Week 12 - Alternating Current
November 21, 2012
The city lights of Prince George, British
Columbia viewed in a motion blurred
exposure. The AC blinking causes the lines
to be dotted rather than continuous.
Exercise 12.1: Discussion Questions
1
a) The current in an ac power line changes direction 120 times per second, and its average value is zero.
Explain how it’s possible for power to be transmitted in such a system.
Answer:
Power can be transmitted because the power delivered to another circuit is
p = iv,
(1)
I.e. the product between the curent and the voltage, and the average power is not necessarily zero.
For example if the circuit to which the ac power line is connected is a pure resistor, then v = iR and
the instantaneous power is
p = i2 R.
1 Figure
is from http://en.wikipedia.org/wiki/File:City_lights_in_motion.jpg
1
(2)
This is always positive which means that power is dissipated in the resistor at any moment. The
average power is given by
Pav =
1 2
I R.
2
(3)
b) Fluorescent lights often use an inductor, known as ballast to limit current through the tubes. Why
is it better to use inductor than a resistor for this purpose?
Answer:
One reason that the inductor is preferable over the resistor is that the average power dissipated in
the inductor is practically zero. This means that we dont waste precious energy on just limiting the
current so that everything can go into producing visible light.
The applied voltage does work on the current as the back emf resists a rising current, this energy is
then stored in the magnetic fields of the inductor and given back when the emf of the inductor does
work on a decreasing current.
c) A light bulb and a parallel-plate capacitor with air between the plates are connected in series to an ac
source. What happens to the brightness of the bulb when a dielectric is inserted between the plates?
Explain.
Answer:
The impedance of the circuit goes down as the dielectric is inserted so the current amplitude goes up
and more power is delivered to the bulb. Therefore the bulb glows brighter. Mathematically this is
seen by looking at the expression for the impedance of the circuit
s
Z=
q
R2
+
XC2
=
R2
+
1
ωC0
2
(4)
Now as the dielectric is inserted the capacitane of the goes to C = KC0 whereK is the dielecric
constant and since K > 1 the capacitive reactance goes down. Physically what happens is that
when the dielectric is inserted, the dipoles in the dielectric is aligned in such a way that they create
an additional elecric field which opposes the original. The electric field weakens which allows more
charge to accumulate on the plates for the voltage provided by the ac source. The consequence of the
capacitor allowing for a greater amount of charge to accumulate is that a greater current can flow in
the circuit.
d) Can a transformer be used with dc? Explain. What happens when a transformer designed for 230 V
ac is conneced to a 230 V dc line?
Answer:
No it can not! One of the great advantages of ac over dc is that voltages can be stepped up and down
using a transformer. This allows for high voltage transmission which very effectively reduces energy
loss due to heating of transmission wires.
If you connect a dc line to any transformer it will not do anything for you. The primary will not
even deliver an emf to the secondary. This is because the emf in the secondary was created by a
2
flux change trough it’s coil. With no changing current we won’t have changing magnetic fields and
therefore no changing flux.
Exercise 12.2: Off to the USA!
You take your hair blower off the USA, where the electrical outlets puts out 120 V (rms) instead of the
230 V (rms) found here in Norway. The blower puts out 1600 W (average) at 230 V.
a) What could you do to operate your blower via the 120 V line in USA?
Solution:
We have to get our hands at a step down transformer. It will need to have a ratio
V2
230 V
23
N2
=
=
=
N1
V1
120 V
12
(5)
of windings in the primary and secondary.
Answer: Get a transformer with the ratio
V2
23
N2
=
=
.
N1
V1
12
(6)
b) What current will your blower draw from an outlet in USA?
Solution:
The current it will draw in USA is determined by the average power,
I=
Pav
1600
A = 13.3 A
=
V1
120
(7)
Answer:
I = 13.3 A
c) What resistance will your blower appear to have when operated at 120 V?
Solution:
The original resistance of the hair blower is
3
(8)
R=
V12
(230V )2
=
Ω = 33.1 Ω,
Pav
1600
(9)
and the transformed ressitance of hair blower is given by equation (31.37) in the book.
Ref f =
33.1
R
=
Ω = 9 Ω.
2
(N2 /N1 )
(23/12)2
(10)
Answer:
Ref f = 9 Ω.
(11)
Figure 1
Exercise 12.3: The L-R-C Parallel Circuit
A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude V
and angular frequency ω. This is shown in figure 1. Let the source voltage be given by v = V cos ωt.
a) Show that the instantaneous voltages vR , vL and vC at any instant are equal to v and that i =
iR + iL + iC , where i is the current through the source and iR , iL and iC are the currents trough the
resistor, inductor and capacitor, respectively.
Solution:
Applying Kirchoff’s loop rule to the three loops including the source voltage v we find that v = vR =
vL = vC , and applying the junction rule to one of the leftmost junctions we get i = iR + iL + iC .
b) What are the phases of iR , iL and iC with respect to v? Use current phasors to represent i, iR , iL
and iC . In a phasor diagram, show the phases of these four currents with respect to v.
Solution:
The resistor voltage has a zero phase compared to the voltage, so since iR = vR /R, this also holds
for the current iR . Similarly the inductor voltage leads the current by 90◦ , so this means that the
current iL lags behind the voltage by 90◦ . Finally the capacitor voltage lags behind it’s current by
90◦ , so the capacitor current iC leads the voltage by 90◦ . The phasor diagram of this situation is
shown in figure 3.
4
Answer:
Figure 2
Figure 3
c) Use the phasor diagram in (b) to show that the current amplitude I for the current i trough the
source is given by
q
2 + (I − I )2 .
I = IR
(12)
C
L
Solution:
The phasor diagram and the pythagorean theorem implies that the current amplitude is
I=
q
2 + (I − I )2
IR
C
L
5
(13)
d) Show that the result of part (d) can be written as I = V /Z with
s
1
=
Z
2
1
1
+ ωC −
.
R2
ωL
(14)
Solution: The current amplitudes is related to their respective reactances by
IR =
V
,
R
IL =
V
V
=
,
XL
ωL
IC =
V
= ωCV
XC
(15)
so a substitution into equation 12 then yields
s
I=
2
1
1
+
ωC
−
V
R2
ωL
(16)
where we have factored out the voltage V from under the square root sign. Now from the definition
of impedance I = V /Z so therefore
s
I=
2
V
1
1
V =
+
ωC
−
R2
ωL
Z
(17)
implying
1
=
Z
s
2
1
1
+ ωC −
.
R2
ωL
(18)
2
1
1
+
ωC
−
.
R2
ωL
(19)
Answer:
1
=
Z
s
Exercise 12.4:
An L-R-C parallel circuit is connected to an ac source of constant amplitude V and variable angular
frequency ω.
a) Find expressions for the amplitudes IR , IL and IC of the currents through the resistor, inductor and
capacitor as functions of ω.
Solution:
In the solution of d) we pointed out how these ampliudes was related to their reactances. We have
IR =
V
,
R
IL =
V
,
ωL
6
IC = ωCV.
(20)
Answer:
IR =
V
,
R
IL =
V
,
ωL
IC = ωCV.
(21)
b) Graph IR , IL and IC as functions of ω for V = 200 V, R = 50 Ω, L = 1.0 H and C = 20 µF.
Solution:
A sketch of the behaviour of the different currents as a function of ω is shown in figure 5.
Figure 4
Answer:
c) Discuss the behaviour of IL and IC in the limits ω → 0 and ω → ∞. Explain why IL and IC behave
why they do in these limits.
Solution:
In the limit where ω → 0, IL → ∞ and IC → 0. This happes because when the frequency becomes
small, there almost no rate of change in the current which means the back emf of the inductor is
really small, so a lot of current is let trough. The capacitor, on the other hand, ’fills up’ quickly and
stops the current. In the limit ω → ∞, IC → ∞ and IL → 0 the rate of change of current trough the
inductor is huge which causes it to set up a huge back emf which almost stops the current entirely.
The capacitor has no time to ’fill up’ now so almost all the current can flow back and forth freely.
d) Calculate the resonance frequency (in Hz) of the circuit, and sketch the phasor diagram at resonance.
What is the current amplitude trough the source at this frequency?
Solution:
7
Figure 5
f0 =
ω0
1
1
√
√
=
=
Hz = 36.6 Hz
2π
2π 1 × 20 × 10−6
2π LC
(22)
Figure 6
We’ve calculated the impedance in the previous problem and found that
s
I=
2
1
1
+ ωC −
V.
R2
ωL
(23)
At resonance, XL = XC which means that the term within the parentheses vanishes. The current
amplitude trough the source is therefore
8
I=
V
= 4.0 A.
R
(24)
The capacitor and inductor are out of phase by 180◦ as shown in figure 7.
Answer:
f0 = 36.6 Hz
(25)
Figure 7
I=
V
= 4.0 A.
R
(26)
e) At the resonance frequency, what is the current amplitude trough the resistor, through the inductor
and trough the capacitor?
Solution:
The current trough the inductor and capacitor is equal because their reactances are equal and they
have the same voltage applied across them
IL = IC = ω0 CV = 36.6 × 20 × 10−6 × 200 A = 146 mA
(27)
and the current amplitude trough the resistor is
IR =
V
= 4 A.
R
(28)
Because that the capacitor and inductor current are out of phase with eachother by 180◦ they exactly
cancel eachother out, and all the current runs trough the resistor at resonance.
9
Answer:
IL = IC = 146 mA
IR =
V
= 4 A.
R
(29)
(30)
Figure 8
Exercise 12.5: A High-Pass Filter
One application of of L-C-R series circuits is to high-pass or low-pass filters, which filter out either the
low or the high frequencies in a signal. A high-pass filter is shown in figure 8, where the output voltage
(filtered signal) is taken out between the a and b terminals. Derive an expression for Vout /V , the ratio
of the output and source voltage amplitudes, as a function of the angular frequency ω of the source.
Show that when ω is small, this ratio is proportional to ω and thus also is small. Show that the ratio
approaches unity as the frequency gets large (The signal passes trough unaffected at large frequencies).
Solution:
Let’s first express the source voltage amplitude in terms of R, L, C, I and ω. We draw a phasor diagram
as shown in figure 9 and from this we find that the voltage equals
Figure 9
10
s
V =
p
R2
+ (XL − XC
)2 I
=
R2
2
1
+ ωL −
I.
ωC
(31)
By the exact same procedure we find the voltage amplidue across the a and b terminals which equals the
voltage across the RL-combination
Vout =
q
p
R2 + XL2 I = R2 + ω 2 L2 I.
(32)
Thus their ratio is
Vout
=
V
s
R2 + ω 2 L2
R2 + ωL −
.
1 2
ωC
(33)
Now when ω is small the 1/ωC term domiates completely in the denominator, while in the numerator
the R terms dominates. Therefore
Vout
→
V
s
R2
= ωCR
1 2
(34)
ωC
showning that the output voltage indeed is proportional to the frequency of oscillation at low frequencies.
At high frequencies however the 1/ωC term in the denominator becomes negigble and
Vout
→
V
r
R2 + ω 2 L2
= 1.
R2 + ω 2 L2
11
(35)