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P2.3.2 Electrical Circuits
P2 Physics
KS4 ADDITIONAL SCIENCEMr D Powell
Mr Powell 2012Index
Connection• Connect your learning to the
content of the lesson• Share the process by which the
learning will actually take place• Explore the outcomes of the
learning, emphasising why this will be beneficial for the learner
Demonstration• Use formative feedback – Assessment for
Learning• Vary the groupings within the classroom
for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex
• Offer different ways for the students to demonstrate their understanding
• Allow the students to “show off” their learning
Activation• Construct problem-solving
challenges for the students• Use a multi-sensory approach – VAK• Promote a language of learning to
enable the students to talk about their progress or obstacles to it
• Learning as an active process, so the students aren’t passive receptors
Consolidation• Structure active reflection on the lesson
content and the process of learning• Seek transfer between “subjects”• Review the learning from this lesson and
preview the learning for the next• Promote ways in which the students will
remember• A “news broadcast” approach to learning
Mr Powell 2012Index
P2.3.2 Electrical circuits (Part A)
a) Electric current is a flow of electric charge. The size of the electric current is the rate of flow of electric charge. The size of the current is given by the equation:
I = Q/t (or Q = It )
I is the current in amperes (amps), AQ is the charge in coulombs, Ct is the time in seconds, s
b) The potential difference (voltage) between two points in an electric circuit is the work done (energy transferred) per coulomb of charge that passes between the points.
V = W/Q
V is the potential difference in volts, VW is the work done in joules, JQ is the charge in coulombs, C
Mr Powell 2012Index
a) Current flow...
Mr Powell 2012Index
a) Coulombs
Electrons are charged particles and each of them have a charge of 1.9 x 10-19 C.
It is a simple property which cannot be removed or changed. It is a useful to us as charges make particles move i.e. opposites attract.
If we add a load of them together and think of them as a single “sphere of charge” or ball we get a whole coulomb of charge and can think about defining the ampere or amp 1A = 1C/s
C
1.9 x 10-19 C x 6.25 x 1018 electrons = 1C
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Mr Powell 2012Index
a) What is an electric current.... MWhen a torch lamp is on, millions of electrons pass through it every second. The electric current through the lamp is due to electrons passing through it. Each electron carries a tiny negative charge. The rate of flow of electrical charge is called the current. The filament of the torch lamp is a fine metal wire. Metals conduct electricity because they contain conduction (or sea of delocalised) electrons. These electrons move about freely inside the metal. They are not confined to a single atom. When the torch is switched on, the battery pushes electrons through the filament. Insulators can’t conduct electricity because all the electrons are held in atoms.
TASK: explain what an electric current is. Why do metals conduct and plastics do not?
draw an atomic structure diagram to help you compare and model the idea..
B
C
A
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When electrons move through a wire we call it an electrical current. The electrons move as there is a potential difference.
The larger the p.d. the higher the current flow or Coulombs per second.
1A = 1Cs-1
A simple graph of this process would be where a steady current has flowed for 20s seconds;
The number of Coulombs of charge that have flowed is 100C
a) Q = It
Mr Powell 2012Index
Potential Difference: is defined as the work done (or energy transfer) per unit charge
V (Volt) = W (work done, J) Q (charge, C)
B
A +
+ If 1J of work is done in moving 1 C of positive charge from A to B then the Pd is 1V
1V = 1 J / C
Potential Difference Theory
Mr Powell 2012Index
This process whereby ions exchange electrons through a molten liquid or dissolved solid is also a way in which a “current” flows. Also a PD between ground and cloud causing a spark to move.
As charge carriers are moving. Hence we can say that energy converted is;
E = QV
E = 5C x 3000V = 15000J
or
E = VIt
E = 3V x 2A x 3sE = 18J
b) V = W/Q - Example
Mr Powell 2012Index
a& b) Circuits Theory Summary..... CVoltage is really a measure of the difference of energy before and after a component. It is measured in volts, symbol V. 1 Volt is equal to 1 joule of energy for each coulomb of charge which passes through the circuit. A voltmeter is connected in parallel with a component.
1V = 1J/C
Current is the flow of groups of electrons. One group has a coulomb of charge (since each electron has a little bit of charge). We call a flow of one coulomb per second an ampere, amp, symbol A. An ammeter is connected in series.
1A = 1C/s
P2.3.2 Electrical circuits (Part A)
a) Electric current is a flow of electric charge. The size of the electric current is the rate of flow of electric charge. The size of the current is given by the equation:
I = Q/t (or Q = It )
I is the current in amperes (amps), AQ is the charge in coulombs, Ct is the time in seconds, s
b) The potential difference (voltage) between two points in an electric circuit is the work done (energy transferred) per coulomb of charge that passes between the points.
V = W/Q
V is the potential difference in volts, VW is the work done in joules, JQ is the charge in coulombs, C
P2.3.2 Electrical circuits (Part A)
a) Electric current is a flow of electric charge. The size of the electric current is the rate of flow of electric charge. The size of the current is given by the equation:
I = Q/t (or Q = It )
I is the current in amperes (amps), AQ is the charge in coulombs, Ct is the time in seconds, s
b) The potential difference (voltage) between two points in an electric circuit is the work done (energy transferred) per coulomb of charge that passes between the points.
V = W/Q
V is the potential difference in volts, VW is the work done in joules, JQ is the charge in coulombs, C
P2.3.2 Electrical circuits (Part A)
a) Electric current is a flow of electric charge. The size of the electric current is the rate of flow of electric charge. The size of the current is given by the equation:
I = Q/t (or Q = It )
I is the current in amperes (amps), AQ is the charge in coulombs, Ct is the time in seconds, s
b) The potential difference (voltage) between two points in an electric circuit is the work done (energy transferred) per coulomb of charge that passes between the points.
V = W/Q
V is the potential difference in volts, VW is the work done in joules, JQ is the charge in coulombs, C
Mr Powell 2012Index
P2.3.2 Electrical circuits (Part B)
f) The resistance of a component can be found by measuring the current through, and potential difference across, the component.
g) The current through a resistor (at a constant temperature) is directly proportional to the potential difference across the resistor.
h) Calculate current, potential difference or resistance using the equation:
V = IR
V is the potential difference in volts, VI is the current in amperes (amps), AR is the resistance in ohms,
Mr Powell 2012Index
f/g/h ) Investigating Components PBuild the circuit as shown on the slide.
It may be that the switch and variable resistor is inside the power pack.
Investigate how the resistance of a;
1. Resistor2. Filament lamp
Changes as you change the potential difference of the power pack from -6V -> 0 -> 6V in steps of 1V
Draw a graph to show your results (line graph with 0,0 in the centre of the page.
Voltage (V) Current (A) Resistance (Ω)
-6-4......+6
I
VR
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f/g/h ) Graphing Resistance MNow you have your raw data work out the resistance of the device by the following formula;
R = resistance (Ohms - )I = Current (Amperes - A)V = potential difference (Volts - V)
Split a sheet of graph paper into 4. Then construct and comment on these graphs......
I
VR
X- Axis Y-Axis
Voltage Current
Current Voltage
Length Resistance
Worked example
The current through a wire is 2.0 A when the potential difference across it is 12V.
Solution
Mr Powell 2012Index
f/g/h ) Analysis
Look at the data on the this graph and answer the following questions...
1) When the p.d. Is 1V then 2V what are the corresponding currents?
2) What can you say about the relationship between current and voltage?
3) What does the gradient of the graph represent?
4) Can you work out the resistance of this wire?
Mr Powell 2012Index
f/g/h ) Summary Questions P
P2.3.2 Electrical circuits (Part B)
f) The resistance of a component can be found by measuring the current through, and potential difference across, the component.
g) The current through a resistor (at a constant temperature) is directly proportional to the potential difference across the resistor.
h) Calculate current, potential difference or resistance using the equation:
V = IR
V is the potential difference in volts, VI is the current in amperes (amps), AR is the resistance in ohms,
P2.3.2 Electrical circuits (Part B)
f) The resistance of a component can be found by measuring the current through, and potential difference across, the component.
g) The current through a resistor (at a constant temperature) is directly proportional to the potential difference across the resistor.
h) Calculate current, potential difference or resistance using the equation:
V = IR
V is the potential difference in volts, VI is the current in amperes (amps), AR is the resistance in ohms,
P2.3.2 Electrical circuits (Part B)
f) The resistance of a component can be found by measuring the current through, and potential difference across, the component.
g) The current through a resistor (at a constant temperature) is directly proportional to the potential difference across the resistor.
h) Calculate current, potential difference or resistance using the equation:
V = IR
V is the potential difference in volts, VI is the current in amperes (amps), AR is the resistance in ohms,
Mr Powell 2012Index
P2.3.2 Electrical circuits (Part C)c) Circuit diagrams using standard symbols (see next slide):
d) VI graphs are used to show how the current through a component varies with the potential difference across it.
e) The VI graphs for a resistor at constant temperature.
m) The resistance of a filament bulb increases as the temperature of the filament increases. (explain resistance change in terms of ions and electrons.)
o) & n) The current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction. LED as example turn on in forwards direction only.
p) The resistance of a light-dependent resistor (LDR) decreases as light intensity increases.
q) The resistance of a thermistor decreases as the temperature increases.
Mr Powell 2012Index
Circuit Symbols
Candidates will be required to interpret and draw circuitdiagrams.
Knowledge and understanding of the use of thermistors in circuits e.g. thermostats is required.
Knowledge and understanding of the applications of light-dependent resistors (LDRs) is required, eg switching lights on when it gets dark
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Simple Circuits SElectrical circuits of many different types are found around us in nearly every device we have in the home, at work and in school.
We can draw a picture like this for each one but a real circuit diagram is a very helpful way of showing how the components in a circuit are connected together.
Each component has its own symbol which makes them simple to understand when they get complex.
1) Can you draw a circuit diagram with the proper symbols (use a ruler)
2) Can you remember any special rules about the potential difference or current from you previous studies?
Mr Powell 2012Index
1
2
3
4
5
What is the symbol? P
Symbols II6
7
8
9
10
ammeter
voltmeter
cell
Indicator / light source
diode
Light emitting diode
resistor
Variable resistor
thermistor
Light dependent resistor
heater
Electric motor
Many cells = a battery
No more semi circles!
of a specific value
Resistance falls as temp rises
Low resistance, connect in series
High resistance, connect in parallel
Emits light when forward biased
Resistance falls as light level rises
Conducts when forward biased
Mr Powell 2012Index
e) Resistor
As the voltage increases the current also increases at the same rate. The resistance is constant. (Directly Proportional)
This is what is called “ohms law”
True only for a resistor at a constant temperature
Mr Powell 2012Index
Mr Powell 2012Index
m) Filament Lamps
The resistance of a filament lamp increases as the temperature of the filament increases. The atoms get very hot and vibrate so slow the electrons down.
The resistance & gradient changes as the temperature of the wire changes
D
Mr Powell 2012Index
Mr Powell 2012Index
O & n) Diode DThe current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction.
Often used in mobile phone transformers to change A.C. to D.C. currents.
Mr Powell 2012Index
The resistance of a light-dependent resistor (LDR) decreases as light intensity increases. This is weird as it is the opposite of what you might expect of a normal resistor. The light frees the electrons.
You might see two types of graph one normal and one in logarithmic form to make a trend easier to see!
p) LDR
Mr Powell 2012Index
q) ThermistorThe resistance of a thermistor decreases as the temperature increases. So the extra heat allows electrons to flow.
This is weird and opposite to normal resistors!
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Simple Circuits Practical
Connect a variable resistor in series with the torch lamp and a battery, as shown in the diagram. Adjusting the slider of the variable resistor alters the amount of current flowing through the bulb and therefore affects its brightness. The torch lamp goes dim when the slider is moved one way. (You can add an ammeter to check the flow.)
We can also control currents to only allow them to flow one way. The diode prevents flow in one direction to protect the radio if the cell it put in the wrong way.
1) What happens if the slider is moved back again?
2) What happens if you have two bulbs in your circuit?
2) What happens if you include a diode in the circuit, try it both ways around?
4) Why might you put a diode in a radio circuit?
Mr Powell 2012Index
Summary Questions P
1. Cell2. Switch3. Bulb4. Fuse
Diode should have the arrow in the direction of pos to neg
Variable resistor or fixed resistor
P2.3.2 Electrical circuits (Part C)
c) Circuit diagrams using standard symbols (see next slide):
d) VI graphs are used to show how the current through a component varies with the potential difference across it.
e) The VI graphs for a resistor at constant temperature.
m) The resistance of a filament bulb increases as the temperature of the filament increases. (explain resistance change in terms of ions and electrons.)
o) & n) The current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction. LED as example turn on in forwards direction only.
p) The resistance of a light-dependent resistor (LDR) decreases as light intensity increases.
q) The resistance of a thermistor decreases as the temperature increases.
P2.3.2 Electrical circuits (Part C)
c) Circuit diagrams using standard symbols (see next slide):
d) VI graphs are used to show how the current through a component varies with the potential difference across it.
e) The VI graphs for a resistor at constant temperature.
m) The resistance of a filament bulb increases as the temperature of the filament increases. (explain resistance change in terms of ions and electrons.)
o) & n) The current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction. LED as example turn on in forwards direction only.
p) The resistance of a light-dependent resistor (LDR) decreases as light intensity increases.
q) The resistance of a thermistor decreases as the temperature increases.
P2.3.2 Electrical circuits (Part C)
c) Circuit diagrams using standard symbols (see next slide):
d) VI graphs are used to show how the current through a component varies with the potential difference across it.
e) The VI graphs for a resistor at constant temperature.
m) The resistance of a filament bulb increases as the temperature of the filament increases. (explain resistance change in terms of ions and electrons.)
o) & n) The current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction. LED as example turn on in forwards direction only.
p) The resistance of a light-dependent resistor (LDR) decreases as light intensity increases.
q) The resistance of a thermistor decreases as the temperature increases.
Mr Powell 2012Index
Starter: What am I...
I like to live in a low pressure argon atmosphere
I am made of metal
I can conduct electricity easily
When I am skinny I like to resist current
I got hot in the right conditions
I glow a lot in the right conditions
I am a very curly
Mr Powell 2012Index
P2.3.2 Electrical circuits (Part D)
i) The current through a component depends on its resistance. The greater the resistance the smaller the current for a given p.d. across the component.
j) The p.d. provided by cells connected in series adds up.
k) For components connected in series: the resistance adds up. there is the same current through each component the total p.d. is shared
I) For components connected in parallel: the p.d. across each component is the same The current splits at branches
Mr Powell 2012Index
1 Series Questions A1) A cell, a resistor, a lamp and an ammeter are connected in series;
a) The current through the battery is …………………… the current through the ammeter.
b) The potential difference across the battery is …………………. the potential difference across the resistor.
c) The current through the lamp is ……………… the current through the resistor.
d) The potential difference across the lamp is ……………… the potential difference across the battery.
greater than less than the same as
Mr Powell 2012Index
A
V V
3.0 V V
V V
3 V4
3 V4
3 V4
3 V4
0.1 A
Series PD
In a series circuit the current is the
same everywhere
In a series circuit the PD is shared
among the components
Mr Powell 2012Index
2 Potential Difference Series..... A2) the cell has a potential difference of 3.0 V and the resistor has a resistance of 8.0 . The ammeter reading is 0.2 A.
a) Calculate the potential difference across the resistor?
b) Calculate the potential difference across the lamp?
Energy / PD Rule
In a series circuit the energy is shared between components
V = IR
Answers
a) V = IR so 0.2A x 8 = 1.6V
b) 3V – 1.6V = 1.4V
Mr Powell 2012Index
3 Potential Difference Series..... A3) A battery, an ammeter, a 10 resistor and a 15 resistor are connected in series. The ammeter reading is 0.36 A. Calculate the potential difference across:
a) the 10 resistor.
b) the 15 resistor.
c) the battery.
d) What would be the resistance of a single resistor that would have the same current if it was connected on its own to the same battery?
Answers
a) V = IR so 0.36A x 10 = 3.6V
b) V = IR so 0.36A x 15 = 5.4V
c) Vs = 3.6V + 5.4V = 9V
d) V=IR or V/I = R 9V/0.36A = 25
Mr Powell 2012Index
4 Potential Difference Series..... A4) A 6.0 V battery, a 10 resistor and a 20 resistor are connected in series with each other.
a) Draw out a circuit diagram
b) Calculate the total resistance of the two resistors in series.
c) Calculate the current in the circuit.
d) Calculate the potential difference across the 20 resistor.
Answers
a) All connected in series
b) RT = R1 + R2 = 10 +20 = 30
c) V=IR or V/R = I I = 6V / 30 = 0.2A
d) V = IR , 0.2A x 20 = 4V
Mr Powell 2012Index
5 Parallel Questions..... A2) the cell has a potential difference of 2.0 V and the resistor has a resistance of 5.0 . The ammeter reading is 0.9 A.
a) Show that the current through the resistor is 0.4 A.
b) The ammeter reading is 0.9 A. Calculate the current through the lamp.
c) Calculate the resistance of the lamp in this circuit.
Resistors in Parallel Rules
Each branch has P.D. of cell
Current splits at branches according to resistance
Answers
a) V = IR so V/R = I so 2V /5 = 0.4A
b) 0.9A – 0.4A = 0.5A
c) V = IR so V/I = R 2V/0.5A = 4
Mr Powell 2012Index
6 Parallel Questions..... A3) A 12 V battery is connected to a 10 resistor in parallel with a 15 resistor, as shown. Calculate the current through:
a) the 10 resistor.
b) the 15 resistor.
c) the battery.
d) What would be the resistance of a single resistor that would have the same current if it was connected on its own to the same battery?
Answers
a) V = IR so V/R = I so 12V /10 = 1.2A
b) V = IR so V/R = I so 12V /15 = 0.8A
c) 0.8A + 1.2A = 2 A
d) V = IR , V/I = R 12V / 2A = 6
Or 1/R T=1/10+1/15 = 0.167 RT = 6 21
111
RRRT
P2.3.2 Electrical circuits (Part D)
i) The current through a component depends on its resistance. The greater the resistance the smaller the current for a given p.d. across the component.
j) The p.d. provided by cells connected in series adds up.
k) For components connected in series: the resistance adds up. there is the same current through each
component the total p.d. is shared
I) For components connected in parallel: the p.d. across each component is the
same The current splits at branches
P2.3.2 Electrical circuits (Part D)
i) The current through a component depends on its resistance. The greater the resistance the smaller the current for a given p.d. across the component.
j) The p.d. provided by cells connected in series adds up.
k) For components connected in series: the resistance adds up. there is the same current through each
component the total p.d. is shared
I) For components connected in parallel: the p.d. across each component is the
same The current splits at branches
P2.3.2 Electrical circuits (Part D)
i) The current through a component depends on its resistance. The greater the resistance the smaller the current for a given p.d. across the component.
j) The p.d. provided by cells connected in series adds up.
k) For components connected in series: the resistance adds up. there is the same current through each
component the total p.d. is shared
I) For components connected in parallel: the p.d. across each component is the
same The current splits at branches
P2.3.2 Electrical circuits (Part D)
i) The current through a component depends on its resistance. The greater the resistance the smaller the current for a given p.d. across the component.
j) The p.d. provided by cells connected in series adds up.
k) For components connected in series: the resistance adds up. there is the same current through each
component the total p.d. is shared
I) For components connected in parallel: the p.d. across each component is the
same The current splits at branches
Mr Powell 2012Index
Pracs / Demos
Suggested ideas for practical work to develop skills and understanding include the following:
using filament bulbs and resistors to investigate potential difference/current characteristics
investigating potential difference/current characteristics for LDRs and thermistors
setting up series and parallel circuits to investigate current and potential difference
plan and carry out an investigation to find the relationship between the resistance of thermistors and their temperature
investigating the change of resistance of LDRs with light intensity.