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**3.3.1 Series and Parallel Circuits – Kirchhoff’s second law**

3 DC Circuits G482 Electricity, Waves & Photons 3.3.1 Series and Parallel Circuits – Kirchhoff’s second law 3.3.2 Practical Circuits Ks5 OCR Physics H158/H558 Index Mr Powell 2012

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Practical Notes.... Practical Skills are assessed using OCR set tasks. The practical work suggested below may be carried out as part of skill development. Centres are not required to carry out all of these experiments. 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 parallel circuits. 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 in Design 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 a data-logger.

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Circuit Rules For resistance calculations at AS you can use the premise of the following rules but they are not sorted can you do this…..

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Series Resistance

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**Resistors in series: I I I I**

RT V1 V2 V3 VT Resistors in series can be replaced by one single resistor: VT = V V V3 But V = IR I RT = I R I R I R3 I RT = I R I R I R3 RT = R R R3

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Parallel

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**Resistors in Parallel I**

RT IT I2 R2 V I3 R3 V Resistors in series can be replaced by one single resistor: But V = IR and I = V R IT = I I I3 V = V V V RT R1 R2 R3 V = V V V RT R1 R2 R3 1 1 1 1 = + + RT R1 R2 R3

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Generalised Formulae So 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!

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Test it out…. If all the bulbs are of equal resistance answer the following; Find all the meter readings What is the total resistance of the bottom branch If 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)

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**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 = 9.2Ω 9.2 Ω and 30 Ω in series gives 39Ω (1) (allow e.c.f. from value of R) 30Ω 2) (V = IR gives) 12 = I × 39 I = 0.31 A (1) (allow e.c.f. from (a)) Basic

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**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) 30Ω Basic

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**3.3.1 Series and Parallel Circuits – Kirchhoff’s second law – Part I**

Assessable learning outcomes state 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 emf An example of more than one source of e.m.f. Is a battery charger. (HSW 6a) Students can verify the rules for resistors experimentally.

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b) Kirchoffs Law I The “current law” states that at a junction all the currents should add up. I3 = I1 + I2 or I1 + I2 - I3 = 0 Current 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!

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**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 A Also remember to make sure you work out current in Amps and time in seconds in your final answers!

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**b) Kirchoffs Law I – Questions?**

Work out the currents and directions missing on these two junctions? 7A 3A

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**a) Simple e.m.f example If I = 100 mA what is that in amps?**

100 mA = 0.1 A 2) What is the current in each resistor? 0.1 A 3) Work out the voltage across each resistor. V = IR so V1 = 0.1A x 30 = 3 V V2 = 0.1A x 40 = 4 V ; V3 = 0.1A x 50 = 5V 4) What is the total resistance? RT = 30 + 40+ 50 = 120 5) What is the battery voltage? V = 0.1A x 120 = 12V

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a) PD Rules - Series

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**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

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**If the variable resistor is adjusted so that there**

a) PD Rules - Parallel battery pd = 12V This means: each coulomb of charge leaves the battery with 12J of electrical energy If the variable resistor is adjusted so that there is a pd of 4V across it This means: each coulomb of charge uses 4J of energy passing through it

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a) Kirchoffs Law II The “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.

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**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 resistor You may want to redraw the circuit then apply simple ideas of additive resistance and ratios to find out the currents?

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**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.

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**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 R1 Answer…. BASIC

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**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 R1 Answer…. BASIC

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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

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e) Complex Examples Hint1: 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 - I1 If we apply Kirchhoffs laws (previous slide) about pd = 0 in closed loop we can say the following two things Starting at Point “P” and going clockwise around the left-hand loop; -I3R2 + E1 - I1R1 = 0 Starting at Point “P” and going clockwise around the right-hand loop; -E2 - I2R3 + I3R2 = 0 Hint2: Solve this using Simultaneous Eq method. You will not be asked to do anything so complex in the real exam! R1 I2 P I3 E2 E1 R2 I1 R3 EXTENSION WORK

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**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 = 0 Starting at Point “P” and going clockwise around the right-hand loop; - E2 - I2R3 +I3R2 = 0 p.d. is lowered with flow I1 PD against flow of current R1 I2 P I3 E2 E1 R2 p.d. is increased with flow I1 I1 R3 PD against flow of current p.d. down with flow I3 left loop p.d. is up with flow I3 right loop EXTENSION WORK

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**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.d Explanation as to why each contribution is + or – Equation for each loop Answer for I3 E1 = 6V E2 = 2V R1=10 R2=10 R3=2 R1 I2 P I3 E2 E1 R2 I1 R3 Answer: I3 = 0.4A

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**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 = 0 I3+I2 = -6V/10 I3+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 = 0 Sub Eq 2 into 2 to eliminate I3 -2V - I210 + (0.6A-I2) 10 = 0 4V/20 = I2 = 0.2A Hence – feed back into Eq2 to yield I3 = 0.4A Answer: I3 = 0.4A E1 = 6V E2 = 2V R1=10 R2=10 R3=2

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**e) Complex example.. EXTENSION WORK Kirchoffs 1st Law; I1 = I2 + I3**

Kirchoffs 2nd law; Sum PD Loop AEDBA 30V = 20I3 + 5I1 - Eqn 1 Sum PD Loop FEDCF 10V = 20I3 - 10I2 10V = 20I3 - 10(I1 - I3) 10V = 30I3 - 10I1 - Eqn 2 Add 2 x Eqn 1 + Eqn 2 70V = 70 I3 I3 = 1 A Substitute this in Eqn 1 I1 = 2A so I2 = 1 A This worked example relies on two equations found from two loops. Each defined for a separate power source. Solve simultaneously to find I’s EXTENSION WORK

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**3.3.1 Series and Parallel Circuits – Kirchhoff’s second law – Part II**

Assessable learning outcomes explain 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 .

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What is the link.....

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**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 resistor Stored chemical energy transfer to heat energy in wires

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**What is Internal Resistance....**

= I(R + r) 6V = I(4 +8) I = 0.5A We also know that; Vload = - Ir Vload = 6V - 0.5A x 4 Vload = 4V So 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;

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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:

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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…. A Vload = (-r)I + y = (m)x + c Vload = y -r = gradient I = x = c

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**Panasonic Cell Example Results**

Current /Amps +/-0.01A Potential Difference /V +/-0.01V 0.27 6.17 0.50 5.90 0.72 5.63 0.90 5.40 1.09 5.15 1.24 4.92 1.37 4.76

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**Potential Difference /V +/-0.01V**

D Cell Sample results.... Variable resistor method using 3 chemical cell or 3*r in series. Current /Amps +/-0.01A Potential Difference /V +/-0.01V 0.01 4.32 0.03 4.26 0.05 4.17 0.07 4.08 0.09 4.01 0.11 3.94 0.13 3.89 0.15 3.8 0.17 3.74 0.20 3.63

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**D Cell Results Hence; 4.35V = emf -r = -3.638 But this is of 3 cells**

So r = 1.213

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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 through or electrical energy produced per coulomb or unit charge or pd when no current passes through/or open circuit Basic iSlice

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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…. Basic iSlice

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**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 + Pload or P= I = I2r + I2R = Ir + IR = I(R + r) but we know IR = Vload then Vload = - Ir So 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.

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**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

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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<<R r=R r>>R Extension Work

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**Cool Calculus... Power delivered Gradient of curve Place at maximum**

Extension Work We 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 delivered Gradient of curve Place at maximum NB: you only need to know the outcome not the maths for AS

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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

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