Name ………………………………………………………………………
Electromagnetism
1
Phenomena, Concepts and Principles:
DC Circuits
Internal resistance; simple application of Kirchhoff’s Laws;
Capacitance
parallel plate capacitor; capacitance; dielectrics; series and parallel
capacitors; charge/discharge characteristics of capacitors in DC RC
circuits; voltage/time and current/time graphs for a capacitor; time
constant; energy stored in a capacitor.
Electromagnetic Induction Magnetic flux; magnetic flux density; Faraday’s Law; Lenz’s Law;
voltage/time and current/time graphs for an inductor; time constant;
self inductance; the inductor; energy stored in an inductor; mutual
inductance; the transformer;
AC Circuits
Electromagnetism
The comparison of the energy dissipation in a resistor carrying direct
current and alternating current; peak and rms voltage and current;
phase; phasors in AC; reactance and impedance and their frequency
dependence in a series circuit; voltage and current and their phase
relationship in LR and CR series circuits; resonance in LCR circuits.
2
Relationships:
V = Ed
C=
εoεr A
d
ΔE = Vq
CT = C1 + C2 + C3 + m
1
1
1
1
=
+
+
+m
CT C1 C 2 C 3
φ = BA
Np
Ns
=
Vp
Vs
E=
ΔΙ
Δt
ε=−
1 2
LΙ
2
V = VMAX sin ωt
Ι MAX = 2 Ι rms
X L = ωL
V = ΙZ
Electromagnetism
τ = RC
RT = R1 + R2 + ...
ε = −L
Q = CV
E cap = 21 QV
τ=
1
1
1
=
+
+ ...
RT R1 R 2
Δφ
Δt
L
R
VMAX = 2 Vrms
3
ε −M
ΔΙ
Δt
Ι = Ι MAX sin ωt
XC =
1
ωC
P = VI
Revision from year 12
Series Circuits.
Label the individual and total voltages. Label the three currents.
This is a series circuit
-V2-
I1 = I 2 = I 3
-V1-
R2
I3
R1
I2
I1
Vtotal = V1 + V2
Rtotal = R1 + R2
Note this, it’s important. If you add a resistor in series, the total resistance increases
and the current becomes smaller (you can usually assume the battery voltage is constant . )
The current anywhere in a series circuit is the same.
Explain the first diagram and complete the second:
Vs
If Vs = 1.5 V,
V1= 0.90 V,
V1
V2
Two lamps ~ energy from cell is shared between the lamps is shared.
Voltage is energy of each charge so voltage is shared
Electromagnetism
4
V2 = 0.6 V
Parallel Circuits
Label the individual and total voltages. Label the three currents.
Vtotal = V1 = V2 .
Itotal
= I 1 + I 2.
1
= 1/R1 + 1/R2 or RT = R1R2/R1 + R2
RT
I1
R1
-V1-
If Vs = 1.5 V,
V1= 1.5
IT
R2
V2 = 1.5
-V2-
I2
Note that when resistors are added in parallel, the total resistance decreases
because there are more pathways for the current) and the total current in the circuit increases.
(You can usually assume the battery voltage is constant unless told otherwise.
You now need to do a few examples from your course book or revision book.
Internal Resistance.
If you connect 1.5 V dry cell to a wire and ammeter with a total resistance of 0.01 Ω, how much
current will flow?
I = V/R = 1.5/0.01 = 150A!!
The reason you won’t really get a current this big is that the dry cell itself has got some
resistance. It is called the internal resistance. We have ignored this internal resistance up to now,
but it is significant.
(nb. The low voltage power supplies we use in the lab are designed to produce a constant voltage so
that they effectively have no internal resistance. They are called constant voltage power supplies.)
A real dry cell can be thought of as an ideal cell (no resistance) with a resistor in series.
Label the diagram.
ε
Electromagnetism
R
5
Describe what happens to the cell voltage when we close the switch.
V
The charge gains energy due to the EMF but loses some energy in
the internal resistance before it comes out.
The EMF (Electro motive force) of a cell is its output voltage when it is not producing any current.
(this is the maximum voltage the cell can produce.)
The internal resistance is the resistance of the cell.
As charge flows through the dry cell, it gains energy due to a chemical reaction, but then it loses
some due to the internal resistance.
Gains energy
Loses energy
ε
The voltage (or energy) gained is represented by the EMF ( ).
The voltage (or energy) drop in the internal resistance is
So the output voltage of a cell is given by:
VR = Ir
Vout = ε - Ir
Energy lost
in resistance
Ir
ε
Energy gained
in cell
Energy of incoming
charge
Electromagnetism
6
Energy of outgoing
charge
•
Notice that when the current is zero, the voltage across the internal resistance is 0
•
As the current through the cell gets bigger, the voltage across the internal resistance will
get greater. And the output voltage will get less.
Experiment
We’ll now connect the following circuit and make measurements.
One lamp
I = 1.0A. Vcell = 1.5V
Now we’ll connect a second lamp in parallel.
Predict what will happen to the current from the cell and the cell voltage.
The current will increase
The cell voltage will decrease
two lamps:
I = 1.8…………………………………. Vcell = 1.4V…………………………
three lamps:
I = 1.6…………………………………. Vcell = 1.3V…………………………
•
Sketch a graph of cell voltage versus current.
•
Write an equation for the line.
VC = -rI +
εMF
Vcell
ε
I
•
Explain the graph.
ε
When I = 0, the output voltage (V0 ) is the same as the EMF ( ). As the current (I increases, the
voltage across the internal resistance increases, and the output voltage VC decreases.
Homework exercise:
Define EMF
………………………………………………………………………………………………………………………………………………………………
Electromagnetism
7
Explain why the output voltage of a battery drops when a lamp is connected to it.
As the current (I) starts to flow, there is a voltage drop across the internal resistance, and so
the output voltage VC decreases.
Experiment note: Any battery and bulbs can be used. It is useful to compare
new and old batteries.
Electromagnetism
8
Homework Experiment
Do a search for “ phet simulations”.
Open “circuit construction”
Construct this circuit:
Start with both switches open.
Determine the EMF:
= ……………………….
Right click on the cell.
Set the internal resistance to about 5 .
Close one switch. Measure the current.
I1 = ………….
Predict the current through the ammeter when
the second switch is closed.
Predicted
Try it and see.
Actual
I2 = ……………………….
I2 = ……………………….
Explain why your predicted current was (or was not) different to the measured current.
The battery current does not double because the resistance has not halved because the battery
also has resistance
Now open the bottom switch. Use the values given to calculate the resistance of the lamp.
…………………………………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………………………………….
Right click on the lamp to see if your answer is correct.
Electromagnetism
9
Ex1 The diagram shows Millie’s motorbike electrical light circuit. The 0.10 Ω resistor represents
the internal resistance.
0.10 Ω
i)
12.0 Ω
1.0 Ω
6.0 V
12.0 Ω
Calculate the total resistance.
R = 12/14 + .1 = 0.96
ii)
Calculate the battery current
I = 6.3 A
iii)
Calculate the output voltage of the battery.
V = 6.0 – 0.63 x 0.10 = 5.9 V
iv)
Calculate the voltage across the headlamp.
5.9 V
v)
Why do the sidelights have higher resistance than the headlight?
They have the same voltage (all in parallel) but sidelights are dimmer (less power) so must have a
smaller current (P = VI) this requires a larger resistance.
Homework
Exercise
0.50 Ω
0.10 Ω
12 V
12.0 Ω
12.0 Ω
This show part of a car’s circuit with the starter motor and the headlights. Why do the light go
dim when the starter motor is turned on?
The lamps draw a small current so there is a small voltage drop across the internal resistance.
The starter motor has very low resistance so a large current flows through the internal
resistance.
This causes a lower output voltage, lamps get less voltage so are dimmer
Electromagnetism
10
Multi Loop Circuits and Kirchoff’s Laws
Life becomes more complicated if you have voltage sources in linked loops like this:
V1
V2
R3
I3
I1
R1
I2
R2
To solve these things, we use Kirchoff’s two rules
Point Rule
“What goes in must come out”.
Or
“current into a point = current out”
Write the point equation for the bottom centre point.
I1 + I2 = I3
Loop Rule What goes up must come down. Or “The total voltage around a loop is zero.”
(Or total voltage gain = total voltage drop in a loop)
This is how we use the loop rule
I
6.0V
4.0V
6.0V
start
start
2.0V
L
H
H
I
If you go from High energy to Low energy, this is a voltage drop
If you go from Low energy to High energy, this is a voltage gain
Electromagnetism
11
L
Explain the diagram:
Charge gains 6V in battery, no energy change until it reaches the resistor, then it drops 4.0V and
2V. Voltage gained equals voltage lost.
Write a loop equation, solve to find I:
6V - 4I – 2I = 0
I = 1A
Now we’ll look at the multi loop circuit from the previous page.
V1
V2
R3
The Left loop goes:
I3
+V1 - I1R1 - I3R3 = 0
R1
I1
I2
R2
The Right Loop goes:
+V2 – i2R2 – i3R3.
I1 = 2.0 A and I3 =3.0 A. Use the point rule to calculate the size of the current in the right loop.
1A
Use the loop rule to calculate the two battery voltages.
R1 = 5.0 Ω
R2 = 10Ω
R3 = 10Ω
V1 = V2 = 40V
Electromagnetism
12
Remember:
If you go past a resistor in the same direction as
the current, this is a voltage drop
I
If you go past a resistor in the opposite direction
to the current, this is a voltage gain
I
If you go past a battery in the same direction
as it’s pushing, this is a voltage gain
If you go past a battery in the opposite direction
that it’s pushing, this is a voltage drop
Homework Experiment.
Do a search for “ phet simulations”.
Open “circuit construction”
Construct this circuit:
1
Using the Loop Rule
2
1. Start with the switch in the bottom branch
open.
5
4
7
2. Use the voltmeter to add the voltages around
the top loop.
n.b
Always keep the leads in the same order
Include voltage drop in the wires.
3
V12 = …………………
V23 = …………………
V34 = …………………
V45 = …………………
V51 = ………………… Total Voltage = …………………………
6
Explain your result:
Total voltage = 0 in a closed loop
Now close the switch in the bottom branch.
3.
Go round the bottom loop clockwise, starting from “5” adding the voltages.
Electromagnetism
13
………………… ………………… ………………… ………………… ………………… ………………… = …………………
4.
Now go round the bottom loop anti clockwise, starting from “5” adding the voltages.
………………… ………………… ………………… ………………… ………………… ………………… = …………………
What do you notice?
……………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………
Notice that,
When you go through a cell in the same direction as the charge flow this is a voltage ……………………
When you go through a cell in the opposite direction as the charge flow this is a voltage ………………
When you go through a resistor in the same direction as the charge flow this is a voltage ……………
When you go through a resistor in the opposite direction as the charge flow this is a voltage ………
Using the point rule.
Look at the current in and out of the left hand junction. What do you notice?
……………………………………………………………………………………………………………………………………………………………
Right click on the bottom lamp, increase the resistance. Repeat the above observation.
……………………………………………………………………………………………………………………………………………………………
Look at the current in and out of the right hand junction. What do you notice?
……………………………………………………………………………………………………………………………………………………………
Electromagnetism
14
Ex1. Sophie has a four cell torch (
i)
= 6 V). A current of 0.50 A flows through the lamp.
A voltmeter across the bulb reads 5.0 V. Calculate the voltage across the internal
resistance. (Sketch a diagram)
1V
ii)
Calculate the size of the internal resistance.
2.0
iii)
Calculate the resistance of the bulb.
10
iv)
How will the size of the internal resistance change as it gets older? How will this affect
the current?
The internal resistance increases.
The current decreases
Ex2. Hillary’s parents ask her to connect the batteries of their yacht to a battery charger. She
sets up this circuit:
battery 1
battery 2
i)
The charger produces 18 V, and battery 1 has 14 V across it. Apply Kirchoff’s Law to the
left loop and calculate the voltage across the resistor.
+ 18 – VR – 14
VR = 4V
ii)
The resistor has a resistance of 15 Ω, calculate the current through it.
I = VR/R = 4/15 = 0.27A
iii)
Apply Kirchoff’s Law to the right hand loop, and determine the voltage across battery 2
VB2 = 14V
iv)
Battery 1 has an EMF of 12 V, calculate the voltage across its internal resistance.
VIR = 2V
Electromagnetism
15
10V
Ex3
I1
4.0Ώ
8.0Ώ
A
I2
8Ώ
The ammeter reads 1.0A.
i) Write a loop equation for the top loop and calculate the value of I1
0.50 A
ii) Calculate the value of I2
0.50 A
iii) Calculate the voltage across the second battery.
12 V
Extension
12 V
I1
3.0Ώ
2.0Ώ
A
2.0Ώ
I2
8V
Calculate the three currents.
Electromagnetism
16
Self Review:
Explain what is meant by:
Potential Difference
Current
EMF
Internal resistance
Output voltage
Explain the equation V =
– IR
State Kirchoff’s laws
Explain the conservation laws that Kirchoff’s Laws are based on.
Electromagnetism
17
When you want to store electrical energy and release it fast, you need a capacitor.
e.g.
~
Heart defibrillator
~
Camera flash
A rechargeable battery can store energy and release it as an electric current, but it can only go in
and out slowly.
A capacitor also stores electrical energy but it can store it and release it really quickly. This is
important when you want to deliver the energy rapidly, like a strobe light.
A capacitor is usually two metal plates close together, separated by an insulating material called a
dielectric.
switch
Simple Circuit
When the switch closes, electrons from the negative terminal of the cell flow to the left. hand
plate of the capacitor. Other electrons are attracted from the right hand plate to the positive
terminal of the battery. The capacitor is then “charged up”. It has negative charge on the left
side and positive charge on the right side.
n.b. Electrons flow around the circuit. They haven’t flowed through the capacitor, but it appears
that they have.
Camera Flash
Switch position 1 Electrons flow from the battery to the left plate
and from right plate to battery. This continues until the capacitor is
fully charged.
(this happens when the camera is turned on)
switch
1
Switch position 2
Electrons flow from the left plate through the
flash tube to right plate until the capacitor is discharged. A large
current causes the flash tube to produce light. (this happens when the
picture is taken)
2
flash
tube
Remember:
A resistor has resistance.
Likewise a capacitor has
Electromagnetism
capacitance
18
Capacitance:
this is a measure of how much charge can be stored in a capacitor when
it is connected to a 1 Volt battery.
It is measure in Farads.
A one farad capacitor can store one coulomb of charge if connected to a one volt battery
Consider a divers tank.
The amount of air it contains depends on :
Volume and Pressure.
The amount of charge the capacitor stores depends on:
Capacitance (this is like volume)
and
Voltage (this is like pressure)
The amount of charged stored is:
Q
= C x V
Capacitance is measured in Farads (named after Michael Faraday )
(A farad is very big, so we often use µF, nF or pF)
micro, nano, pico 10-6, 10-9, 10-12
Homework exercises
ex 1. A capacitor that stores 1 Coulomb ( 6 x 10
capacitance of 1F
18
electrons) when connected to 1 Volt has a
ex2. Calculate the capacitance of a capacitor that stores 3 x10
2.5 x 10-13 F
-12
C when connected to 12 V.
ex3. How much charge is stored in a 10 µF capacitor connected to a 2.0 V battery?
2.0 x 10-5 C
Charging Up
Consider an empty bottle as it is filled with air:
Sketch graphs of pressure/time and airflow/time
on.
Airflow
Pressure
time
Electromagnetism
after the compressor is turned
time
19
Air flows in fast initially, but the flow rate decreases as the pressure builds up and it becomes
more difficult to shove more air in.
The pressure increases fast initially, but then rises more slowly until it remains constant
Consider the capacitor as it is charged and discharged.
Predict the shape of the current flow/ time graph
as it charges.
1
Predict the shape of the capacitor voltage/time
graph ( V= Q/C ) as it charges.
2
Experiment note. This can be shown easily with a datalogger with
voltage probes connected across the resistor and capacitor. (probes
across resistor are detecting the current.
time
time
Switch
to 1
Switch
to 2
Current: The capacitor is initially empty (uncharged) so it starts filling fast (the current is big).
As it fills, it becomes more difficult to push more charge in, so the current decreases until it
reaches maximum when the capacitor is fully charged.
Experiment Note: A resistor and capacitor with a time constant around 1 second is useful. A
datalogger and computer are used to show the graphs. The current graph can be obtained with a
voltage probe across the resistor.
Electromagnetism
20
Voltage: The capacitor voltage is initially zero . It rises quickly at the start because the current
is large but then it rises more slowly until it reaches a maximum.
When fully charged, the capacitor voltage is the same as the battery voltage .
Time Constant
How long does it take to fully charge a capacitor? Forever
Vmax
fully charged??
0.63 Vpm
Voltage
1 time constant
time
This is very difficult to determine as the capacitor voltage gets closer and closer to the cell
voltage, but never quite reaches it. So the time taken to charge a capacitor is described by the
time constant.
This is the time taken to reach 0.63 times the maximum voltage.
Show this on the graph.
When the capacitor is discharging, the time constant is the time taken to drop to 0.37 times the
maximum voltage. Show this on the graph.
Vmax
1 time constant
0.37 Vpm
Time Constant Equation:
Think above the dive tank. What two things determine the time it takes to fill? (reach maximum
pressure)
volume ~ thickness of pipe
Consider the RC circuit above. What two things determine how long it takes to fill ? (reach
maximum voltage )
Capacitance, resistance.
Electromagnetism
21
The time constant is given by:
T = RC
Changing C and R
On the following graph, sketch the new charging curve if the resistance increased.
V
Explain:
Larger resistance
Smaller Current
Takes longer to charge
t
On the following graph, sketch the new charging curve if the capacitance increased.
V
Explain:
Larger capacitance
Stores more charge
Takes longer to fill
t
ex. A 100 µF capacitor is connected to a 10 kΩ resistor and a 2.0 V battery.
Calculate the time constant: T = 1 s
Sketch the charging curve (V/t) (include 3 data points)
CHARGING
1.73V
2.0V
1.26V
1
2
Homework exercise:
A 150 F capacitor discharges into a resistor
As shown.
Calculate the resistance.
Electromagnetism
10 .0 V
6.3 V
22
5.0 ms
R = 33.3
Electromagnetism
23
Voltages in an RC Circuit
Label the Voltages in the circuit. (VS, VR , VC )
VS
At any time,
VS = VR + VC
VS = IR + VC
VR
VC
When the switch is first closed:
VC = 0
because cap has no charge so it has no voltage across it
VR = VS
because Kirchoff’s loop eqn.
When the capacitor is fully charged:
I = 0
because current flows until capacitor voltage is equal and opposite to battery
VC = VS
voltage. The current is then zero
VR = 0
Homework Experiment.
Go to the site
The
The
The
The
http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=48
left hand loop contains a battery, the right hand loop contains a capacitor and resistor.
bars represent the voltages across the three components.
yellow dots represent electrons.
capacitor is originally uncharged.
Click the red switch to charge the capacitor.
Explain what happens to each voltage:
Battery voltage: ……………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
Resistor voltage: ……………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
Capacitor voltage:
…………………………………………………………………………………………………………………………………………………………………………………..
……………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
Electromagnetism
24
Explain what happens to
the rate of charge flow
after the switch is closed.
…………………………….
……………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
The red graph represents …………………………………… . Explain its shape
……………………………. ……………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
The blue graph represents …………………………………… . Explain its shape
……………………………. ……………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………………
Electromagnetism
25
Capacitors in Series and Parallel
Experiment note. Many digital meters have a capacitance function.
We’ll connect a capacitor to a capacitance meter, then connect a second one in parallel.
Predict what will happen to the total capacitance.
(Don’t forget, capacitance is a measure of the amount of charge that can be stored.
Prediction: …………………………………………………………………………………………………………………………………
Result: C doubles
Now we’ll do it again but add a second one in series.
Prediction: …………………………………………………………………………………………………………………………………
Result: C halves
Parallel
(capacitance increases )
When two or more capacitors are added in parallel, their areas are
greater, so the capacitance adds, so
CTot = C1 + C2
(proof)
QTOT = Q1 + Q2
CTOT V = C1 V + C2V
(Q = C V)
CTOT = C1 + C2
n.b When capacitors are in parallel, they always have the same voltage.
Series
(capacitance decreases )
I
When two or more capacitors are added in series, they share
the applied voltage so the capacitance decreases
C1
C2
C2 gets its charge from C1
so Q1 = Q2 = QTOT
VTOT = V1 + V2
Q
CTOT
= Q + Q
C1 C2
1 = 1/C1 + 1/C2
CTOT
n.b When capacitors are in series, they always have the same charge.
Electromagnetism
26
Examples
1) A 2.0 µF capacitor is connected to a 2.0 kΩ resistor.
(a) Calculate the time constant.
T = 0.0040s
(b) A 4.0 µF capacitor is connected in series with it. Calculate the time constant.
T = 0.0026s
(c) A 4.0 µF capacitor is connected in parallel with it. Calculate the time constant.
T = 0.012s
2) A 200 µF capacitor is connected in series to a 300 µF capacitor and a 1.5 V cell.
(a) Explain what quantity is the same for both capacitors
Same charge. C1 gets charge from the cell, C2 gets charge from C1.
(b) Calculate the total charge stored
1.8 x 10-4C
(c) Calculate the voltage across each capacitor
200 µF – 0.9V
300 µF – 0.6V
(d) Repeat the above but the capacitors are connected in parallel.
(a) Same voltage
(b) 7.5 x 10-4C
(c) 1.5V
3) A 100 µF capacitor is connected to a 3.0 V battery.
(a) Calculate the charge stored.
3 x 10-4 C
The capacitor is disconnected and then connected to a 200 µF capacitor.
(b) Determine the charge stored
3 x 10-4 C
(c) What must be the same for both capacitors?
Voltage
(kirchoff’s loop)
(d) Calculate the charge stored in each capacitor
capacitance is how much charge can be stored when connected to 1 volt
200 µF capacitor can store twice as much charge for the same voltage
so 200 µF capacitor stores 2/3 of the charge C = 2.0 x 10-4C
100 µF capacitor stores 1/3 of the charge C = 1.0 x 10-4C
Electromagnetism
27
(d) Calculate the energy stored in each capacitor
E200 = 3.0 x 10-4 J
E100 =1.5 x 10-4 J
Homework exercise
A 200 nF capacitor is connected to a 1.5 V cell.
The capacitor is disconnected and connected to an uncharged 300 nF capacitor. Calculate the
charge on each one.
200nF
QTOT = 3 x 10-7 C
Both same voltage
2/5 Q = 2/5 x 3 x 10-7
= 1.2 x 10-7C
300nF
3/5Q = 1.8 x 10-7C
Energy in a Capacitor
Capacitors store electrical energy. The energy stored equals the work
done storing it.
charging a capacitor
Recall
V
Energy = Voltage x Charge
As charge increases, voltage increases.
Energy stored = ½ VQ = ½ CV2
Q
(Q = CV)
eg. Calculate the energy stored in a 200 µF camera capacitor connected to a 2.0V battery.
E = 400μJ
Capacitor Construction
This relates to an experiment in the Experiment folder
By experiment, we have found that the capacitance of an air filled capacitor depends on:
1) Area
so
2) Separation
Electromagnetism
28
C α A/d
C=
A/d
ε0
is a constant, it is called the permittivity of free space. (8.84 x 10-12 Fm-1)
Describe how you could measure it experimentally.
Measure area and separation of two metal plates. Measure capacitance with a meter.
eg.
A capacitor has a plate area of 30 cm x 20 cm. The plates are 10 mm apart.
Calculate its capacitance.
5.3 x 10-8F
Dielectric: We also found that inserting an insulator affected the capacitance. The insulator is
called a dielectric. It does two things. It keeps the capacitor plates apart, and also increases the
capacitance.
ε
The dielectric constant ( R ) of a material is a measure of how much the capacitance increases
εR
= capacitance with dielectric
capacitance without dielectric
so the formula for capacitance is:
C=
ε 0ε r A
d
Making a Capacitor
Most cheap capacitors are made by getting two metal plates with wax paper between, and rolling
them into a tight cylinder.
What is the purpose of the paper?
Keep plates apart
Dielectric (increases the capacitance)
Examples
(1) An air filled capacitor has a capacitance of 5.0 µF. Plastic is inserted that increases the
capacitance to 15.0 µF. Calculate the dieletric constant.
εR = 3.0
Electromagnetism
29
(2) Two metal plates A4 sized plates are separated by 0.10mm of paper with a dielectric constant
of 1.3.
Calculate the Capacitance in pF.
C = 6900 pF
(3) What could you do to increase the capacitance?
Increase area, thicker paper
(4) One capacitor is marked:
220 nF 6.0 V
Another is marked:
440 nF 2.0 V.
Why does it have a voltage marked? Which can store more charge?
Shows maximum voltage that can be applied. 200nF
220 nF stores more charge as Q = C x V - it has half the capacitance but 3 x the voltage.
Using Capacitors
Storing energy
Capacitors are used for storing energy eg in a camera flash. Computers have them as back up
power for the memory chips.
Stud Finders
Ever tried to hang a heavy picture on the wall.
You need to find the wood behind the wall board.
This is called a stud.
The stud finder has two capacitor plates side by
side. The stud finder detects a change in the
capacitance. How will it change if there is wood
behind the wallboard? Capacitance increases.
Capacitors are also used for measuring.
Capacitor Fuel Gauges
e.g. Airplanes often measure fuel level with a
capacitor. Two metal plates are put in the fuel tank and connected to a capacitance meter that is
the fuel gauge.
As the fuel level drops, the capacitance decreases.
The capacitance meter measure this and is
calibrated to measure the level of fuel.
Measuring Blood Pressure
Electromagnetism
30
C
The capacitance transducer used the fact that the capacitance of a parallel plate capacitor
changes according to the rule
C =
εR εO A
d
movable
plate
The different arrangements that are used are
shown below. The diaphragm is moved by the
blood pressure in the artery.
diaphragm
fixed
plate
d~increases
diaphragm
movable
dielectric
fixed
plate
diaphragm
fixed
plate
fixed
plate
movable
plate
εR ~ increases
A ~ increases
(i)
For each of the above arrangements, say which of the variables on the right hand side of
the equation above is being changed by the movement of the diaphragm.
(ii)
How will the capacitance of each change if the diaphragm moves LEFT?
Electromagnetism
31
Homework examples
i) A laptop has a 0.50 F capacitor to maintain power when the battery is being changed. The
battery is 14 V. When the battery is removed, the capacitor slowly discharges as shown.
14
V(V)
0
5
10
15
t(h)
a) Estimate the time constant of the RC combination.
T = 7 hours
b) Calculate the resistance of the laptop circuit.
R = 50 000 Ω
c) Calculate the maximum charge stored in the capacitor.
Q = 7C
d) The laptop takes 20 hours to almost completely discharge. What is the average
this time? (I = Q/t)
I = 9.6 x 10-5A
current during
e) Explain why increasing the capacitance would mean you had more time to change the batteries.
C increases, therefore Q increases
I = Q/t
t = Q/I therefore t increases
ii) A second 0.5F capacitor is connected in parallel. What does this do to:
The charge stored?
Increases
The energy stored?
Increases
The total capacitance?
Increases
The time constant?
Increases
Electromagnetism
32
iii) A 0.5F capacitor is connected in series. What does this do to:
The charge stored?
Decreases
The energy stored?
Decreases
The total capacitance?
Decreases
The time constant?
Decreases
Electromagnetism
33
Electromagnetic induction is the production of an electric current by a changing magnetic field
We first need to review our year 12 work on magnetic fields.
Magnetic Fields
A magnetic field is a region in space where a non contact force acts on a ferrous metal such as
iron. They are best detected by compasses.
Magnetic fields are created by moving charges. This is obvious when we see the magnetic field
around a current carrying wire.
Produced by
moving charge
Caused by
charge
Magnetic fields
Detected by a
compass
Produced by
any charge
Electric fields
Non-contact
force
produced
Draw the magnetic field shape around this wire.
Detected by
electric dipole
Label N and S poles
+
+
-
-
A wire coil carrying a current produces a strong concentrated magnetic field the same shape as a
Bar magnet
Electromagnetism
34
Force on a Charge in a Magnetic Field
When a charge moves across a magnetic field, it experiences a force that is perpendicular to the
velocity and to the magnetic field.
•
•
•
The magnetic field is in the x direction
The eletrons are moving in the y direction.
The electrons are therefore pushed in the z direction.
Force
z
S
y
N
x
electron
beam
electron
gun
The magnetic field is to the right.
Electrons are coming out of the page.
They experience a force to the ……………………
F
The magnetic field is out of the page.
Electrons are travelling to the ……………… of the page.
They experience a force to the ……………………
Draw or describe the path of the electron beams:
Experiment note: some sort of electron tube is needed
to show the force on the moving electron.
Electromagnetism
35
Field out
Electromagnetic Induction in a Wire:
When a wire cuts across a magnetic field, all the electrons inside it experience a force and move
to one end.
One end of the wire becomes positive the other becomes negative i.e. a voltage is induced across
the ends of the wire. (If you were going to say voltage goes through the wire, wash your mouth
out with soap!!)
Force
S
Show the direction of the
force on the electrons.
z
Wire's
velocity
y
N
x
wire
Which end becomes -ve?
Top
Why is there no current?
No circuit
The size of the induced voltage (or EMF) is given by:
ε
=Bvℓ
Experiment note: A very sensitive galvanometer (eg scalamp) and a strong magnet are used.
nb. The voltage is maximum when the wire cuts the field at right angles.
The voltage is zero when the wire moves parallel to the field
If the wire is connected to a closed loop, a current will flow around the loop.
This is the basic principle behind different devices such as: generators, microphones.
Self Review:
When does a magnetic force act on an electron?
How does a magnetic field induce a voltage in a wire?
Can you get a voltage but no current? Explain.
What is the formula for the voltage induced across a wire?
Electromagnetism
36
Magnetic Flux
A very useful concept is the idea of magnetic flux. It can be thought of as the quantity of
magnetic field.
The amount of grass in a lawn is:
amount of grass = Area (m2) X grass density (Nm-2)
eg a 30 m2 lawn has a density of 120
plants per m2. How many plants are
there? 3600
Likewise the quantity of magnetic field is:
magnetic flux = Area x Flux density (field strength)
B is the symbol for field strength or flux density.
A is the symbol for the area the field passes through.
φ = BA
OR
Units are T X m2
⊥
OR Weber (Wb)
The strength of a magnetic field is shown by density of the field lines.
A
B
In which case is the field strength highest,
C or D? D
In which case is the field strength highest,
C or A? A
C
In which case is the flux highest, A or D?
D
In which case is the flux highest, A or C ?
same
D
Electromagnetism
37
ex A circular loop, radius 3.0 cm is perpendicular to a magnetic
( B = 0.020T).
(i) Calculate the magnetic flux through it.
φ = BA = 5.65 x 10
-5
field
Wb
(ii) find the magnetic flux through the loop when the loop is rotated (1) 900 (2) 450
φ =0
φ = 5.65 x 10
-5
x
sin45º
= 4 x 10-5 Wb
Faraday’s Law
Michael Faraday was an experimental physicist and a smart guy. He noticed that when you shoved a
magnet into a coil, you got a voltage induced. He looked at this using the idea of magnetic flux. He
said it’s quite simple.
(1) If you change the magnetic flux in a loop, you get an induced emf (voltage)
(2) The faster you change the magnetic flux, the greater the induced voltage.
This is called Faraday’s Law:
Induced Voltage = Rate of Change of magnetic flux
V=
Or
Δφ
Δt
There are various ways of changing the flux in a loop:
(i)
(ii)
(A)
Or the magnetic field strength (B) .
We can change the effective area
Experiment note: A sensitive galvanometer with a wire attached and a powerful magnet are
used. Make a loop in the wire, place it in the magnetic field and change the area and angle
(i) Area:
You can change the actual area
x
x
x
Electromagnetism
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
38
Or the area perpendicular to the magnetic field.
rotate
Experiment note: Use a coil connected to a galvanometer
(ii)
Magnetic field strength:
N
OR
△
φ
Explain
Pushing the magnet in or closing the switch both causes a flux change in the
coil producing an induced EMF.
Ex. A 100 turn coil has a X sectional area of 10cm2 . The magnetic field through it changes from
0.55T to 0.25T in 0.1s.
(i)
Calculate the flux change.
3 x 10-4 Wb
(ii)
Calculate the induced voltage in one loop.
V = 0.003 V
(i)
Calculate the induced voltage in the coil.
V = 0.3V
Homework exercise
What is flux? See notes
State Faraday's Law in words.
Units?
Describe ways of changing flux.
Electromagnetism
39
Lenz’s Law.
Experiment Note: These are light
aluminium rings mounted on a swivel
that is free to rotate. One ring has a
gap in it. A strong neodymium
magnet is needed.
Look at this demonstration:
The loop is producing a magnetic field that opposes the incoming magnet.
Lenz was also a smart guy too. His is the shortest law in physics. “ – “
He stuck a negative sign into Faraday’s Law to make it:
−Δφ
ε=
Δt
Essentially what this means is that whenever you change the magnetic flux in a loop, you get a
voltage that opposes the flux change.
2 induced current
3 induced mag field
1 increasing flux
e.g. When you push a magnet into a loop, a voltage is produced that causes a current to flow..
Where does this electrical energy come from?
N
S
N
You must do work on the magnet. The loop produces an opposing force on the magnet.
When you push the magnet towards the coil, the changing magnetic field (flux change) causes an
induced current. The induced current creates a magnetic field. According to Lenz’s law, this must
Electromagnetism
40
oppose the motion of the magnet, so the coil pushes the magnet away because the left end of the
coil becomes a N pole.
Likewise if the magnet moves away to the left, the induced magnetic field opposes the motion and
tries to pull it back. The left end of the coil becomes a S pole.
These examples demonstrate
Faraday’s law and Lenz’s law.
Magnet in the Tube
Draw an arrow showing the magnetic field caused by the tube.
Draw an arrow showing the current in the tube
S
I
N
Experiment Note: PASCO and Professor Bunsen supply a
magnet and aluminium tube. Alternatively, use a strong
neodymium magnet. Drop it down the tube (≃ 1m), it will fall
very slowly.
B
The magnet
moves into
the tube
This
produces a
FLUX
change
This
produces
an induced
CURRENT
This
produces a
MAGNETIC
field.
Magc field
opposes
the falling
magnet
Coil and Loop.
When the switch is closed, the current suddenly increases. This
causes the coil’s magnetic field to increase. This causes a FLUX
change in the loop, producing a CURRENT in the loop. The current
produces a magnetic field in the opposite direction to the coil’s field.
The loop jumps up.
Experiment note: a big coil with an iron core is needed with light aluminium rings
Homework Exercise;
Explain why a magnet falling through a tube experiences an upward force:
See above
What does the size of the force depend on?
strength of magnet
conductivity of pipe
Electromagnetism
41
Metal
loop
loop
speed of magnet
gravity field
Electromagnetism
42
Magnet in the Coil
A magnet is dropped through a coil connected to a data logger. Predict the shape of the
S
current/time graph over the page.
N
Experiment Note: connect a coil to a
data logger (use voltage probe) drop
the magnet through the coil.
Predicton
Experiment
current
current
time
time
As the magnet falls into the coil, the induced current flows one way and the top of the coil
becomes a N pole. (The coil is opposing the motion of the magnet) As it falls out, the bottom
becomes a N pole because a S pole is moving Away. (Again the coil is opposing the motion of the
magnet)
The magnet moves out faster than it moves in, so the second induced current is larger and less
time.
Applictions.
There are lots of examples of things that use electromagnetic induction.
e.g. microphone, generator
Moving coil microphone:
Sound waves cause the diaphragm and coil to vibrate.
As the coil vibrates near the magnet, there is a
continuous flux change. This induces an alternating
current in the coil that has the same frequency as
sound wave. This alternating current goes to an
amplifier
Electromagnetism
43
Flexible diaphragm
fixed to coil
magnet
AC Generator
B
A
C
D
When the Coil is rotated from the position shown, there is a flux change in the coil. Firstly, as it
moves clockwise to the vertical position, the magnetic flux through the coil increases. In the next
900 turn the flux is decreasing. The voltage is induced in the opposite direction.
Coil position
(End on)
Flux
Angle
Voltage
Angle
Electromagnetism
44
Another way of thinking about it is the speed that the wire coil cuts across the magnetic field.
B
Draw in the sketch the flux though the coil.
Explain why the voltage is maximum when the coil is parallel to the field.
Wire is cutting field at maximum speed OR flux is changing at maximum rate
Explain why the voltage changes direction every 1/2 turn.
Wire changes direction every ½ turn OR flux increases then decreases every half turn
Generator Voltage Equation
What four things does the maximum voltage depend on ?
Vmax= = BANw
The voltage varies sinusoidally,
V = Vmax sin θ
or
(where θ = 0 when the coil is perpendicular to field
V = BANw sinwt
(remember θ = wt
Mutual Induction
A changing magnetic flux in one coil can cause an induced EMF in a second coil.
Discuss how electric toothbrushes charge up.
Coil in the base carries AC. This causes alternating flux in coil inside handle. This causes an
induced emf that pushes a current through the battery. (AC is first converted to DC)
Electromagnetism
45
Transformers
Transformers are usually used to increase or decrease a voltage.
There are three types of transformer:
(1)
Step down transformers, decreasethe voltage
e.g
iphone
These are used to reduce the voltage so low voltage devices such as cell phone chargers can
operate off the mains
(2)
Step up transformers, increase the voltage
Many devices such as TVs need a high voltage.
(3)
Isolating transformers, don’t change the voltage
These are used when an appliance such as a circular saw or concrete mixers are used,
(especially outside) for safety, to prevent electric shock.
How they work
Iron core
V in
V out
primary
secondary
An alternating voltage across the primary causes an alternating current in the primary
This produces an alternating magnetic field in the core.
This produces a continuous flux change in the secondary coil
This produces an induced voltage in the secondary coil.
Electromagnetism
46
IP
Is
Vp
Vs
Np
Ns
The output voltage depends on the input voltage and the ratio of turns
Vs = VP x NS/NP
Or
Vs = NS/NP
Vp
If Ns > Np this is a step up transformer and the output voltage increases
If Ns < Np this is a step down transformer and the output voltage decreases
If Ns = Np this is an isolating transformer and the output voltage doesn’t change
Efficiency
In theory the energy output equals the energy input. In practice some energy is converted to heat.
This is due to:
1)
energy loss in the Coils. This is reduced by making the wire out of thick copper.
2)
Eddy currents in the iron core, this is reduced by making the core out of many thin
insulated sheets of iron.
In an ideal transformer,
Power in = Power out.
VpIp
=
VS IS
Jess suggests that if a transformer can increase the voltage, you can get out more energy than
you put in. Use the equation above to discuss if this is true.
No, if you increase V you decrease I (E = VIt)
Example: Priya wants to make a transformer to charge her cell phone from the mains. (………….
Volts) The cell phone battery is 4.0V.
(i)
Calculate the turns ratio
240: 4
= 60:1
(ii)
The cell phone battery draws 100mA. What current flows into the primary coil?
1/60 x 100mA = 1.7mA
Electromagnetism
47
Mutual Inductance
The induced voltage in the secondary coil is proportional to the rate of change of current in the
primary coil.
Vs α
ΔI
Δt
or
Vs = M
ΔI
Δt
M is called the mutual inductance and is a measure of how well the changing
flux in the first coil
induces a voltage in the second coil. T
If the right hand coil (secondary) is moved closer
to the left hand coil, the mutual inductance
increases.
If an iron core is inserted, mutual inductance
increases
Both these things increase the mutual inductance
Mutual Inductance is measured in Henrys (H )
High Voltage Transmission Lines
generato
r
Ir
Ir
V
V
V
pon s
pon s
rico e
rico e
i
o
i
o
re
c
mrestation
c
In a power station,mgenerators
are turned, usually by water in a hydro
or by steam in a
n
u
n
u
a
o generator typically produces 3000V AC.a
o
thermal power station.
The
t
t
2
r
n
r
n
The power lost as heat in a wire is:
P = I R.
y
d
y
d
The power transmitted adown the wire is
P = VI
a
r
r
To reduce power loss, the
resistance
and
current
must
be
as
small
as
possible.
To reduce the
y
y
current in the transmission lines, the voltage is stepped up to around 500,000V thus reducing the
power loss.
V
At the other end, a substation transformer reduces the voltage
to around 3000V.
This is reticulated around the streets. Local transformers like
the one shown reduce the voltage to 240V.
Electromagnetism
48
Ex: A transmission line has a resistance of 5.0 Ω. You want to transmit 1000W of power. You could
do this either using high voltage (eg.1000V and 1.0A) or low voltage
(eg. 100V and 10A.) We’ll see which wastes most power.
Experiment Note: This can be demonstrated. Connect a 12V AC supply to a 1:10 ratio transformer.
Connect this to about 1m of nichrome wire, then a 10: ratio transformer. Connect a 12V lamp to
this. CAUTION – the nichrome will have a high voltage.
Calculate the power loss using 1000V and 1A
P = I2 R
P = 5W
Now calculate the power loss using 100V and 10A
P = I2 R
P = 500W
Comment on the difference.
By increasing the voltage 10 times, the power loss is reduced 100 times.
Induction Coils
In many situations, you need to produce a high voltage from DC current e.g.
car spark plugs. Normally transformers need to run off AC because they need a constantly
changing magnetic field. Induction coils are pretty cunning. They are like transformers but they
operate off DC. They produce the constantly changing magnetic, by switching the current on and
off
Primary
urPrimary
Secondary
C Switch opens or closes
When the switch closes, the current through the primary suddenly increases
This produces a rapidly changing magnetic field in the core
This causes a large flux change in the secondary coil.
This causes a large voltage pulse across the secondary coil.
The size of the voltage is increased by having thousands of turns on the secondary coil.
A high voltage pulse occurs across the secondary coil every time the switch opens or closes.
Electromagnetism
49
Switch opens
or closes
Sudden change
in
Magnetic flux
Large flux
change in
secondary coil
Large voltage
(EMF)
induced in
secondary coil
Self Inductance
A coil can induce an EMF in a second coil. It can also induce an EMF in itself!!
coil
Inductors
An inductor is a single coil with an iron core.
core
Describe what happens when the switch is closed.
Current through coil increases slowly because
coil opposes current change – bulb lights slowly.
Second bulb lights up rapidly.
Experiment Note: Use a large value inductor or a PSSC coil full of iron bars to show the time delay
in lighting the bulb.
An Inductor is a device that always opposes a change of current. This is why there is a delay
before the bulb lights up.
(nb. An ideal inductor has no resistance and so does not oppose a constant current.)
The diagram below explains how they work. If electric current is coming from the left and is
increasing, there is an increasing magnetic field inside the coil. i.e. There is a flux change in the
coil. This flux change causes an induced voltage. According to Lenz’s law, the direction of the
induced voltage must oppose the change that caused it.
This induced voltage is called a Back EMF
Electromagnetism
50
Increasing
current
Decreasing
current
Induced
Voltage
Induced
Voltage
Experiment Note: Use a large value inductor or a PSSC coil full of iron bars to show the time delay
in lighting the bulb.
Inductors produce a Back EMF that always opposes a change in current. This causes a delay in the
bulb glowing when the switch is closed.
(A real inductor also has some resistance and so it also opposes the flow of continuous current.
This is why there is a rheostat in the lower branch so the lamps are equally bight.)
Electromagnetism
51
Sketch a graph of Current / time and Inductor Voltage /time from when the switch is closed
(Assume its an ideal inductor)
I = V/R
I
t
V
The time constant is τ = L
R
Explain what this means
t
Larger inductance – longer time taken to reach maximum current
Larger resistance – maximum current smaller – takes less time to reach maximum current
Extension. Sketch a graph of Inductor Voltage /time for a real inductor.
V
t
A real inductor has resistance so there is a voltage across it even when the current is constant
The size of the induced voltage (or back EMF) depends on two things.
(i)
The number of turns of wire and the amount of iron in the core. Together this is
called the selfinductance. It is measured in Henries (H)
(ii)
The rate at which the current changes. (see Faraday’s law)
Induced Voltage is:
ε = −L
ΔI
Δt
The –ve sign means the Induced Voltage opposes the change in current
Self Inductance is a measure of how much induced voltage is produced when the current changes.
Ex. The circuit above is connected. A short time after the switch is closed, the current is rising at
3.0 A/s. There is back EMF of 0.50 V. Calculate the inductance of the inductor.
ΔI
ε = −L
Δt
Electromagnetism
0.5 = L x 3.0
L = 0.5/3 = 0.17H
52
Energy
There is a lot of energy stored in the magnetic field of an inductor.
If the current suddenly drops to zero, there is a very large flux change and a very large voltage is
induced
The energy stored in an inductor is given by:
E = ½ LI2
Compare with a capacitor,
E = ½ CV2
Extension. Discuss what would happen if a charged capacitor was connected to an inductor.
Charged capacitor discharges through inductor. Electrical energy converted to magnetic. Magnetic
field then reduces causing a current that charges the capacitor. This cycle then reverses –
current oscillates.
Homework exercise:
Ex1 what is an inductor?
Ex 2 what is inductance?
Ex 3 what does the inductance of an inductor depend on?
Ex3 In the circuit above, the switch is closed, the current starts rising at 6.0 A/s. There is back
EMF of 0.250 V. Calculate the inductance of the inductor.
L = 0.042 H
Ex4 A1.0 mH inductor is connected to a battery and a switch. The switch is closed after 0.10 s,
the current is 0.10 A. Calculate the average EMF across the inductor during this time..
ε = 10−3V
Ex 4 Explain why this voltage is much less than when the switch is opened.
When the switch closes the current rises slowly. When the switch opens, the current suddenly
drops to zero. This rapid change in current causes a large EMF. There will be a spark across the
switch.
Electromagnetism
53
Electromagnetism Summary
Flux is the quantity of magnetic field
φ = BA
Faraday’s Law:
Induced Voltage in a wire is ε =
Δφ
Δt
For a coil, multiply by the number of turns.
Lenz’s Law
The Induced Voltage (EMF) opposes the flux change that caused it
A coil can induce a voltage in another coil.
This is called Mutual Inductance.
Closing the switch causes the primary current to increase
This causes a flux change in both coils.
This causes an induced voltage in the secondary
The size of the voltage induced in the second coil is:
ε =M
ΔI
Δt
M is the mutual inductance of the two coils. It is a measure of how well the current change in the
primary induces a voltage in the secondary.
A single coil can also induce a voltage in itself. This is called Self Inductance.
Closing the switch causes the current to increase
This causes a flux change in the coil
This causes an induced voltage in the coil.
The size of the voltage induced in the coil is:
ε =L
ΔI
Δt
L is the self inductance of the coil. It is a measure of how well the current change in a coil induces
a voltage in itself.
If the switch closes, the bulb brightens slowly because the induced voltage across the coil opposes
the battery voltage.
When the current is constant, only the resistance in the circuit affects the size of the current.
When the switch is opened, the current suddenly drops to zero
This causes a rapid flux change.
This causes a very large induced voltage across the coil (causing a spark across the switch)
ε =L
ΔI
Δt
Electromagnetism
54
So when
ΔI
Δt
is big, the induced voltage is big
Another way of explaining it is…..
The time constant determines how quickly the current rises and falls. τ =
L
R
So when the switch opens, R is very big (infinite?) so the time constant τ is very small.
REVIEW QUESTIONS
What is an Inductor?
What does it do?
What is Inductance?
Units?
What does the size of the back voltage depend on?
What is a transformer?
What is it designed for?
How does it work?
Explain how a generator works.
Why does the voltage vary?
What does the voltage depend on?
Electromagnetism
55
Batteries produce DC or direct current.
This means that the current flows in one direction.
Generators produce AC or alternating current. This means that the current reverses direction
continuously.
Experiment Note: RMS Voltage – connect identical bulbs to AC and DC on power pack – connect
voltage probes of computer interface or dual beam oscilloscope across each lamp.
Voltage
DC
AC
AC
DC
Experiment note. Connect lamps to AC/DC supply. Connect CRO OR datalogger across both lamps
Both lamps have the same brightness. Therefore they are producing the same power.
However the voltages across each are very different. The DC voltage is constant.
The AC voltage changes size and direction.
How do you think we define the size of an alternating current or voltage??
Why can’t we use the average voltage?
Average is zero.
The above voltages both produce the same heating effects, so we can say that the
AC Voltage is effectively the same as the DC Voltage. We can see from the graph that the DC
Voltage is approximately
0.71 x the maximum AC Voltage.
i.e
effective voltage = 0.71 X
and effective current = 0.71 X
Vpeak
I
peak
Mathematically, this is called the “root mean square” voltage (or RMS voltage)
To calculate it, you square the voltage function, find the mean for one cycle, then square root it!!
……………… Complicated??
Electromagnetism
56
Don’t worry, for a sine function you just multiply the maximum value by
i.e VRMS = 1 VPeak
√2
(nb as a decimal, 1
= 0.71)
√2
IRMS = 1
√
1
2
IPeak
ex.
The “mains voltage in NZ is 240V. This is the RMS value. ie a lamp will glow with the same
brightness connected to the mains as to a 240V battery.
Calculate the instantaneous maximum voltage across a lamp connected to the mains.
Vmax = √2 x VRMS = 340V
Extension: (for the mathematics wizards)
P = I 2R
= (Imaxsinωt)2 R
= I2 max x sin2ωt x R (average value of sin2ωt is ½.)
Average power = ( Imax)2 x ½ x R = IRMS2 R
( Imax)2 x ½ x = IRMS2
IRMS = Imav
√2
ex. Emma’s car generator puts out 18 Volts peak. Calculate the RMS voltage.
Vpeak = 18V
VRMS = √2 x Vpeak = 13V
Summary: RMS voltage is the AC voltage that is equivalent to the same DC voltage.
VRMS =
1
2
x VPeak
i.e. 14V peak = 10 VRMS = 10V DC
ex. The mains voltage in NZ is …………… V. Calculate the peak voltage across a heater.
Homework Questions
What does RMS voltage mean? See notes
How does it relate to peak voltage?
Electromagnetism
57
How do you convert RMS voltage to peak voltage and vice versa?
AC in a Resistor
If we compare the current through the resistor
and the voltage across it, we can see that they
are in phase
A
V
=
1
VP
i.e. The current is maximum when the voltage is
maximum
I
VR
eak
AC in a Capacitor
Phase
The ammeter and voltmeter show that the current and voltage are
not in phase
V
=
1
V
A
When the voltage is maximum (capacitor fully charged) the current
is zero.
When the capacitor voltage is minimum (capacitor fully discharged)
the current is maximum.
The current leads the voltage by 90º ~ ¼ cycle
Pe
Experiment note:
ak for both these demos, you need a signal generator that produces slow (0.1Hz)
AC. Use low current output, use galvanometers for A and V . To use the galvanometer as a
voltmeter connect high resistance in series. The value can be chosen by trial and error.
A suitable value for R and C would be 5kΩx 200μF
The graphs can be produced using a computer interface. Voltage and current probes can be used,
or the current graph for the capacitor can be obtained by putting a resistor in series and
connecting a voltage probe across it.
Reactance of a capacitor – 50Hz 6V AC supply. Use a 6V light bulb, add 100μF, 300μF, 1000μF
caps in series.
Electromagnetism
58
Revision
Relate this to what you know about the DC voltage across a capacitor, and the current “through”
it. Sketch graphs of V/t and I/t for a capacitor being charged.
I is maximum when VC = 0
I is zero when VC = maximum
I
VC
t
t
Voltage/Current for a capacitor
I
Vc
Reactance
When we put a resistor in the circuit, the current decreases. It converts electrical energy to
heat. A capacitor also reduces the current, but it doesn’t convert electrical energy to heat. A
capacitor stores the energy that goes into it, then releases it back to the circuit.
We say a capacitor has REACTANCE. This is a measure of how much it opposes the flow of
current. (It is equivalent to resistance and is measured in Ohms).
Electromagnetism
59
For a resistor,
R=
V
I
For a capacitor
Xc = VC/I
The size of the reactance depends on:
1) Capacitance: A smaller capacitor makes the light dimmer (because the reactance is more)
i.e.
Xc∝ 1/C
(Reason: A bigger capacitor can store more charge, so more current
flows into a bigger capacitor every quarter cycle)
Electromagnetism
60
Bigger capacitor
Bigger current
Smaller Reactance
2) Frequency: Increase the frequency of the current and the light brightens ( because the
reactance is less)
i.e.
Xc∝
1/W
(Reason.: If the frequency increases the capacitor charges and
discharges more often so more charge flows on and off in a second,
so the current is more)
Combining the above gives:
Xc =
1
ωC
Homework exercise:
What happens to the AC current in a circuit if a capacitor is put in the circuit?
becomes smaller
What happens to the AC current in a circuit if a smaller capacitor is put in the circuit?
Becomes even smaller
What happens to the current in the above circuit if the frequency of the AC current increases?
becomes greater
NB
DC can’t flow through the circuit (except when the
circuit is initially turned on)
AC can flow through the circuit, so a capacitor can pass
AC and block DC.
What’s the frequency of a DC current? Zero.
What’s the reactance of a capacitor when connected in a DC circuit? Infinite.
Homework questions
What does reactance mean?
See notes
What does the reactance of a capacitor depend on?
Capacitance and frequency of current
Why can’t DC pass through a capacitor, but AC can?
For DC, the frequency is zero, so reactance is infinite
What is the equation for a capacitor’s reactance?
See notes
Electromagnetism
61
What is the phase relationship between current and voltage for a resistor?
See notes
What is the phase relationship between current and voltage for a capacitor?
See notes
Electromagnetism
62
Resistor/Capacitor Circuit (RC)
Measure Vs Vc and VR with a voltmeter.
Vs
VR= 3V Vc
Vc
VR
= 4V
my prediction for
Measured
Vs = ………………..
Vs = 5V
Explain why. ……………………………………………………………………………………………………………………………………………
Experiment Note: A suitable combination is R = 33Ω C = 47μF. By connecting digital multimeters
to R and C the voltages can be measured. (Adjust the frequency so VC and VR are similar.) Then
measure voltage across both.
Reason:
So
Vs
VR and Vc
Label the diagram
are 90º out of phase.
is the vector sum of
i.e.
Vs =
When
VR is zero, Vc is max
VR
Vc and VR
√VC2 + VR2
VC
and when
VR is maximum, Vc is zero
Impedance
Impedance is the total of resistance and reactance. (Symbol Z, Units Ω)
Label the diagram
Impedance Phasors
Voltage Phasors
VR
R
XC
Z
VC
Z = √R
2
+ X C2
Vs =
Electromagnetism
63
VS
Vs
I
ex. A 20 Ω resistor and a variable capacitor are connected in series to a variable frequency power
supply.
(i)
Explain what happens to the current if the capacitance increases.
Xc =
(ii)
Xc =
(iii)
Xc =
1
ωC
C increases therefore XC decreases. I = V/XC
Explain what happens to the current if the frequency of the current increases.
1
w increases therefore
ωC
XC decreases
therefore I increases
The capacitor’s reactance is 15 Ω. Calculate the angular frequency of the current if the
capacitance of the capacitor is 200 F
1
ωC
w = 1/XCC = 330 rads-1
(iv)
Calculate the circuit’s impedance.
Z = 25Ω
(v)
If the supply voltage is 15V, calculate the current.
I = 0.60A
(vi)
37º
therefore I increases
By what angle does the supply voltage lag the current?
Electromagnetism
64
AC in Inductors
Remember that an inductor is a coil that always opposes a change of current.
(It doesn’t affect a constant current)
It produces a back emf whenever the current changes.
Phase
The voltage across the inductor leads the current by
90º When the current is zero (it is changing most
A
V
rapidly) the induced voltage is maximum
=
Experiment Note: 1The easiest way of seeing the phase relationship for the inductor is using a
datalogger with a current
and voltage probe. The primary coil of a large transformer can be used.
VP
(Because of the inductors resistance, the phase relationship is not exactly 90º).
eak
When the current is maximum, (it is not changing) the induced voltage is zero.
I
VL
i.e. Voltage is maximum when
i.e. Voltage is zero when
ΔI
is max
Δt
ΔI
is zero
Δt
At t= 0, the current is increasing rapidly, so the inductor voltage is maximum
At t =
T
,
4
Δ I is momentarily zero, so the inductor voltage is zero
Current lags inductor voltage by 90º ¼ cycle
Electromagnetism
65
Reactance of an Inductor
When an inductor is introduced into an AC circuit, the current reduces because the current is
continually changing and the inductor opposes changing current.
The inductor opposes the flow of current but doesn’t convert electrical energy to heat. (it stores
it in a magnetic Field)
so we say an inductor has reactance measured in Ω
XL =
VL/I (R = VR/I)
Experiment Note: Use a PSSC coil stuffed with iron bars. Remove bars to change the inductance.
The size of the reactance depends on:
(i) Inductance:
Increasing the inductance by putting more iron
into the core decreases the size of the
current so the reactance has decreases
XL ∝ L
(ii) Frequency :
Increasing the frequency of the current reduces the size of the current so the reactance has
increased (this is because the rate of change of current increases)
XL ∝ w
Combining the above gives:
XL = wL
(Compare the reactance of a capacitor,
Xc =
1/wC )
n.b. A “real” inductor has got some “ohmic resistance”. So we can think of it as an ideal inductor
and a resistor in series. (ideal… perfect …no resistance). You can think of it like this:
Electromagnetism
66
Homework Questions
Why can’t you simply add voltages in an RC circuit?
What does the reactance of an inductor depend on?
What is the phase relationship between V and I in an inductor?
What is a “real” inductor?
Why does frequency affect reactance?
What is impedence?
e.g. A real 1.0 mH inductor has a resistance of 1.0 Ω. It carries 10 mA at 1000 rads-1.
Calculate its reactance. 1.0Ω
Calculate its impedence. 1.4Ω
Calculate the voltage across it. 0.014V
Calculate the phase relationship between the current through it and the voltage across it.
45º
Application:
Loudspeakers have big bass speakers called woofers and little treble speakers called tweeters. To
stop high frequency notes going to the woofer, it has an inductor in series. To stop the low
frequencies going to the tweeter it has a capacitor in series.
STEREO
Electromagnetism
67
Extension: An alternative is to have the inductor and capacitor in series. The speakers are
connected to them in parallel with the inductor and the capacitor. Which goes where?
RLC Circuits
When you put an inductor, a capacitor and a resistor all together in a circuit, some curious things
happen.
As the frequency of the current increases:
the reactance of the capacitor decreases
XL
the reactance of the inductor increases
XC
reactance
Z
fo
frequency
The reactance of the inductor and capacitor are always in opposite directions.
At one particular frequency they are equal (and opposite), so they cancel out
When this happens, the current should become maximum (Only the resistance affects it)
VS
Vc
VL
current
VR
fo
frequency
Starting at low frequency, the current steadily rises to a maximum
This is called the resonance frequency. This occurs when the voltage across the capacitor and
inductor is equal and opposite
As the frequency continues to increase, the current decreases
Experiment Note: A resonant circuit can be easily set up using a PSSC coil stuffed with iron bars
(L) a lamp (R) and a capacitor (about 10μF). The current can be monitored via the bulb, or an AC
ammeter can be used.
Electromagnetism
68
Reactance Phasors
Draw the reactance phasors in each example
1. Low Frequency
2. Resonant Frequency
3. High Frequency
XL
XL
XL
∑X
∑X=0
R
∑X
Z
R=Z
XC
R
XC
Z
XC
1. At low frequency, the capacitor has a high reactance this causes the circuit to have a high
impedence and the current is small.
2. At the resonant frequency the reactances of the capacitor and inductor are equal and opposite.
Only the resistor affects the current. The current is maximum because the impedence is minimum
3. At high frequency, the inductor has a high reactance this causes the circuit to have a high
impedence and the current is small.
Resonant Frequency.
A Capacitor, Inductor combination will have a natural or resonant frequency. When the frequency
of the power supply matches the natural frequency maximum current is produced .
At Resonance,
XL = XC
ωL= 1
ω=
1/√LC
ωC
This is like resonance in SHM. The amplitude of a swing is maximum when the driving frequency
matches the natural frequency.
Electromagnetism
69
Homework exercises
What is the phase relationship between the voltages across the capacitor and inductor?
What happens to the reactance of the capacitor and inductor at resonance?
What happens to the circuit current at resonance?
What is meant by resonance?
What is meant by reactance?
What is meant by impedence?
Applications
Metal Detectors
A metal detector usually contains an LRC circuit which is resonating. (i.e. the reactances of the
inductor and capacitor are equal and opposite. The current is maximum. The inductor has air in the
core and this is the detector that you walk through.
When metal passes through the detecting coil
the inductance increases so the circuit
is no longer resonating.
The current will decrease.
Vc
VL
VR
Detecting coil
Radios
The tuning circuit of a radio is an LRC circuit.
Radio waves cause electrons in the aerial and
tuner to oscillate. i.e an alternating
current is produced in the circuit. The current is
maximum when the frequency of the radio waves
(and current)matches the resonant frequency of the
circuit.
Different stations use different frequency
waves. The resonant frequency can be changed by
altering the capacitance.
Increasing the capacitance will reduce the resonant frequency
Electromagnetism
70
aerial
To amplifier
Review questions.
(1)
XC
(i) Sketch graphs of:
Capacitor reactance / frequency
Inductor reactance / frequency
Resistance / frequency
XL
R
(ii)
State what happens to the Capacitor reactance when the frequency increases decrease
(iii)
State what happens to the Inductor reactance when the frequency increases increase
(iv)
Explain what happens at resonance.
Capacitor reactance and inductor reactance are equal and opposite – impedence = resistance. I =
max.
(2) Sketch graphs of:
Current / time
Resistor Voltage / time
Capacitor Voltage / time
Inductor Voltage / time
I
VL
VC
VR
For one cycle of AC in an LCR
circuit
(3)
R = 300 Ω
C = 7.5 µF
L = 1.2 H
f = 50 Hz
Vc
VL
(i)
Calculate the reactance of the capacitor and inductor at this frequency.
XC = 420Ω
XL = 377Ω
(ii)
Calculate the voltage across each component if a current of 1.0 A flows.
VC = 420V
VL = 377V
VR = 300 V
(iii)
Calculate the voltage across:
(a) VL and VR combined V = 481 V
Electromagnetism
71
VR
(b) VL and VC combined 43 V
(c) Vc, VL and VR combined 303 V
(iii)
Calculate the resonant frequency.
ω=
1
= 333
LC
f =
ω
53Hz
2π
(iv)
(Calculate the current at resonance if the supply voltage is 15.0 V
I= 15/300 A
(iii)
What happens to the resonant frequency if the capacitance decreased?
C decreases therefore XC increases and resonant frequency increases W = 1/√LC
(iv)
Explain what happens to the resonant frequency if the inductor’s iron core was
removed
L decreases
R.F. increases
(4)
An RC circuit contains a 100 Ω resistor and a capacitor connected to a 141 V, 50 Hz
supply. A current of 1.0 A rms flows.
(i)
Calculate the impedence of the circuit.
Z = V/I = 141Ω
(ii)
Draw a resistance/reactance/impedence phasor diagram
R
XC
(iii)
Calculate the reactance of the capacitor.
100Ω
(iv)
Calculate the capacitance of the capacitor.
3.0 x 10-4 F
(v)
Calculate the phase angle between the supply voltage and the current.
Electromagnetism
72
Z
45º
(vi)
What happens to the reactance if the frequency increases?
XC = 1/WC XC decreases because capacitor is charging and discharging more rapidly ~ △Q is
greater
(vii)
What happens to the current if the frequency increases?
I = VC/XC XC decreases so current
increases
R
φ
Z
XC
(viii)
What happens to the capacitor reactance if the frequency increases?
decreases
(ix)
What happens to the phase angle if the frequency increases?
Decreases
Electromagnetism
73
Electromagnetism
74