Presentation on theme: "3.3.1 Series and Parallel Circuits – Kirchhoff’s second law"— Presentation transcript:
1 3.3.1 Series and Parallel Circuits – Kirchhoff’s second law 3 DC CircuitsG482 Electricity, Waves & Photons3.3.1 Series and Parallel Circuits – Kirchhoff’s second law3.3.2 Practical CircuitsKs5 OCR Physics H158/H558IndexMr Powell 2012
2 Practical Notes....Practical Skills are assessed using OCR set tasks. The practical work suggested below maybe carried out as part of skill development. Centres are not required to carry out all of theseexperiments.In this module, there is much potential for experimental work and improving instrumentation skills.Use a multimeter as an ohmmeter to ‘verify’ the rules for total resistance for series and parallelcircuits.Use a calibrated light-meter to plot the variation of resistance of an LDR against intensity.Determine the internal resistance of a chemical cell.Use a potential divider circuit to show the validity of the potential divider equation inDesign a light-sensing circuit based on a potential divider with a light-dependent resistor.Design a temperature-sensor based on a potential divider with a thermistor.Monitor the output potential difference from either light-sensors or temperature-sensors using adata-logger.
3 Circuit RulesFor resistance calculations at AS you can use the premise of the following rules but they are not sorted can you do this…..
7 Resistors in Parallel I RTITI2R2VI3R3VResistors in series can be replaced by one single resistor:But V = IR and I = VRIT =III3V =VVVRTR1R2R3V =VVVRTR1R2R31111=++RTR1R2R3
8 Generalised FormulaeSo all this leads us to a generalised formulae for any number of resistance.In Series resistance is simple but for Parallel resistances we must use the reciprocal.NB: when multiplying out the second formula make sure you treat all terms in the same way!
9 Test it out….If all the bulbs are of equal resistance answer the following;Find all the meter readingsWhat is the total resistance of the bottom branchIf I swapped all three bulbs for one bulb what resistance should it be so that the current flow in the main branch is still 3A (i.e. combine all resistances using formulae)
10 Plenary Question…. Answer…. A battery of e.m.f 12 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram.Calculate the total resistance of the circuit. (3)Calculate the current through the 30Ω resistor. (1)1) (three parallel resistors) (1)R = 9.23Ω (1)R = 9.2Ω9.2 Ω and 30 Ω in series gives 39Ω (1)(allow e.c.f. from value of R)30Ω2) (V = IR gives) 12 = I × 39I = 0.31 A (1)(allow e.c.f. from (a))Basic
11 Plenary Question…. Answer…. A battery of e.m.f 12 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram.Calculate the total resistance of the circuit. (3)Calculate the current through the 30Ω resistor. (1)30ΩBasic
12 3.3.1 Series and Parallel Circuits – Kirchhoff’s second law – Part I Assessable learning outcomesstate Kirchhoff’s second law and appreciate that this is a consequence of conservation of energy;apply Kirchhoff’s first and second laws to circuits;select and use the equation for the total resistance of two or more resistors in series;select and use the equation for the total resistance of two or more resistors in parallel;solve circuit problems involving series and parallel circuits with one or more sources of emfAn example of more than one source of e.m.f. Is a battery charger. (HSW 6a)Students can verify the rules for resistors experimentally.
13 b) Kirchoffs Law IThe “current law” states that at a junction all the currents should add up.I3 = I1 + I2 or I1 + I2 - I3 = 0Current towards a point is designated as positive.Current away from a point is negative.In other words the sum of all currents entering a junction must equal the sum of those leaving it.Imagine it like water in a system of canals!
14 b) Kirchoffs Law I Examples; If I1 = 0.1A, I2 = 0.2A, I3 = 0.3A There are some important multipliers for current:1 microamp (1 A) = 1 x 10-6 A 1 milliamp (mA) = 1 x 10-3 AAlso remember to make sure you work out current in Amps and time in seconds in your final answers!
15 b) Kirchoffs Law I – Questions? Work out the currents and directions missing on these two junctions?7A3A
16 a) Simple e.m.f example If I = 100 mA what is that in amps? 100 mA = 0.1 A2) What is the current in each resistor?0.1 A3) Work out the voltage across each resistor.V = IR soV1 = 0.1A x 30 = 3 VV2 = 0.1A x 40 = 4 V ;V3 = 0.1A x 50 = 5V4) What is the total resistance?RT = 30 + 40+ 50 = 1205) What is the battery voltage?V = 0.1A x 120 = 12V
18 a) Kirchhoff's Voltage Law For any complete loop of a circuit, the sum of the emfs equal the sum of potential drops round the loop.This follows from the law of conservation of energy:The total energy per coulomb produced = the total energy per coulomb delivered
19 If the variable resistor is adjusted so that there a) PD Rules - Parallelbattery pd = 12VThis means:each coulomb of chargeleaves the batterywith 12J of electricalenergyIf the variable resistoris adjusted so that thereis a pd of 4V across itThis means:each coulomb of chargeuses 4J of energypassing through it
20 a) Kirchoffs Law IIThe “voltage law” states that the sum of e.m.f’s around a circuit or loop is equal to zero. i.e. all energy is transferred before you return back to the cell.In other words the sum of all voltage sources must equal the sum of all voltages dropped across resistances in the circuit, or part of circuit.Think of it like walking around a series of hills and returning back to point of origin – you are then at the same height!For more complex examples we must note the following rules;There is a potential rise whenever we go through a source of e.m.f from the – to the + side.There is a potential fall whenever we go through a resistance in the same direction as the flow of conventional current. i.e. + to -NB. Both laws become obvious when you start applying them to problems. Just use these sheets as a reference point.
21 e) Kirchoff Laws – More Complex Questions Given the following circuit can you pick out how the current might behave using kirchoffs current laws. We are looking to find the current in the main branch or 0.75 resistor and also through the 1 resistorYou may want to redraw the circuit then apply simple ideas of additive resistance and ratios to find out the currents?
22 e) Kirchoff Law I - Answers In detail this means can rewrite the circuit and fill in the values for V and I as such…Energy (p.d.) is shared simply according to resistance.
23 Extension… if current = 0.16A what is R1 c) Plenary Question….A battery of emf 24 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram in the diagram below.Show that the resistance of the single equivalent resistor that could replace the four resistors between the points A and B is 50Ω.Extension… if current = 0.16A what is R1Answer….BASIC
24 Extension… if current = 0.16A what is R1 c) Plenary Question….A battery of emf 24 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram in the diagram below.Show that the resistance of the single equivalent resistor that could replace the four resistors between the points A and B is 50Ω.Extension… if current = 0.16A what is R1Answer….BASIC
25 e) Answers......This example is more simple than it looks. In fact you have one resistor on its own (0.75)Then the other three are in parallel with each other. With the 1 on one branch and the two 1.5 resistors on the other.You can then simply use resistance ratios to determine the current flow.The ratio for the parallel part is 1:3 so we say that the least current flows through the most resistive part. Hence: current through 1 resistor must be 0.75A.The current through main branch must be the sum of these i.e. 1A
26 e) Complex ExamplesHint1: This is complex and you must try and be consistent in your calculations in direction and which way the current is flowing or p.d. is lost!If we apply Kirchhoffs laws (previous slide) about current we can say that;I1 = I2 + I3 or 0 = I2 + I3 - I1If we apply Kirchhoffs laws (previous slide) about pd = 0 in closed loop we can say the following two thingsStarting at Point “P” and going clockwise around the left-hand loop;-I3R2 + E1 - I1R1 = 0Starting at Point “P” and going clockwise around the right-hand loop;-E2 - I2R3 + I3R2 = 0Hint2: Solve this using Simultaneous Eq method. You will not be asked to do anything so complex in the real exam!R1I2PI3E2E1R2I1R3EXTENSION WORK
27 e) Complex Examples Hint: consider only E direction not I’s Starting at Point “P” and going clockwise around the left-hand loop;-I3R2 + E1 - I1R1 = 0Starting at Point “P” and going clockwise around the right-hand loop;- E2 - I2R3 +I3R2 = 0p.d. is lowered with flow I1PD against flow of currentR1I2PI3E2E1R2p.d. is increased with flow I1I1R3PD against flow of currentp.d. down with flow I3 left loopp.d. is up with flow I3 right loopEXTENSION WORK
28 e) Complex question…. EXTENSION WORK Use the theory from the previous slide to answer the question below working out the current flow we have called I3.Hint use the technique shown on the previous slide. But try and reason it out yourself with the rules you have been given. This may take some time!Your answer should include the following;Reference to the rules of current flow & p.dExplanation as to why each contribution is + or –Equation for each loopAnswer for I3E1 = 6VE2 = 2VR1=10R2=10 R3=2 R1I2PI3E2E1R2I1R3Answer: I3 = 0.4A
29 e) Answers to question.... P E1 E2 R1 R2 I2 R3 I3 I1 Starting at Point “P” and going clockwise around the left-hand loop;-I3R2 + E1 - I1R1 = 0 Eq 1-I310 + 6V - I110 = 0I3+I2 = -6V/10I3+I2 = -0.6A Eq 2 or I3 = (0.6A-I2)Starting at Point “P” and going clockwise around the right-hand loop;-E2 - I2R3 +I3R2 = 0 Eq 3-2V - I210 +I310 = 0Sub Eq 2 into 2 to eliminate I3-2V - I210 + (0.6A-I2) 10 = 04V/20 = I2 = 0.2AHence – feed back into Eq2 to yield I3 = 0.4AAnswer: I3 = 0.4AE1 = 6VE2 = 2VR1=10R2=10 R3=2
30 e) Complex example.. EXTENSION WORK Kirchoffs 1st Law; I1 = I2 + I3 Kirchoffs 2nd law;Sum PD Loop AEDBA30V = 20I3 + 5I1 - Eqn 1Sum PD Loop FEDCF10V = 20I3 - 10I210V = 20I3 - 10(I1 - I3)10V = 30I3 - 10I1 - Eqn 2Add 2 x Eqn 1 + Eqn 270V = 70 I3I3 = 1 ASubstitute this in Eqn 1I1 = 2A so I2 = 1 AThis worked example relies on two equations found from two loops. Each defined for a separate power source.Solve simultaneously to find I’sEXTENSION WORK
33 3.3.1 Series and Parallel Circuits – Kirchhoff’s second law – Part II Assessable learning outcomesexplain that all sources of e.m.f. have an internal resistance;explain the meaning of the term terminal p.d.;select and use the equations e.m.f. = I (R + r), and e.m.f. = V + Ir .
35 Modelling Electricity One idea to help us explain electricity is to think of a electricity like gravitational potential energy. When you are up high you have lots of it.We can even use diagrams to help us to understand what is going on...transfer to heat energy in resistorStored chemical energytransfer to heat energy in wires
36 What is Internal Resistance.... = I(R + r) 6V = I(4 +8)I = 0.5AWe also know that;Vload = - IrVload = 6V - 0.5A x 4Vload = 4VSo 4 Volts is measured across the terminals on the cell.Let us picture a circuit with a real chemical power source;We expect 6V or 6JC-1 from a cell or the e.m.f. (electromotive force) but because our power source is real some of this energy will be lost inside the cell. Through resistance “r”The energy is converted into heat when a current is actually drawn from the cell through a circuit.We can use our R1+R2 = RT formula to work out the emf .For example if the internal resistance r = 4 and the load resistance R = 8. By using our formula;
37 What is EMF?The term electromotive force is due to Alessandro Volta (1745–1827), who invented the battery, or voltaic pile."Electromotive force" originally referred to the 'force' with which positive and negative charges could be separated (that is, moved, hence "electromotive"), and was also called "electromotive power" (although it is not a power in the modern sense).Maxwell's 1865 explanation of what are now called Maxwell's equations used the term "electromotive force" for what is now called the electric field strength. But, in his later textbook he uses the term "electromotive force" both for "voltage-like" causes of current flow in an electric circuit, and (inconsistently) for contact potential difference (which is a form of electrostatic potential difference).Given that Maxwell's textbook was written before the discovery of the electron, it is understandable that Maxwell exhibits what (in terms of modern knowledge) is inconsistency in the use of the term "electromotive force". The word "force" in "electromotive force" is a misnomer:
38 Practical…There are various methods for exploring this idea but they all end up in creating a graph.You can use a variable resistor method or the method of adding bulbs in parallel.Both will give similar readings.Try the adding bulbs one today.Just work through the exp and create a quality graphs and write a conclusions….AVload = (-r)I + y = (m)x + cVload = y-r = gradientI = x = c
39 Panasonic Cell Example Results Current /Amps +/-0.01APotential Difference /V +/-0.01V0.276.170.505.900.725.630.905.401.095.151.244.921.374.76
40 Potential Difference /V +/-0.01V D Cell Sample results....Variable resistor method using 3 chemical cell or 3*r in series.Current /Amps +/-0.01APotential Difference /V +/-0.01V0.014.320.034.260.054.170.074.080.094.010.113.940.133.822.214.171.124.740.203.63
41 D Cell Results Hence; 4.35V = emf -r = -3.638 But this is of 3 cells So r = 1.213
42 Plenary Question….2) When the switch S is open, the voltmeter, which has infinite resistance, reads 8.0 V. When the switch is closed, the voltmeter reads 6.0 V.a) Determine the current in the circuit when the switch is closed….. (1)b) Show that r = 0.80 Ω. (2)In the circuit shown the battery has emf and internal resistance r.1) State what is meant by the emf of a battery….energy changed to electrical energy per unit charge/coulomb passing throughor electrical energy produced per coulomb or unit chargeor pd when no current passes through/or open circuitBasiciSlice
43 Plenary Question….2) When the switch S is open, the voltmeter, which has infinite resistance, reads 8.0 V. When the switch is closed, the voltmeter reads 6.0 V.a) Determine the current in the circuit when the switch is closed….. (1)b) Show that r = 0.80 Ω. (2)In the circuit shown the battery has emf and internal resistance r.1) State what is meant by the emf of a battery….BasiciSlice
44 Power Transfer Ideas.... Pcircuit = Pcell + Pload or P= I = I2r + I2R We can also think about the situation as a power transfer and then rearrange…..Pcircuit = Pcell + PloadorP= I = I2r + I2R = Ir + IR = I(R + r)but we know IR = Vload thenVload = - IrSo we now have a formula which gives us the POTENTIAL DIFFERENCE across our power supply. Obviously this will be less than the emf , since you drop some voltage over the internal resistor, r. The Ir part is the volts lost in the power supply.
45 Load Resistance Matching… This is very useful as then we can derive and plot a graph of Power against load resistance.This peaks when R=r….Basic Power
46 Fudge It Maths...There are 3 scenarios to consider from the data plotted in this graph about the Power delivery....We can think (using fudge maths) about three cases to give us an idea of the value of Power….But this is not neat and hard to see the answer as you can see!r<<Rr=Rr>>RExtension Work
47 Cool Calculus... Power delivered Gradient of curve Place at maximum Extension WorkWe can also look at the idea of the maximum of the graph and differentiate the function.You will never have to do this but you can look at the function as a quotient differential form (AS Maths)This means you differentiate using the concept of u & v to yield a complex derivative. You cannot do simple differentiation but have to do a complex sum as shown below......Power deliveredGradient of curvePlace at maximumNB: you only need to know the outcome not the maths for AS
48 ConnectionConnect your learning to the content of the lessonShare the process by which the learning will actually take placeExplore the outcomes of the learning, emphasising why this will be beneficial for the learnerDemonstrationUse formative feedback – Assessment for LearningVary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sexOffer different ways for the students to demonstrate their understandingAllow the students to “show off” their learningConsolidationStructure active reflection on the lesson content and the process of learningSeek transfer between “subjects”Review the learning from this lesson and preview the learning for the nextPromote ways in which the students will rememberA “news broadcast” approach to learningActivationConstruct problem-solving challenges for the studentsUse a multi-sensory approach – VAKPromote a language of learning to enable the students to talk about their progress or obstacles to itLearning as an active process, so the students aren’t passive receptors