Topic Nine: Alternating Current (AC)
Electricity (AS 90523, Physics 3.6)
RMS Values
Objectives
By the end of this section you should be able to:
1 state the relationship between RMS and maximum value for AC
2 calculate the power dissipated in a resistance when AC flows through it.
Mains electricity
Mains electricity is produced by alternators. In these devices, coils rotate in a magnetic field.
As we have previously seen, this produces an alternating voltage given by V = BANω sin ωt.
For simplicity, we write this as V = Vmax sin ωt with Vmax = BANω.
When this voltage is applied to a resistive circuit, the resulting current will also be alternating
and sinusoidal.
V
I = V = max sin ωt which we can write as I = Imax sin ωt
R
R
P
Power
O
one cycle
T
t
Most mains electricity is used for its heating effect. The power dissipated in
a resistance is given by P = VI. We can also write this as P = I2R. Because
the current varies with time, it follows that the power does too.
Clearly, using maximum values in a power formula will not give the right
answer for power. To get the right answer we must use special average values
known as RMS (root mean square) values.
QUESTION
1 Explain why using maximum values of V and I to calculate power does not produce
the correct value.
Also explain why using average values for V and I does not give the correct answer:
first decide what the average values for V and I would be.
RMS (root mean square) values
RMS values of voltage and current may be used directly in power formulae like those above.
P = VRMSIRMS
P =I 2RMSR
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Topic Nine: Alternating Current (AC) Electricity
For example, an alternating current of RMS value 6 A would produce the same power in a
resistance as a constant 6 A direct current.
For a sinusoidal alternating current, it can be shown that:
I max
I RMS =
2
and
VRMS =
V max
2
RMS values are so useful that, unless specified otherwise, values quoted will be RMS rather
than maximum values.
Thus, the New Zealand mains voltage is 230 V RMS. The maximum value would be given
by Vmax = 2 × 230 = 325 V.
Similarly, as a rule, meters read RMS values.
Example
A heater connected to the mains draws a maximum current of 16 A from the mains.
(a) Calculate the electrical power delivered to the heater.
To calculate the power we use RMS values.
I max
= 16 = 11.3 A
2
2
P = VRMS I RMS =230 × 11.3 = 2600 W
I RMS =
(b) Calculate the resistance of the heater.
When using Ohm’s Law we must use either just RMS values or just maximum values.
We here use RMS values.
R=
VRMS
I RMS
= 230 = 20 Ω
11.3
QUESTION
2 A 10 Ω resistance is connected to an alternator. The maximum output of the
alternator is 20 V.
Find:
(a) The RMS output.
(b) The maximum current through the resistance.
(c) The RMS current through the resistance.
(d) The power dissipated in the resistance.
(e) The factor by which the power dissipated would increase if the alternator
output voltage doubled.
General calculation of RMS values
The RMS value of a current is found by averaging the squares of the current values and then
taking the square root of this average (RMS = square root of the mean of the squares).
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AC in a Capacitance
Objectives
By the end of this section you should be able to:
1 explain how it is that AC can pass through a capacitance
2 define the reactance of a capacitance
3 explain how the reactance of a capacitance depends on its capacitance value and on the
frequency of the AC
4 state the phase relationship between current and voltage for a capacitance
5 illustrate this phase relationship graphically and by phasor diagram.
Introduction
I
A
A
+
q
–
–q
C
–
+
I
–q
C
q
The circuit shown consists of an alternating voltage source (alternator) connected to an
ammeter and a capacitor in series. Initially, some charge flows through the meter in order to
charge the capacitor. When the voltage reverses in the second half of the cycle, charge flows
from the capacitor and through the meter again until the capacitor is charged in the reverse
sense. While no current flows from one capacitor plate directly to the other, the meter would
record an alternating current at any other point in the circuit. Because of this, we say that an
alternating current can pass through a capacitor.
Capacitors are often used in circuits because they block DC and pass AC, especially highfrequency AC. Unlike resistances, capacitances do not dissipate energy; they simply store it
in the electric field between the plates and give it back to the circuit when they discharge.
Reactance of a capacitance
The size of the current in a capacitance depends on three things:
1 the frequency
I∝f
With a higher frequency, the same amount of charge flows through the meter more
often.
2 the capacitance I ∝ C
A larger capacitance stores more charge, so a bigger current flows through the meter.
3 the voltage
I∝V
With a bigger voltage, more charge is stored and so more current flows through the
meter.
In fact, we find that I = 2πfCV or I = ωCV
We write this as
⎛
⎞
V = ⎜ 1 ⎟ I or V = X C I with X C = 1
⎝ ωC ⎠
ωC
In this equation, V and I are both maximum or both RMS values.
The equation is similar to Ohm’s law V = RI for a resistance.
XC is called reactance (units: ohm, Ω)
It is not called resistance. This is because the current through a resistance is in phase with
the voltage across it, whereas this is not true for a capacitance.
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Topic Nine: Alternating Current (AC) Electricity
Phase relationship for a capacitance
Suppose the current is a sine function of time, as
the graph shows. For the first half of the cycle,
current flows into the capacitance. At time T/2
the current stops flowing in and starts flowing
out. Therefore, at time T/2 the charge stored
must be a maximum.
For a capacitance V = Q/C, so the voltage across
the capacitance must also be a maximum at
T/2. However, the current reached its maximum
earlier at time T/4. Thus the current is not in
phase with the voltage but leads it by 90°. This
is logical, since the current would have to flow
into a capacitance for a while before the voltage
built up. The phasor diagram shows this phase
relationship.
V
I
T
4
Q
T
2
3T
4
T
t
X
I
V
Example
A 40 μF capacitance is connected to an AC power supply that delivers 30 V at a frequency
of 50 Hz. Find:
(a) the reactance of the capacitance at this frequency
1
XC = 1 =
= 80 Ω
ωC 2 π × 50 × 40 × 10−6
(b) the RMS current through the capacitance
V = XC I ⇒ I = V = 30 = 0.38 A
Xc 80
(c) the RMS current that would flow through the capacitance if the power supply
frequency increased to 100 Hz.
If the frequency doubled, then the reactance would halve. This would double the
current through the capacitance. I = 0.75 A.
QUESTION
3 A capacitance is connected to a 20 V AC power supply which delivers 3.0 A when
the frequency is 200 Hz. Find:
(a) the reactance of this capacitance at this frequency
(b) the value of the capacitance
(c) the current that would flow in the capacitance if the frequency were changed
to 50 Hz.
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AC in an Inductance
Objectives
By the end of this section you should be able to:
1 define the reactance of an inductance
2 explain how the reactance of an inductance depends on its inductance value and on the
frequency of the AC
3 state the phase relationship between current and voltage for an inductance
4 illustrate this phase relationship graphically and by phasor diagram.
Introduction
An inductance opposes any change in the current flowing through it. An inductance offers
no opposition to constant DC, but it will oppose AC. This is because an alternating current is
continually changing. To bring about a change in the current through the inductance shown,
we must apply a voltage V given by:
V = L ΔI
Δt
I
V
This is to oppose the voltage V = −L ΔI
Δt
L
produced by the inductance.
Inductances are often used in circuits because they pass DC but oppose AC, especially highfrequency AC. Resistors dissipate energy, but inductances simply store it (in the magnetic
field), and then return it to the circuit as the current decreases.
Reactance of an inductance
For an inductance we find V = ωLI or V = XLI with XL = ωL.
In this equation, V and I are both maximum or both RMS values.
This equation is similar to Ohm’s Law V = RI for a resistance. XL is called reactance
(unit: ohm, Ω) . It is not called resistance. This is because the current through the inductance
is not in phase with the voltage across it. This is discussed below.
We would expect the reactance of an inductance to increase with frequency, because the
greater the frequency the greater the rate of change of current and inductance opposes
current change.
Phase relationship for an inductance
Suppose the current is a sine function of time as the graph shows. Since V = L ΔI , the voltage
Δt
is proportional to the rate of change of current, to the slope of this current–time graph.
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Topic Nine: Alternating Current (AC) Electricity
At times T/4 and 3T/4 the current is not changing, and so the voltage must be zero. The slope
of the current graph is a maximum at time 0 and so the voltage is a maximum then. The
slope of the current graph is maximum negative at time T/2 and so the voltage is a maximum
negative then. We see that the current is not in phase with the voltage but lags it by 90°. The
phasor diagram shows this relationship.
QUESTION
4 A 3.0 H inductance is connected to an alternator which delivers a current of
20 mA at a frequency of 100 Hz. Find:
(a) the reactance of the inductance at this frequency
(b) the voltage output of the alternator
(c) the current which would be delivered by the alternator if a second 3.0 H
inductance were connected in series with the first
(d) the current which would be delivered by the alternator if the second 3.0 H
inductance were connected in parallel with the first.
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Addition of Alternating Voltages
Objectives
By the end of this section you should be able to:
1 state the phase relationship between current and voltage for a resistance
2 illustrate this phase relationship graphically and by phasor diagram
3 construct phasor diagrams for series RC and RL circuits
4 sketch the corresponding graphs for these phasor diagrams
5 define impedance for the above circuits
6 carry out calculations involving the addition of the voltages in the above circuits
7 carry out calculations involving the phase angle between the alternator voltage and current in
the above circuits.
In the circuit shown we find that:
R
C
VA max ≠ VR max + VC max
This is because VR max and VC max happen at different times. The
explanation for this is that in a series circuit, each component carries
A
the same current (I, say), but while VR is in phase with I, VC lags
behind I by 90°.
Thus VR reaches a maximum when I does, but VC reaches a maximum one quarter of a period
later. Of course, it is still true that at any instant of time VA = VR + VC holds.
To find VA max and the phase relation between VA and I we draw a phasor diagram. First, we
place on the diagram the phasor for the quantity common to all components. In a parallel
circuit it would be the voltage phasor; here, in a series circuit, it is the current phasor. Then
the other phasors are drawn, their lengths giving the maximum value of the quantity they
represent, and the angle they make with the I phasor giving their phase relation with I. Here
the result is like that shown below on the left.
V
X
VA
I
VC
VC
VR
VA= VC + VR
VR
t
The phasors rotate with angular velocity ω as shown. At any instant in time, the actual value
of a quantity is given by the vertical component of its phasor. This principle may be used to
derive the graphs shown.
The sum of VR and VC may be found by point by point addition of the graphs, or by finding
the vector sum of the phasors VR and VC. Note that the sum of sinusoidal waves of the same
frequency is always another sinusoidal wave of that same frequency.
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Topic Nine: Alternating Current (AC) Electricity
Alternating current in an RC circuit
For the circuit on the previous page, we have, from the phasor diagram:
VA max = VR2 max + VC2 max
2 R2 +
VA max = I max
Im2 ax
ω2C 2
⎛
⎞
VA max = ⎜ R 2 + 21 2 ⎟ I max
ωC ⎠
⎝
1
is like resistance. It is called the
ω2C 2
impedance of the circuit. The symbol for impedance is Z. It is measured in ohms.
Comparing this with V = IR, we see that
R2 +
VA = ZIA
VA lags the current in the circuit by angle θ where:
θ = tan −1
= tan −1
VC max
VR max
Imax // ωC
Imax R
= tan −1 1
ωC R
Note that the above equations are for maximum quantities but similar equations hold for
RMS quantities. This is because RMS quantities are obtained from maximum quantities by
dividing by 2 .
Example
In the circuit illustrated, the voltages across the components are as shown in the margin
diagram.
I
(a) Draw a phasor diagram for the circuit.
VR
This is drawn to the right.
VC
VA
(b) Find the alternator voltage.
Since VR leads VC by 90° we add them “by Pythagoras” to get VA.
VA = VR 2 + VC 2 = 4.02 + 5.02 = 6.4 V
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Continuing Physics
(c) Find the phase angle by which the current supplied by the alternator leads the
alternator voltage.
This angle is given by:
θ = tan −1
VC
VR
= tan −1 5.0 = 51°
4.0
(d) Find the current supplied by the alternator.
To deduce this we just look at a single component, the resistance:
I = V = 4.0 = 0.40A
R 10
QUESTION
5 For the worked example shown:
(a) Calculate the impedance of the circuit. To do this, you will need to just look
at a single component, the alternator.
(b) Calculate the frequency of the AC. To do this, you will need to just look at a
single component, the capacitance.
(c) Explain what would happen to the size of the phase angle between the alternator
current and the alternator voltage if the alternator frequency increased.
(d) At what frequency would the voltage across the capacitance equal the voltage
across the resistance?
(e) What would be the value of the phase angle for the situation in (d) above?
Alternating current in an RL circuit
I
VA
VL
R
X
I
L
VR
In the series circuit shown, the resistance and the inductance carry the same current. The
voltage across the resistance is in phase with this current, while the voltage across the
inductance leads the current by 90°. Thus VL leads VR by 90°. The phasor diagram shows
these phase relationships. From this diagram we see that:
VA = VR 2 + VL 2
⇒ VA = R 2 I 2 + ω2 L 2 I 2
⇒ VA =
(
)
R 2 + ω2 L 2 I
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Topic Nine: Alternating Current (AC) Electricity
R 2 + ω2 L2 is like resistance. It is known as the impedance of the circuit.
VA leads the current I in the circuit by an angle θ, where:
θ = tan −1
VL
VR
⇒
⇒
⎡ ωLI ⎤
θ = tan −1 ⎢
⎣ IR ⎥⎦
⎡ ωL ⎤
θ = tan −1 ⎢
⎣ R ⎥⎦
In the above equations, V and I may both be taken as maximum values, or they may both
be taken as RMS values.
Recall that often it is necessary to think of an inductor as a pure inductance in series with a
pure resistance. The above analysis applies to this situation.
At very high frequencies the reactance of the inductance (XL = ωL) would be very high and
the resistive part of the inductor would make a scarcely noticeable difference.
At very low frequencies the reactance of the inductance would be very small and the resistive
part of the inductor would be dominant.
In particular, for direct current (f = 0) we would have VL = ωLI = 0, and the inductance can
be ignored.
As the frequency of the alternating current increases, the inductance in the circuit becomes
more important. This is because the reactance (ωL) of the inductance is proportional to
frequency while the resistance R does not vary with frequency.
Example
When an inductance is connected to a 12 V car battery, a current of 4.0 A flows. When
the inductance is connected to an alternating voltage source which supplies 12 V at
50 Hz, the current is 2.0 A. Find the resistance and the inductance of this inductor.
Batteries supply direct current. This is unaffected by inductance:
V = IR
⇒ 12 = 4.0 × R
⇒ R = 3.0 Ω
For the alternating voltage source:
ω = 2πf = 314 s–1
V=
(
)
R 2 + ω2 L 2 I
⇒ 12 =
(
)
32 + 3142 L2 2.0
⇒ L = 0.017 H
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Continuing Physics
QUESTION
6 This question is about the phase angle between the alternator current and the
alternator voltage in a series RL circuit like the one illustrated.
(a) What value of the phase angle would you expect for very high frequencies?
Explain your answer.
(b) What value of the phase angle would you expect for very low frequencies?
Explain your answer.
(c) Find an equation for the frequency of the AC that gives a phase angle of 45°.
Your answer should be in terms of R and L.
7 Frances has an electrical device which needs to be supplied with 90 V AC. The
only supply available is a 230 V, 50 Hz supply. She thinks about putting a resistance
in series with the device. The device may be regarded as a pure resistance, value
180 Ω. The series resistance would need to have a voltage of 140 V across it.
(a) What value of series resistance would be required?
(b) After more thought, Frances decides that a series inductance would be better
than a series resistance. Why would it be better?
(c) What voltage should be dropped across the series inductance?
(d) What phase angle would there be between the supply voltage and the supply
current?
(e) Find the value of the inductance that would be required.
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Resonance
Objectives
By the end of this section you should be able to:
1 construct a phasor diagram for a series RCL circuit
2 sketch the corresponding graph for this phasor diagram
3 define the impedance for the above circuit
4 explain what is meant by resonance
5 state the impedance of a series RCL circuit at resonance
6 calculate the resonant frequency of an RCL circuit
7 compare VC and VL below resonance, at resonance and above resonance
8 state the phase relation between VA and I below resonance, at resonance and above
resonance
9 carry out calculations involving phase angle and addition of voltages in a series RCL circuit
10 discuss practical applications of resonance.
In the series circuit shown, the same current
flows through each component.
I
VA
VL
L
VA
VR is in phase with I
VL is 90° ahead of I
VC is 90° behind I
8
R
C
I
VR
VC
Thus VL and VC are 180° out of phase and “add by subtraction”.
The phasor diagram shows these phase relationships. From the diagram we see how to
sum the voltages across the three components to find VA.
VA =
(V
L
− VC
)
2
+ VR 2
2
⎞
⎛
⎛
⎞
= ⎜ ⎜ ωL − 1 ⎟ + R 2 ⎟ I
ωC ⎠
⎟⎠
⎜⎝ ⎝
2
The quantity
⎛
1 ⎞ + R 2 is the impedance of this circuit. We see that it is a
⎜⎝ ωL −
⎟
ωC ⎠
minimum when VL = VC
i.e. when ωL = 1
ωC
⇒
ω2 = 1
LC
⇒
ω=
1
LC
⇒
f =
1
2 π LC
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Continuing Physics
This frequency is called the resonant frequency. At resonance, the effects of the inductance
and the capacitance cancel (VL = VC) and the circuit impedance equals R. Thus VA = RI at
resonance.
At resonance, the alternator voltage is in phase with the current. The circuit behaves as a
purely resistive one. Since the impedance is at a minimum, it follows that, for a constant
voltage alternator, the current is a maximum. If R is small, the current can be very large, and
therefore so can VL and VC. Each could be much larger than the supply voltage VA.
When the alternator frequency is lower than the resonant frequency, VC > VL and capacitance
effects dominate in the circuit. For example, VA lags I and at very low frequencies would
do so by almost 90°.
For frequencies above the resonant frequency, VL > VC and inductance effects dominate. For
example, VA leads I and at very high frequencies would do so by almost 90°.
There are many similarities between this electrical resonance and mechanical resonance. In
particular, note that the frequency of oscillation in the circuit equals the driving frequency
(the alternator frequency) rather than the resonant frequency.
V
VL
VL
VR
VR
VC
t
VC
Example
For the circuit shown, find:
(a) the resonant frequency
1
1
⇒ f =
=
= 160 H
2 π LC 2 π × 4 × 10−2 × 25 × 10−6
(b) the current at resonance
At resonance the impedance is just R
⇒ V = IR
⇒ I = V = 20 = 20 A
R 1
(c) the voltage across each component
At resonance VR = VA = 20 V
We also could use VR = IR = 20 × 1 = 20 V
VL = ωLI = 2π × 160 × 4 × 10–2 × 20 = 800 V
At resonance VC = VL = 800 V
We could also use:
20
VC = 1 =
= 800 V
ωC 2 π × 160 × 25 × 10−6
I
20 V
40 mH
18
25NF
Note that the voltages across the inductance and the capacitance are much larger than
the voltage supplied by the alternator. This can happen at resonance, since the current
is only limited by the resistance. With a small resistance, a large current can flow which
can lead to large voltages across the other two components.
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Topic Nine: Alternating Current (AC) Electricity
QUESTION
8
70 V 30 V
25 V
I
A
In the series LCR circuit illustrated, the voltages across the components are
shown.
(a) Draw a phasor diagram for the circuit. Show I, VA, VL, VC and VR.
(b) Find the alternator voltage.
(c) Is the alternator current ahead or behind the alternator voltage?
(d) Find the size of the phase angle between the alternator current and the alternator
voltage.
(e) Is the supply frequency above or below the resonant frequency?
Applications
TV and radio receivers
These receivers make use of resonant circuits to tune in on a particular
channel or radio station. Each channel or radio station has its own
broadcast frequency which is unique in its geographic area. To receive
a particular signal, component values are adjusted until the resonant
frequency of the circuit matches the frequency of that signal.
In a series LCR circuit, the current is largest when the signal is at the
resonant frequency. The circuit effectively screens out signals not at
the resonant frequency, since these produce much smaller currents.
The graph shows how the current in a series LCR circuit depends on
the frequency.
I
f0
f
Metal detectors
Metal detectors also contain a series LCR circuit. An alternating voltage is applied to this
circuit. The frequency of the alternating voltage is just below the resonant frequency of the
circuit. The inductor is in the form of a coil which is held near where the metal is expected.
If metal is nearby, eddy currents are produced in the metal and this increases the inductance
of the coil. This increased inductance lowers the resonant frequency of the circuit. As the
resonant frequency of the circuit is now closer to the frequency of the alternating voltage, a
larger current results. This increase in current is easily detected.
Weapons detectors used in airports and other places operate on a similar principle. More
sophisticated detectors may scan the LCR circuit with a range of frequencies. The shape of the
resulting I against f graph can be interpreted to reveal the nature of the particular metal item
which produced it. Thus a handgun produces a “fingerprint” different to that of a coin.
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Continuing Physics
Filters
Filters are electrical circuits which allow only some frequencies to pass through. They are
often used in tone controls to control the treble and bass frequencies in stereos, radios,
etc. In the filters illustrated below, a variable resistance could be used to vary the filter
characteristics.
A filter would also be used in a heart-rate monitor used in a hospital. The heart-rate monitor
would pick up some signal from the mains (50 Hz) as well as the heart-beat itself. Since
the mains frequency is much higher than the heart-beat frequency (typically about 1 Hz),
filtering it out would be easily done with a low-pass filter (see below).
The low-pass filter
This filter allows only low frequencies to pass through. A simple example is illustrated. If
the incoming signal is of low frequency, then the reactance of the capacitance will be high
and the voltage across it, the output voltage, will be high.
At high frequencies, the reactance of the capacitance will be low. Thus the voltage across
it, the output voltage, will be low. In this way, we see that a low frequency produces a high
output while a high frequency produces a low output.
The high-pass filter
This filter allows only high frequencies to pass through. A simple example is illustrated.
If the incoming signal is of high frequency, then the reactance of the capacitance will be
low and the voltage across it will be low. Thus the voltage across the resistance, the output
voltage, will be relatively high.
At low frequencies, the reactance of the capacitance will be high. Thus the voltage across
it will be high. Thus the voltage across the resistance, the output voltage, will be relatively
low. In this way, we see that a high frequency produces a high output while a low frequency
produces a low output.
Extension: DC power supplies
The electricity delivered to our homes is 230 V AC.
Many electronic circuits, for example those in
stereos, videos and computers, require DC electricity
at some different voltage.
A power supply typically contains:
1 a transformer to convert the voltage to the required voltage
2 one or more diodes to rectify the AC to DC
3 a capacitor to smooth the output voltage.
Some power supplies also contain regulation circuitry which
further smooths the output voltage and also limits the current
drawn from the supply. This protects the power supply in the
event of a short-circuit.
Diodes recap:
Diodes allow current flow in one direction
only
This
Not this
Sometimes a device will continue working
for a few seconds after the mains switch is
turned off. The energy stored in the smoothing
capacitor in its power supply enables this to
happen.
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Topic Nine: Alternating Current (AC) Electricity
Vin
Vout
Vin
V
t
Vout
The diode allows the
secondary current to
flow in one direction
only. The voltage
that results is direct,
but not at all smooth.
Input
Output
t
t
The capacitor smooths
the voltage. When the
capacitor charges, the
waveform is sinusoidal.
As it discharges, the
waveform is exponential.
The above illustrates half-wave rectification. Half of the input wave-form is lost. In full-wave rectification, the
negative part of the input waveform is inverted and made positive. One way of doing this is to use a bridge rectifier.
This is shown below:
Input
Output
A
Vin
Vout
B
Vin
V
t
Vout
t
t
The diodes are arranged so that current from the secondary of the transformer can only flow to A. Any current
returning to the secondary must do so from B.
The capacitor discharges for a shorter time, and so the voltage remains more constant. The smoothing is better and
the full waveform is utilised.
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Continuing Physics
Exercise Nine:
AC Circuits
1 Complete the following table of peak and RMS values:
Peak value
RMS value
Voltage
240 V
Current
10 A
Power
3400 W
12 V
50 Hz
2 In the circuit shown on the right, a 12 VRMS supply is connected
in series to a 500 Ω resistor. The voltage across the resistor is
8.0 VRMS.
a Calculate the RMS current in the circuit.
500
b What is the RMS voltage across the capacitor?
c Calculate the reactance of the capacitor.
d Find the capacitance of the capacitor.
3 When connected in series to a signal generator a 2.4 H ideal
inductor (negligible internal resistance) and a resistor of
200 Ω have the RMS voltages, shown on the right.
Vs
a Calculate the RMS current in the circuit.
2.4 H
4.0 V
3.0 V
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Topic Nine: Alternating Current (AC) Electricity
b Find the RMS supply voltage.
c What is the reactance of the inductor?
e Find the frequency of the signal generator.
f On the axes below, draw the voltage against time, for one complete cycle, of:
i
the supply
ii the inductor
iii the resistor.
V (V)
t (s)
4 A 55 mH inductor, with internal resistance of 20 Ω, is connected in series to a 600 μF
capacitor. The voltage in the circuit is supplied by a 5.0 V signal generator of variable
frequency.
a Calculate the impedance of the circuit when the signal generator is set at 60 Hz.
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Continuing Physics
b What is the current in the circuit when the signal generator is set at 60 Hz?
c Find the resonant frequency of this circuit.
d Calculate the maximum current of this circuit.
e Draw the graph of current vs frequency for this circuit, including the resonant
frequency.
I (A)
f (Hz)
5 (From UB 2003) Jean decides to investigate the role of
inductors and capacitors in AC circuits. She sets up the
following circuit.
a Show that the angular frequency of the AC supply is 314
rad s–1.
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Topic Nine: Alternating Current (AC) Electricity
b Show that the reactance of the capacitor is 1590 Ω.
c Show that the impedance of this circuit is 984 Ω.
d Calculate the current in this circuit.
6 (From UB 2003) Electrical devices often require small DC voltages in order to operate.
Jean wishes to design a 12.0 V DC supply to charge up her cellphone.
She initially constructs the following circuit.
Vout
The supply voltage is 235 VRMS and the operating frequency is 50.0 Hz. The number
of turns in the transformer primary is 8000.
a Show that the peak value of the input voltage is 332 V.
b Calculate the number of turns required to produce a peak value of 12.0 V across
the secondary turns of the transformer.
Jean now places two diodes and a resistor in the circuit to make a full wave rectifier,
as shown in the diagram below. (The supply voltage and frequency are unchanged.)
A
B
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Continuing Physics
c Jean measures the voltage between A and B with an oscilloscope and finds that A
is always at a higher potential than B. Explain why.
d Assuming there is no voltage drop across the diodes, on the axes provided below
sketch a graph of voltage between points A and B against time. Include a maximum
voltage value and a period value.
To improve the circuit, Jean places a capacitor into the circuit as shown in the diagram
below.
Vout
e Explain, using physical principles, the effect of the capacitor on the output
voltage.
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Topic Nine: Alternating Current (AC) Electricity
f On the axes below, sketch a graph of Vout against time. No numerical values need to
be included.
7 (From UB 1999) The radio in Stephanie’s CD player has an LRC antenna
circuit inside it as shown below. The component values are L = 6.0 × 10–7 H,
R = 980 Ω and C is a variable capacitor.
a What is the capacitance of the capacitor when the circuit is resonant at
9.8 × 107 Hz?
L
R
C
b If the value of C found in (a) is doubled, what will be the new frequency
the radio is tuned to?
c If C is kept constant at the value found in (a), on the axes provided
below sketch a graph of current vs frequency for the range 6.0 × 107 to
14.0 × 107 Hz.
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Continuing Physics
In a particular LR circuit, as shown below, the frequency of the AC voltage source is
2.5 × 107 Hz and the inductor has an inductance of 3.3 × 10–6 H. The resistor has a
resistance of 470 Ω.
d What is the phase difference (in radians) between the voltage across the resistor
and the source voltage?
L
0.15 VRMS
R
8 (From UB 2002) The following simple LR series circuit was constructed
by Tobie. It was assumed that the inductor was ideal (i.e. had zero
resistance).
21.0 7
L = 0.190 H
After some investigation, it was found that the inductor L was non-ideal
with a resistance of 12.0 Ω. The above circuit was redrawn by Tobie to
include this resistance.
Non-ideal inductor
21.0 7
12.0 7
L = 0.190 H
Supply voltage = 20.0 VRMS
50.0 Hz
Supply voltage = 20.0 VRMS
50.0 Hz
a Show that the angular frequency of the AC supply was 314 rad s-1.
b Calculate the reactance of the inductor.
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Topic Nine: Alternating Current (AC) Electricity
c Calculate the RMS current that was flowing in the circuit.
Tobie decided that the inductor needed further investigation.
She constructed the following circuit.
A
B
Non-ideal inductor
12.0 7
Tobie connected points A and B to a 15.0 V direct current
(DC) power supply and adjusted the variable resistor so
that bulbs C and D (which were identical) shone with the
same brightness.
L = 0.190 H
C
d State the value of the variable resistance.
D
Tobie then disconnected the DC supply and replaced it with a 15.0 VRMS, 50.0 Hz, AC
supply. Bulb D did not change brightness but bulb C changed considerably.
e Explain why bulb D showed no sign of changed brightness.
f i Would the brightness of bulb C have increased or decreased?
ii Explain your answer to (f)(i).
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