Kirchhoff's Laws - A-level Physics

all right I have to admit that my first kirchoff's law video wasn't the best video ever not least because i mispronounced his name so let's see if we can do a better job this time you don't pronounce it off you pronounce it KIAC off-key a cough came up with two laws for electricity his first law is this total current entering a junction is exactly equal to the total current leaving it or in other words current is conserved at junctions yes you might think that that is pretty darn obvious but it's a very useful law to you think about doesn't matter how many Y's you have going into and out of a junction we can call all of the currents i1 i2 i3 i-4 i-5 in this case and we know that we have two currents going in so we can say i1 plus i2 has to be equal to u i3 plus i-4 plus r5 it's never going to be that complicated though but there are times when you have circuit diagrams and you might be stuck as to what to do next actually when it comes to circuits that have branches in them and so you might need to remember that actually the most obvious thing is that the currents going into a junction is equal to the cars coming out kick off second law is this in any closed loop Network the total EMF is equal to the sum of PD drops and we know that to be true just from when it comes to looking at one cell or battery and one resistor this is the simplest circuit that we can have we have an EMF here and we have a voltage across here we know that if this EMF is six volts then the PD drop across this resistor must also be six volts if we have two resistors in there again the total PD drops across both resistors must be six volts just be aware that when it comes to EMF though batteries and cells do push in a certain direction so if all of these are three volts cells here what is the total EMF of the circuit well pushing this way we have three volts pushing this way we have three volts from the cell in the middle but this one here is trying to push it in the opposite direction so therefore actually the total EMF is just three volts because these two cells are cancelling each other out as it were so therefore the PD drop across this resistor also has to be three volts okay let's have look at a more complicated example of how kickoff second law can be used in this case we have two EMFs but they're not in line with each other as it were they're part of this branching circuit this parallel circuit and so this is a very standard case of when we have to use kickoffs second law we're told the resistance of this resistor is 10 ohms and we're told that this one is 30 ohms now how many loops can you see in this situation well here we have one loop here technically that should be the other way round because current is flowing this way we know that current is going to flow in this way in this way actually it's not always obvious which way currents are going to flow so if you get given questions generally you will be shown which way the currents are flowing okay we have another loop here as well it's just a closed loop so there's one loop we can literally just forget about that we have to consider the circuit as a whole but we can break it into just different loops there's one loop that's the cell and the resistor and here's another loop cell and two resistors but there is one more loop as well it's like one of those puzzles how many squares or triangles can you see and we have one big loop as well now let's try to figure out what the pds across these resistors are going to be so at the minute if we just look at this loop it seems like we have a potential divider don't we because we have this 12 volts and we have 10 and 30 so if this was just a loop by itself then we would assume that the voltages would be 9 and 3 that's just your potential divider equation but the problem is is that that's not the only cell involved we also have this one and so we can't say that that 12 volts is being split between these two resistors according to their resistances but there is one thing that we know to be true and that is that we have one loop here where we have a 10 volt battery find cell it doesn't matter 10 volt EMF and we have one resistor we know therefore that due to kick off second law that because we have a 10 volt EMF the PD drop across this resistor must be 10 volts now we can look at the second loop here we have a 12-volt EMF and we have our two resistances but we know that it has to be a 10 volt drop across this 30 ohm resister therefore now using our knowledge that potential has to be divided across T resistors like this we know therefore that this has to be 2 volts let's consider the big loop forgetting about the resistor in the middle here we have a 10 volt battery pushing this way 12 volt battery pushing this way so overall the total EMF of this big loop is 2 volts so therefore if we're just concerned with this resistor at the top it has to have a PD drop of 2 volts and we've already seen that that is correct you could then get asked a question like what is the current through the 10 ohm resistor well we're gonna use this vehicle as IR I is V over R but what is the V gonna be like we said it's 2 volts and we could fill that out either from these 2 loops or just the big loop divide that by the turn ohms and we have zero point 2 amps tell you what let's call that current there I - let's call this 1 I 1 let's call this 1 I 3 can we figure out what I 1 is straight away no we can't because we have this branch here and we need to know I 3 first let's find out I 3 here we have current flowing through this resistor do we have a voltage yes do we have a resistance yes so therefore we can just 210 divided by 30 and so that gives us zero point 3 3 amps now we know that at this Junction here Kia cost first law has to be satisfied so we can see that I 1 and I 2 are we going in and that has to be equal to I 3 coming out so therefore if we want to find this current here the I 1 we can say that's going to be I 3 take away I - in other words 0.33 take away 0.2 that gives us 0 point 1 3 amps now it gets complicated when we add in maybe a third resistor here let's call that 10 ohms as well I'm not going to do any calculations with it because it will take a while and this a little bit beyond hey level but the thing is by adding this one resistor in that changes the whole circuit again this big loop here we have a total EMF of 2 volts but we can't say that it's just going to be 1 volt across each of these 10 ohm resistors doesn't work like that or what we can say is that the PD across this resistor here is bound to decrease because that 2 volts now has to shared somehow across these two looking at our first loop over here as well we now know that this 10 vault has to be shared across the 10 ohm and the 30 ohm resister again we can't just split that up in a ratio 3 to 1 because it doesn't work that way but we do know that the PD across this one is going to drop because yes the 10 volts does need to be shared somehow across the 10 ohm and 30 and resistor so there we go that's a basic look at care cost loss hope you found that helpful at least more helpful in the last time if you did please leave like and if you have any suggestions on what I can do next all questions then please leave them in a comment down below I'll see you next time

Channel: Science Shorts

Views: 202,901

Keywords: pronounce, specimen, loop, first, entrance, 11th, cie, practical, shorts, past, ap, school, british, english, pronunciation, experiment, 2nd, wjec, solutions, placement, current, college, circuit, drop, emf, grade, sat, sample, lecture, mark scheme, pd, aqa, exam, voltage, revision, electricity, questions, university, paper, second, tutorial, science, 1st, technique, homeschooling, explained, equations, answers, lesson, edexcel, ocr, high school, worked, practice, example, 12th, advanced, spoken

Id: JbJ6aGYa63k

Length: 7min 13sec (433 seconds)

Published: Thu Jan 09 2020