1
The University of Adelaide
Department of Electrical and Electronic Engineering
Electrical Systems B — Electromagnetic Fields
Tutorial 6 Solution
1. (a) When there is a current I through a coil of N turns there is a magnetic flux Φ linked
with the coil where
N Φ = LI
and L is the self inductance of the coil. The induced emf induced in the coil, sensed
in the direction of the current, is given by
E = −N
dΦ
di
= −L
dt
dt
When the current is steady (ie di/dt = 0) the induced emf is zero and the coil does
not affect the current in the circuit (if we ignore any resistance that it may have)
but as it does have an inductance, whenever the current is changing there will be an
induced emf.
(b) There are two ways in which induced emf in an inductance, as defined in magnitude
by Faraday’s law and direction by Lenz’s law, may be given expression in a schematic
circuit. Both are illustrated in Figure 1.
A
i
L
+
L
_
C
A
i
vI = - L di
dt
B
vL = L di
dt
_
+
B
Figure 1: Expressions of Faraday’s and Lenz’s Law.
In the left part of the diagram accessible nodes of the physical inductor are labelled
A and B. The wriggly line drawn between nodes A and C does not represent all
aspects of the inductor, but merely serves as a reminder that an inductor is present,
and that one of its terminals is connected to node A. The induced emf aspect of the
inductor is represented by the voltage generator vI appearing between nodes C and
B. We note that the current i and the voltage generator vI are sensed in the same
direction, downward in this case. The negative sign in the equation
vI = −L
di
dt
2
is an expression of Lenz’s law. It should be noted that node C in not a physically
accessible node, but is introduced so that we can show the induced voltage separately
from the part of the symbol which serves as a reminder that an inductor is present.
We note also that in this representation there is supposed to be no potential difference
between the ends of the wiggly line, i.e. between nodes A and C.
A more compact schematic circuit expression of the same inductor is shown in the
right hand part of Figure 1. Here the wiggly line does represent the entire inductor,
in that firstly it is drawn between nodes A and B to which the ends of the physical
inductor are connected, and secondly in that it has between its ends the same voltage
vL as would appear between nodes A and B in the left hand diagram, interpreted
as described above. We should note that in the right hand diagram the current i
and the inductor voltage vL are sensed in opposite directions, ie. the terminal of vL
which is marked + is the node at which current enters the inductor.
Comparing the two diagrams, we see that the appropriate relation between i and vL
which honours both Faraday’s and Lenz’s laws is
vL = L
di
dt
We note specifically here the absence of a negative sign. Lenz’s law has been acommodated by suitable choices of the relative senses of i and vL .
We see slso that the significance of the wiggly line in the two circuits is quite different;
in one case there is no voltage between its ends and in the other case the inductor
voltage vL , appropriately sensed, appears between its ends.
In the second circuit the expression given for vL should not be thought of as defining
a voltage generator in parallel with the inductor, but instead as making a statement about the relation between the current through the inductor and the potential
difference across it.
In most electrical circuit diagrams the second convention is the one normally chosen
to represent inductors. This choice is made for the reasons firstly that it is more
compact than the other, and secondly that because when compared, as in Figure 2,
with the representations of the two other basic elements of electric circuits, i.e.
the resistor and the capacitor, there is the similarity that when the currents and
voltages are sensed in the same relative directions in the three diagrams, none of the
accompanying equations contain a negative sign. (How nice.)
In answering the question posed in the Tutorial, we will choose here the second
convention, but an alternative answer in which the first convention (i.e. using explicit
voltage generators in series with the inductors) is equally acceptable. What is not
acceptable is either the omission in either convention of a clear statement of the
senses for the voltages and currents, or an incorrect statement of these senses. (Nag,
nag, nag.)
Turning now to the first circuit, it is reproduced in Figure 3 with a reference direction
for current i introduced, and voltages vL and vR , sensed in relation to the current i
so that the equations
vR = Ri
vL = L
di
dt
3
+
vL
i
i
+
vR
L
_
vC
R
C
_
_
vL = L di
dt
i
+
i=C
vR=R i
dvC
dt
Figure 2: Current voltage relations in R, L and C.
_
vR
+
_
R
_
+
vL
i
+
Figure 3: The simple RC circuit.
are both true. The below equation for the voltage change around the loop is obtained
by starting at the top left corner and proceeding counter-clockwise.
E −L
di
− Ri = 0
dt
Turning now to the second circuit, it is reproduced in Figure 4 with variables vC , q
and i already defined in the question, and new variables vR and vL sensed in relation
to i so that the usual equations
vR = Ri
vL = L
di
dt
apply. Before proceeding to apply Kirchhoff’s loop rule, however, we should note
that the relative senses of vC and q and of i and q are such that the equations
q = CvC
and
i=−
dq
dt
4
L
+
_
vR
vL
+
i
R
C
_
-q
q
_
vC
+
Figure 4: The RLC circuit.
apply. Now with the variables defined as above we can apply Kirchhoff’s loop rule
in a counter-clockwise direction around the circuit to obtain
vC − vR − vL = 0
and after a number of substitutions obtain the form
q
dq
d2 q
+ R + L 2 = 0.
C
dt
dt
2. (a) The schematic circuit with apropriate notation is provided in Figure 5.
+
_
i
_
vR
+
R
vL
L
_
+
Figure 5: Excitation of inductor with resistance.
(b) The current flowing in an LR circuit after time t is
p
i = I0 1 − e−t/τ
Q
where I0 is the final i.e. t → ∞ value of the current, i.e.
5
I0 =
100
E
=
= 10 A
R
10
τ=
L
2
=
= 0.2 s
R
10
and the time constant
v
100 V
vR = 100 (1 - e-5t )
vL = 100 e-5t
t
Figure 6: Voltages across L and R vs time.
In the direction sensed, the potential difference vR across the resistor is iR. In the
direction sensed, the potential difference vL across the inductor is E − iR. These are
illusrated in Figure 6.
(c) The rate of change of current is
I0
di
= e−t/τ
dt
τ
At t = 0.1 s
p
Q
i = 10 1 − e−0.5 = 3.93 A
10 −0.5
di
= 30.3 As−1
=
e
dt
0.2
i. The energy stored in the magnetic field of an inductor is UM = 1/2 Li2 .
The rate at which energy is being stored in the field is
PM =
dUM
dt
di
dt
= 2 × 3.93 × 30.3 = 238 W
= Li
ii. The rate of thermal loss due to the resistance of the coil is
PR = i2 R = 3.932 × 10 = 155 W
6
iii. Energy is delivered by the battery at the rate
Ei = 100 × 3.93 = 393 W
Since PM + PR = (238 + 155) = 393 W we see that energy is conserved.
3. (a) The r.m.s. value of the sinusoid v(t) means the square root of the average value of
v 2 . With reference to Figure 7, the average value of v 2 is found by integrating v 2
over one period and dividing that period, T .
v
Vmax2
Vmax2
2
π/2ω
0
3π/2ω
π/ω
2π/ω
t
Figure 7: Plot of v 2 .
Vr.m.s. =
8
T
2
v dt =
0
p
v2
Q1/2
2
Vmax
2
= Vmax
8
=
T
0
v 2 dt
T
„1/2
cos2 ωtdt
0
8
T
1/2 (1 + cos 2ωt) dt
0
2
T
= 1/2Vmax
therefore Vr.m.s. =
^$ T
2
1/2Vmax
=1/2
√
= Vmax / 2
(b) The average power consumption is given by Pav = Vr.m.s. Ir.m.s. in a purely resistive
circuit.
Therefore
Similarly
Ir.m.s.
Ir.m.s.
Imax
Vmax
=
=
=
=
P/Vr.m.s. = 2400 W/240 V = 10 A r.m.s.
√
Imax / 2
√
2I
= 14 A
√ r.m.s.
2Vr.m.s. = 340 V
The resistance, R, can be calculated using R = Vmax /Imax or R = Vr.m.s. /Ir.m.s.
giving R = 240/10 = 24 Ω.
4. (a) With reference to Figures 8 and 9, where T is the period, the energy stored in the
capacitor is given by
1
UE = q 2 /(2C) = CvC2
2
7
Thus the maximum value is
1
2
CVmax
2
w
W
1
−6
2
× 5 × 10 × 20 J
=
2
= 10−3 J
UE =
The energy stored in the inductor is given by
i
ωq0
q
q
The current i is sensed
clockwise here.
q0
0
i
t
0
-q
t
The charge q is on the
left-hand plate.
The capacitor was initially charged so that the left
hand plate was positive relative to the right hand plate.
Different selections for the senses of charge and
current could produce differently shaped curves,
but the answer should be internally consistent.
Figure 8: Current and capacitor charge of a LC circuit.
1
UM = Li2
2
Thus the maximum value is given by UM = 10−3 J (By conservation of energy).
(b) When there is some resistance in the circuit, energy is dissipated in the resistor
whenever the current is non-zero, so the stored energy gradually decreases, the peak
current and charge decrease, and the oscillation is damped.
5. Consider a simple circuit containing a capacitor and an AC generator. The potential
difference across the capacitor changes sign every half period, as the alternating current
carries charge in one direction and then the other. During a half cycle the charge on
one of the capacitor places changes from CE to −CE where E is the peak generator emf.
8
UE
0
t1
t2
t3
t4
0
t1
t2
t3
t4
t
UM
t
Figure 9: Energy in capacitor and inductor of an LC circuit.
This change in charge of ∆Q = 2CE over a half cycle can be thought of as the current
passing “through” the capacitor (although no charge actually crosses the space between
the capacitor plates). If we increase the frequency f of the emf we see that the same
charge ∆Q is transferred per half cycle, but the number of cycles per second increases, and
therefore the current increases. Hence, the greater the frequency, the less the capacitor
impedes the flow of charge. This can be seen mathematically from the expression for
Xc = 1/(ωC) ie as the angular frequency ω increases the reactance decreases and the
current I = EXc increases.
At low frequencies, the opposite happens. The rate of flow of charge “through” the
capacitor tends to zero, so that the capacitor acts as an open circuit. Also Xc tends to
infinity as the angular frequency ω tends to zero.
6. (a) The sum of the given signals is
v = v1 + v2 = A cos ωt + B cos (ωt + φ)
= A cos ωt + B cos ωt cos φ − B sin ωt sin φ
= cos ωt (A + B cos φ) − sin ωt (B sin φ)
This can be written as
v = D cos (ωt + δ) = D [cos ωt cos δ − sin ωt sin δ]
provided that
D cos δ = A + B cos φ
and
9
D sin δ = B sin φ
To find D, square both sides of these two equations, and add, making use of the
identity cos2 δ + sin2 δ = 1. This gives D2 = A2 + B 2 + 2AB cos φ.
(b) Using the cosine rule as illustrated in Figure 10
D2 = A2 + B 2 − 2AB cos (π − φ)
But cos (π − φ) = − cos (φ) so
D2 = A2 + B 2 + 2AB cos φ
D
δ
φ
B
A
Figure 10: Vector addition.
The amplitude of the signal obtained by adding two sinusoidal signals with the same
frequency but different phase and amplitude is found by representing the signals as
phasors, and adding them vectorially. (The phase of the resulting signal, δ, can be
found by applying the sine rule to the triangle.)
(c) Since vr.m.s. = √12 D and the r.m.s. values of v1 and v2 are given by √12 A and
respectively, the r.m.s. values are related in the same way as the amplitudes.
√1 B
2
2
= (V1 )2r.m.s. + (V2 )2r.m.s. + 2 (V1 )r.m.s. (V2 )r.m.s. cos φ
Vr.m.s.
7. Let the peak current in the circuit be I0 . The voltage dropped across the resistor has a
peak value of 300I0 and is in phase with the current.
The inductance has a reactance of
Lω = 2 × 2π × 50 = 628 Ω
The voltage dropped across inductor has a peak value of 628I0 and leads the phase of the
current by 90◦ . From the phasor diagram of Figure 11
therefore
tan φ = 628/300
φ = 64◦
Thus the current lags the applied voltage by 64◦ .
10
VL
= 628 I0
φ
VR= 300 I0
Figure 11: Voltages in an RL circuit.
VL = IωL
VR = IR
VC = I / ωC
Figure 12: Voltages in a series LRC circuit.
8. (a) In the following voltages and currents are assumed to be the r.m.s. values.
i. For a series LRC circuit, the impedance is given by
Z=
R2 + (XL − XC )2
The maximum current flows when the impedance is minimum, i.e. when XL =
XC .
Then Z = R. On the phasor diagram shown in Figure 12, the phase of the
current corresponds to the positive x-axis. Vector addition of potentials shows
that the emf in the circuit when XL = XC has the same phase as the current.
This is a characteristic of a purely resistive circuit.
ii. Since the circuit behaves as a resistance, we can apply Ohm’s Law
R=
V
240
=
= 60 Ω.
I
4
iii. The r.m.s. voltage across the inductance is
VL = IXL = IXC
at resonance
11
4
I
=
ωC
50 × 2π × 1.3 × 10−6
= 9.79 × 103 ≈ 10 kV
=
(b) A phasor diagram (assuming XL > XC ) representing the detuned system with an
unknown current I is shown in Figure 13. Noting the Pythagoras (3-4-5) triangle,
we see that the resultant voltage across the series combination of R and L must be
100I. Thus
240 = 100Ir.m.s.
or
Ir.m.s. = 2.4 A
The power factor is
cos φ =
60I
= 0.6
100I
The dissipated power is
Pav = Er.m.s. Ir.m.s. cos φ = 240 × 2.4 × 0.6 = 346 W
VX = 80 I
= 240 V r.m.s.
φ
VR = 60 I
Figure 13: Power factor from a vector diagram.
9. All voltages and currents in the following discussion are the r.m.s. values. The phase of the
source potential must be such that the potential across the coil and that across the resistor
add to give 60 V r.m.s. as shown in Figure 14. The phase lead of the potential across the
coil is not 90◦ , indicating that there is a resistive component to the coil impedance. This
potential can be resolved into resistive and inductive components VR and VL as shown.
Measurement from a scaled diagram or application of the cosine rule gives
VR = 29.4 V r.m.s.
29.4
VR
=
= 25 Ω
R=
I
1.2
12
VS = 60 V
45 V
VL
φ
20 V
VR
Figure 14: Vector diagram of voltages.
The voltage drop across the inductive part of the coil is,
VL = 34.1 V r.m.s. but VL = XL I = ωLI
34.1
VL
=
= 0.090 H
L =
ωI
100π × 1.2
The power dissipated in the circuit is
Pav = VS,r.m.s. Ir.m.s. cos φ
= (VS,r.m.s. cos φ) Ir.m.s. = (20 + 29.4) × 1.2 = 59 W.
Note that the power is dissipated in the resistive parts of the circuit.