Equipment Electric Circuit kits Ammeters Volt Meters

Recap Series circuit sheets Printable available in lesson folder 1ai) Pd = I x R Pd = 0.5 x 8 Pd = 4v 1aii) Voltage in series circuit is shared. Vtotal = Vcomponent 1 + Vcomponent 2 Vcomponent 2 = Vtotal - Vcomponent 1 + Vcomponent 2 = 9v - 4v Vcomponent 2 = 5v

Lesson intentions: To understand how resistance is added in parallel. Date and title: Parallel Circuits 14 May, 2019 S t e p s t o S u c c e s s Learning outcomes: Identify parallel sections in circuit diagrams. Calculate the total resistance of two resistors placed in parallel. Calculate the current at any point in a circuit. 4 6 12 8 Key words: Series, Parallel, Resistor, Current, Potential Difference

Bigger picture Topic Potential difference and resistance Component characteristics Current and charge Series Circuits Topic Electrical Circuits Parallel Circuits

Build it Results Results Idea A1 = V1 = A2 = V2 = A3 = V3 = Using practical equipment please try and work out/prove a rule for current and potential difference for a Parallel circuit. Using your mind please try and work out how we work out the resistance of a whole circuit if there is more than one resistor in parallel.. Think about pathways electrons can take A1 A3   A2 V1   R1 V2 R2   Quickly run through the basics of building the circuits. They may struggle to put the ammeters in parallel You may need to model how to do this. One side of the ammeter has a cable plugged into the cable that plugs into the ammeter (it is not massively intuitive with the kit we have, do not expect the kids to just be able to do it). Students should build the first 2 circuits (They will have to move the ammeter and voltmeters around the circuit as there are not enough to have 3 of each) Ensure students are clear on how to wire in Ammeters (series) and voltmeters (parallel) They must think about the resistors in parallel.. I like to give an analogy of school corridors.. One corridor only so many people can fit down.. But if you have 2 corridors the same amount of people can move easier. When bringing it back together at the end of the practical see if the students can work out the patterns in their results from what they see. R3 V3 Results A1 = A2 = A3 = Results V1 = V2 = V3 = Idea What do you think happens to total resistance if we put them in parallel?

Rules and why Current is split between the different branches of a parallel circuit Potential Difference of the power source in a parallel circuit is the same through each branch. Resistance decreases in a parallel circuit …because there are the same number of electrons flowing through the circuit, but they have different pathways to take. Current will be higher on the branch with a lower resistance because it is easier for the electrons to move through. ….because the energy is being carried by the electrons. The electrons carry the same amount of energy (provided by the battery). So if they have different pathways to take they still carry the same amount. ….because the current (electrons) have multiple pathways to go through all the components. The electrons do not have to get through all of the resistors and therefore the total resistance of the circuit is lower You can pupil model the current moving around a 2 loops with a few students and how much has to go past each part in the circuit Ask SDe for help with this model if needed.

Parallel Rules Summary Current is SHARED between each branch Potential Difference is the SAME across all branches. Resistance is LOWER overall You can pupil model the current moving around a loop with a few students and how much has to go past each part in the circuit Ask SDe for help with this model if needed.

Calculating series resistance The total resistance is SMALLER than the smallest resistance R1 = 6 Ω R1 R2 R2 = 2 Ω This concept is hard to grasp. It is calculated as 1 over because we know the overall resistance will be SMALLER (you will be asked this) Relate it to fractions in maths.. All they are doing is ADDING FRACTIONS. They will know how to do this especially if you prompt them. They will likely not know what a reciprocal is so they need to flip the fraction to get RT . You can explain this with fraction. Rt/1.. What is anything divided by 1?... Itself. So Rt/1 = Rt If I flip one side.. I have to do it to the other.. Improper fractions we can easily turn into decimals with a calulator Warning: Smaller than the smallest resistor: 1.5 < 2 RT ≠ 1 RT

 Bringing it all together! Don’t Forget the 3 important rules for series circuits! 1) a) Find the combined resistance of the following circuit. = 30Ω = 60Ω 20 v NOT the answer!!! Smaller than the smallest (30Ω)  b) Find the voltage across V1 and V2 Practice questions.. Remind them to use the 3 rules of series circuits and that voltage will need to be calculated individually for each component. Parallel circuits: V1 = V2 = VT = 20v c) Find the currents I1 , I2 and IT. IT = I1 + I2 = ⅔ + ⅓ = 1A

 Bringing it all together 2! 1 1 1 1 1 4 1 From the following circuit, find the combined resistance R1=250 Ω 1 1 1 1 1 4 1 a) RT = R1 + R2 = 250 + 1000 = 1000 + 1000 1 5 1000 R2=1000 Ω RT = 1000 RT = 5 RT = 200Ω  Smaller than the smallest (250 Ω)

The two bulbs are identical. A1 is 4 A. So what are A2, A3 and A4? Answer: A2 – 2 A, A3 – 2 A, A4 – 4 A. Note: I found that most of my students (Year 10, ability range A* - B) were able to answer this and the rest of the questions without having been taught the ‘series and parallel circuits rules’, thanks to their understanding of current and voltage and their practical experience from previous lessons. The two bulbs are identical. A1 is 4 A. So what are A2, A3 and A4?

Answer: V1 – 4 V, V3 – 4V. V2 is 4 V. So what are V1 and V3?

The voltage across the battery is 6 V. Answer: A – 2 V, C – 4 V. This one is trickier (Each pathway needs to add up to 6v A+b = 6v A+c= 6v This may not immediately be obvious The voltage across the battery is 6 V. The voltage across bulb B is 4 V. What are the voltages across bulbs A and C?