P2 Physics P2.3.2 Electrical Circuits Ks4 Additional Science Mr D Powell

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 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 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

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), A Q is the charge in coulombs, C t 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, V W is the work done in joules, J

a) Current flow...

C 1.9 x 10-19 C x 6.25 x 1018 electrons = 1C 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 e- 1.9 x 10-19 C x 6.25 x 1018 electrons = 1C

a) What is an electric current.... M a) What is an electric current.... When 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.. C B A

a) Q = It 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

Potential Difference Theory Potential Difference: is defined as the work done (or energy transfer) per unit charge V (Volt) = W (work done, J) Q (charge, C) + B If 1J of work is done in moving 1 C of positive charge from A to B then the Pd is 1V + A 1V = 1 J / C

E = QV E = 5C x 3000V = 15000J or E = VIt E = 3V x 2A x 3s E = 18J b) V = W/Q - Example 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 3s E = 18J

a& b) Circuits Theory Summary..... Voltage 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), A Q is the charge in coulombs, C t 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, V W is the work done in joules, J 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), A Q is the charge in coulombs, C t 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, V W is the work done in joules, J 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), A Q is the charge in coulombs, C t 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, V W is the work done in joules, J

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, V I is the current in amperes (amps), A R is the resistance in ohms, 

f/g/h ) Investigating Components Build 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; Resistor 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

f/g/h ) Graphing Resistance M f/g/h ) Graphing Resistance Now 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...... Worked example The current through a wire is 2.0 A when the potential difference across it is 12V. Solution X- Axis Y-Axis Voltage Current Length Resistance

f/g/h ) Analysis Look at the data on the this graph and answer the following questions... When the p.d. Is 1V then 2V what are the corresponding currents? What can you say about the relationship between current and voltage? What does the gradient of the graph represent? Can you work out the resistance of this wire?

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, V I is the current in amperes (amps), A R 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, V I is the current in amperes (amps), A R 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, V I is the current in amperes (amps), A R is the resistance in ohms, 

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.

Circuit Symbols Candidates will be required to interpret and draw circuit diagrams. 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

Simple Circuits S Electrical 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. Can you draw a circuit diagram with the proper symbols (use a ruler) Can you remember any special rules about the potential difference or current from you previous studies?

What is the symbol? P 1 2 3 4 5

Symbols II 6 7 8 9 10

Low resistance, connect in series cell Low resistance, connect in series Indicator / light source No more semi circles! resistor High resistance, connect in parallel thermistor Electric motor Resistance falls as temp rises ammeter Resistance falls as light level rises Light emitting diode Emits light when forward biased Light dependent resistor heater of a specific value voltmeter Conducts when forward biased diode Many cells = a battery Variable resistor

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

Q=It Q = 0.16A x 10s = 1.6Cs-1 V=IR V/I= 4V/0.16A = 25

D 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

V=IR 8V/1.85A = 4.32 Or 2V/0.825A = 2.42 If v+/-0.1 V what is % error for each voltage reading 8V -> 1.25%, 2V -> 5%

D O & n) Diode The 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.

p) LDR 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!

q) Thermistor The 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!

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. What happens if the slider is moved back again? 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?

P Summary Questions Cell Switch Bulb 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.

I like to live in a low pressure argon atmosphere I am made of metal 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

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

greater than less than the same as 1 Series Questions 1) A cell, a resistor, a lamp and an ammeter are connected in series; The current through the battery is …………………… the current through the ammeter. The potential difference across the battery is …………………. the potential difference across the resistor. The current through the lamp is ……………… the current through the resistor. The potential difference across the lamp is ……………… the potential difference across the battery. greater than less than the same as

Series PD 3 V 4 3 V 4 V V V A 3.0 V 0.1 A In a series circuit the PD is shared among the components In a series circuit the current is the same everywhere V V 3 V 4 3 V 4

2 Potential Difference Series..... 2) 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. Calculate the potential difference across the resistor? Calculate the potential difference across the lamp? Energy / PD Rule In a series circuit the energy is shared between components V = IR Answers V = IR so 0.2A x 8  = 1.6V 3V – 1.6V = 1.4V

3 Potential Difference Series..... 3) 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: the 10  resistor. the 15  resistor. the battery. 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 V = IR so 0.36A x 10  = 3.6V V = IR so 0.36A x 15  = 5.4V Vs = 3.6V + 5.4V = 9V V=IR or V/I = R 9V/0.36A = 25 

4 Potential Difference Series..... 4) A 6.0 V battery, a 10  resistor and a 20  resistor are connected in series with each other. Draw out a circuit diagram Calculate the total resistance of the two resistors in series. Calculate the current in the circuit. Calculate the potential difference across the 20  resistor. Answers All connected in series RT = R1 + R2 = 10 +20  = 30  V=IR or V/R = I I = 6V / 30  = 0.2A V = IR , 0.2A x 20  = 4V

Resistors in Parallel Rules 5 Parallel Questions..... A 2) 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. Show that the current through the resistor is 0.4 A. The ammeter reading is 0.9 A. Calculate the current through the lamp. 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 V = IR so V/R = I so 2V /5  = 0.4A 0.9A – 0.4A = 0.5A V = IR so V/I = R 2V/0.5A = 4 

A 6 Parallel Questions..... the 10  resistor. the 15  resistor. 3) A 12 V battery is connected to a 10  resistor in parallel with a 15  resistor, as shown. Calculate the current through: the 10  resistor. the 15  resistor. the battery. 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 V = IR so V/R = I so 12V /10  = 1.2A V = IR so V/R = I so 12V /15  = 0.8A 0.8A + 1.2A = 2 A V = IR , V/I = R 12V / 2A = 6  Or 1/R T=1/10+1/15 = 0.167 RT = 6 

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

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.