t + 1) + V2m sin(2t + 2)+….…+ Vnm sin(nt + n)volts
with n being the number of the harmonic. The impedance of a pure capacitor depends on the
frequency, i.e. its reactance XC = 1/C, and the current leads the voltage by 90o. . The graph
shows the variation of reactance with frequency.
Gaph to show the variation of XC with frequency
3.5
Xc (ohms)
3
2.5
2
1.5
1
0.5
1500
1400
1300
1200
1100
1000
900
800
700
600
500
400
300
200
100
50
0
frequency (Hz)
The reactance is  when the frequency is 0 and thus the current due to the Vo term will be 0.
The current due to the first harmonic will be the voltage of that harmonic divided by the
impedance at that frequency and so V1m sin(t + 1 + 900)/(l/C). For the second harmonic
the current will be the voltage of that harmonic divided by the impedance at that frequency
and so V2m sin(2t + 2) + 900)/(l/2C). Thus the current waveform will be:
i = C V1m sin(t + 1 + 900) + 2C V2m sin(2t + 2) + 900) + …….. + nC Vnm sin(nt
+ n + 90o)A
Each of the voltage terms has had their amplitude altered by a different amount but the phase
changed by the same amount. The result of this is that the shape of the current waveform is
different to that of the voltage waveform.
In terms of phasors, since the impedance Z = V/I, then for each of the Fourier series terms we
have:
Zn = Vn = Vn
In
In/2
=
1 -90o
nC
3
1/nC is the reactance Xc for the harmonic concerned. We can represent the above
relationship in complex notation as Zn = -jXC = -j(1/nC) or 1/j nC.
Example
Determine the waveform of the current occurring when a 2F capacitor has connected across
it the half-wave rectified sinusoidal voltage V = 0.32 + 0.5 cos l00t + 0.21 cos 200t V.
There will be no current arising from the d.c. term. For the first harmonic the reactance is
1/(2 x 10-6 x 100)  and so we have current of 0.5 x 2 x 10-6 x 100cos (100t + 90o) A. For the
second harmonic the reactance is 1/(2 x 10-6 x 200)  and so the current 0.21x 2 x 10-6 x 200
cos (200t + 90o). Thus the resulting current is:
i = 0.5 x 2 x 10-6 x 100cos (100t + 90o) + 0.21x 2 x 10-6 x 200 cos (200t + 90o) A
Revision
A voltage of 2.5 + 3.2 sin l00t + 1.6 sin 2001 V is applied across a 10F capacitor. Determine
the current.
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11.3.4 Circuit elements in series or parallel
For circuit elements in series, the total impedance is the sum of the impedances of the
separate elements. Thus if we have an inductance L in series with resistance R when there is
an input of a voltage having harmonics, the impedance Z, of the nth harmonic is the sum of
the impedances for the nth harmonic of the two elements and is thus:
Z~=R+jncoL [29]
If we had a resistance R, an inductance L and a capacitance C in series, then:
Z~=R+jncoL+ 1
[30]
jncoC
For parallel circuits, say a resistance R in parallel with an inductance L, when there is an
input of a voltage having harmonics, the total impedance Zfl of the nth harmonic is given by:
1 1+ 1
[31] Z~~R jncoL
If we had a resistance R, an inductance L and a capacitance C in parallel then the total
impedance for the nth harmonic is given by:
I_1 1
1
[321
+
R+ jncoL lIjncoC
Example
A voltage of V = 100 cos 3 14t + 50 sin(5 x 3 14t - 300) V is applied to a series circuit
consisting of a 10 fl resistor, a 0.02 H inductor and a 50 ~ capacitor. Determine the circuit
current.
For the first harmonic, the resistance is 10 ~, the inductive reactance is coL = 314 :: 0.02 =
6.28 ~ and the capacitive reactance is 1/coC = 11(314:: 50:: 10~ = 63.69 ~. Thus the total
impedance for the first harmonic is:
Zi = 10 +j6.28 -j63.69 = 10 -j57.41 = ,1102+57.412 Ltan~1 -57.41 =58.3L(-80.10)
10
Thus the current due to the first harmonic is:
100L00 -172L80.10
58.3L(-80.1o) -For the fifth harmonic, the resistance is 10 ~, the inductive
reactance is ScoL = 5 :: 314 :: 0.02 = 31.4 fl and the capacitive reactance is 1/ScoC = 11(5 ::
314:: 50:: 10~ = 12.74 ~. Thus the total impedance is:
z5= 10+j31.4-j12.74= 10+j18.66
=
10
1102+18.662 Ltan~1 18.66 =21.2L61.80
Thus the current due to the fifth harmonic is:
5
i5 = 2510.;L~~6310.80~o =2.36L(-91.80)
Thus the current waveform is:
i = 1.72 cos(314t +80.10) +2.36 cos(3 :: 314t- 91.80) A
Revision
7
A voltage of 200 cos 3 14t - 40 sin 2:: 3 14t V is applied to a circuit consisting of a 20
fl resistor in series with a 100 ~AF capacitor. Determine the current in the circuit.
8
A voltage of 1.0 sin SOot + 0.2 sin 3 :: SOOt V is applied to a circuit consisting of
three series components, a resistance of 500 ft, an inductance of 1 H and a capacitance of 1 ~
Determine the current.
9
A triangular voltage waveform given by the Fourier series V = 123 inductance of 0.1 H and a capacitance of 25 ~. Determine the circuit current.
11.3.5 Selective resonance
The presence of harmonics in a waveform may give rise to selective resonance. This is when
a circuit containing both inductance and capacitance resonates at one of the harmonic
frequencies. It thus occurs, for a series circuit or parallel circuit involving inductance and
capacitance, when the inductive reactance of a particular harmonic is equal to the capacitance
reactance of that harmonic, i.e. when:
ncoL = [33]
ncoC
where n is the number of the harmonic. When selective resonance occurs, the magnitude of
the harmonic concerned is greatly increased. This has the effect of distorting the current
waveform. It can also give rise to dangerously high voltage drops across the inductance and
capacitance in the circuit. Selective resonance can also be beneficially used for tuning a
circuit to a particular harmonic if that harmonic is to be enhanced while the others are
reduced.
Example
The voltage 300 sin cot + 100 sin 3cot V is applied to a series circuit consisting of resistance
10 CY, inductance 0.5 H and capacitance 0.2 ~IF. Determine (a) the flindamental frequency
for resonance with the third harmonic, and ~) the current waveform at that frequency.
(a) Resonance with the third harmonic occurs when (equation 33) ncoL = lincoC and so:
co=n'~ =+~0.2'::l0~ = 1054rad/s
and hencef co/2ir = 168 Hz.
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(1))The impedance of the circuit at the flindamental is:
ZI =R+j(coL-~)
= 10+j(1054::0.5
l054::0.2::l0~)
=
1
10 -j4217 = 4217L(-89.90) ~
Thus the current at the flindamental is 300L00I[4217L(-89.~)] = 0.071 L89.~ A. The
imnedance at the third harninni~
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