 To solve various problems from electricity chapter, we need to understand and remember four formulae one is Ohm's law a formula on resistivity series and parallel resistance formula and finally the power formula All the other formulae you might see in your book can be derived just by using these four That's that's what makes these four the most important ones and So in this video, we will look at each one in detail and use them to solve certain board type questions And you can see the index on the right-hand side. So if you want to jump to any specific topic, feel free to do that So let's start with questions on graphs So here's a question where we are given a graph of voltage and current and we're asked to calculate the resistance of this material Can you pause and think about how you will solve this? Okay, let's see we are given values of voltage and current and from that we're asked to figure out what the Resistance is meaning we need some kind of relation between voltage current and resistance and that relation itself is what we call Ohm's law So Ohm's law says voltage v equals IR This basically means that if the voltage across the conductor doubles the current doubles if the voltage halves the current becomes half And the resistance R is a constant which does not depend on the voltage or the current Okay, so how do we find the resistance over here? Well from this equation resistance becomes v divide by I And if you look at the graph the values of voltage and current are given you can pick any set of values you want But in our graph only for four volts. We know the value of the current So we'll take that set and so if we plug in When V is four volt we get the current to be point two volt Sorry point two amps and that gives us four divide by point two which is 20 volt per ampere, which is often called Ohms. So this Material has a resistance of 20 ohms All right, let's try one more a little different one in this we are given two graphs Representing the two different materials and we are asked to figure out which of the two materials has a higher Resistance again pause the video and see if you can give us give this one a shot Okay, again, we are asked to calculate resistance given voltage and current Which uses a clue that we have to go back to ohms law. So again R equals V by I But over here you might say look there are no values of current and voltage given. How do I calculate resistance? Well, the question doesn't ask us to calculate resistance. It only asks us to To compare and tell which one has a higher resistance. So we don't have to do any calculation But we have to look at this formula and see if you can get some insights Well, one way we can think about this is we can say look When we are comparing two resistances if we keep the current in both those materials the same And let's say the voltage is higher Then for that material the resistance would be higher. Does that make sense for the same current higher voltage means higher resistance That that's one way to look at this And so if we look at the graph we can say let's keep the current Same for both of these and the way we can do that is we can draw a horizontal line like this Right, this represents the same value of current for both these graphs And now for each of them, let's figure out what the voltage is for this one for that current The voltage is over here and for this one for that current the voltage is over here And so this means for the same value of current the second one has a higher voltage Compared to the first one And therefore the second one must have a higher resistance compared to the first one And so the second material has the higher resistance With this we can now jump to the next topic questions on resistivity So here's the first one when the length when the length of a wire is doubled What happens to its resistivity? So can you recall that resistivity formula and then think about what happens over here? Okay, let's see since we are asked about resistivity and there's only one formula for us in resistivity That immediately reminds us I may have to use this formula r equals rho into l divided by a And what's given to us what's given to us is the length of the wire doubles Meaning this length becomes two times and we're asked what happens to resistivity Does it double does it become half the answer is nothing happens Why why doesn't anything happen? But because it's important to understand in this formula as the length of the area changes The resistance of the material changes. That's what the formula is saying The resistivity does not depend on the dimensions So whether you double the length or you half the thickness or you do whatever you want to its dimensions the resistivity does not change The resistivity only depends on the material and also it of course depends on the temperature But that's not important over here. So changing the dimensions has no effect on resistivity. So nothing happens to resistivity Okay, let's look at another problem We are given a silver wire has same length and resistance as a gold wire Which is thicker given resistivity of gold is less than that of silver Again, can you give this one a shot? Okay, let's see Again, we are dealing with resistivity and therefore immediately I can I can guess that we may have to use this formula r equals rho into l divided by a Now, let's look at what is asked. We are asked to figure out which one is thicker Right. So what I do immediately because of this is I know I know now I have to figure out what a is a the cross sectional area That's the that's the meaning of thickness over here. So I will immediately rearrange this formula to get a What next well now, let's look at the data the data is a silver wire has the same length The same length and the resistance as the gold wire So if you were to compare the gold wire and the silver wire, they have both the length to be the same And even the resistance to be the same So if I were to calculate the area of gold and the area of the silver wire The value of l and r are the same for them Right. That means this ratio will become the same for them. Whatever that number is Whatever that value is that value turns out to be the same for them And so in this particular problem the area purely depends on the value of resistivity. Does that make sense? So whichever has more resistivity will have more area So which is which which is given to have more resistivity. It's given that silver has a higher resistivity than gold If silver has higher resistivity, it will have higher area And so immediately just by inspection we can say that the silver That's the one that should have The more area of cross-section. So that's the one that should have more thickness With this we can jump to the next set of problems on series and parallel Resistors. So here's a question. What is the equivalent resistance between a and b? Again, graded pause and see if you can try this on your own first Okay, so let's start with understanding when resistors are said to be in series and when they're in parallel So they are said to be in series If they have the same current flowing through them So if you look at these two resistors, whatever charges are flowing through this resistor All of them would also have to flow through this one, right? So the current here and current here must be the same And so we can say these resistors are in series. So I'm going to mark them with one color over here But what about this resistor and this one? Are they in series? It may look like that at first But no because whatever charges are flowing over here Not all of them have to pass over here. Some of them might some of them might flow over here So the charges can split meaning the current can split and therefore they are not in series Similarly, this and this are also not in series This and this are also not in series. This and this are also not in series. So only these are in series On the other hand, if you look at these two resistors, you can see they're connected across each other Such that they're connected across the same point They are connected across the same point and as a result the voltage across them will be the same And when the voltage across them would be the same, we say they are in parallel And so again, let me shade them with a different color These two are in parallel And if there was say for example a resistance over here Then these two wouldn't be in parallel because then they wouldn't be connected across the same point on this side Over here, they are connected but on this side, they wouldn't be so for parallel on both sides They need to be connected across the same point. Okay, but anyways in this problem. They are in parallel So now how do we solve them? Well, we need to remember the formula for series The resistances just add up. So to calculate the equivalent resistance, we just add them up over here But when it comes to parallel resistances, we do the reciprocal the reciprocal of the equivalent resistance Happens to be the sum of the reciprocals of the individual resistances and over here one big mistake Which I used to always do is that after doing all the calculation, I forget to do the reciprocal I'll tell you what I mean when I'm solving this. So anyways, we can go ahead and solve this now We know these two are in series and so we can just add them up. So five plus two gives me seven ohms So I replace these two resistors with just one resistor of seven ohm What about these two? Well, when I calculate for these two since they are in parallel, I will do one over rp Let me do that over here I'll do one over rp Equals one over 10 plus one over 40 And then I take the common denominator of 40. So I'll get four plus five I'll get five over 40 and I might think hey, that's the answer five over 40 or maybe one divided by eight But remember this is one over rp. So do the reciprocal. So the rp that is the Actual value becomes eight ohms. So the equivalent resistance over here becomes eight ohms And so I replace that two resistors with one resistor of eight ohms And now I have two resistors again in series. So I can go ahead and add them seven plus eight is 15 So that gives me my final answer 15 ohms I have reduced this entire circuit to one equivalent resistance and that is 15 ohms I thought of putting one more similar problem Because this is so important. Can you pause and try to solve this one after about five seconds? I'll present the entire solution Okay, so Here's the entire solution You can pause and check the solution yourself. The final answer is over here 15 ohms is the equivalent resistance Okay, let's take another type of problem on series and parallel circuits. So here's a question How many 20 ohm resistors are necessary in parallel? to carry one ampere across five volt Again, great. I had to give it a shot and here's a clue sometimes when things are not very clear I like to draw a diagram diagrams make things very clear Okay, I guess for me a diagram is necessary to make sense of this So we are asked to find how many 20 ohm resistors are necessary to carry one ampere across five volt So here's how I'm thinking So I have a wire across which five volt of voltage is put and it needs to carry one ampere of current All right. Now do that. There must be some resistors over here We are given only 20 ohm resistors and we are asked how many such 20 ohm resistors Need to be put in parallel to get one ampere Now since there is current and voltage and again resistors I'm guessing I need to use ohms law over here So what I do is I look at this and I say hey, there is voltage given there's current given So from that I can immediately use ohms law and figure out what this resistance must be Right. So the resistance according to ohms law must be v over i that has to be five ohms So this means look at how the question changes. So this means we need to figure out how many 20 ohm resistors are necessary in parallel to get Five ohms, right? That's all it means this part of the problem was to get to this part And so we are basically asked how many are needed in parallel to get five ohms in other words This Is rp in other words it's given that equivalent resistance or we figured out the equivalent resistance needed is five ohms And for that equivalent resistance of five ohms How many 20 ohm resistors are needed to be connected in parallel? So now that means I can go ahead and use the parallel formula So one over rp Equals one over r1 Plus one over r2 and so on Again, remember for the parallel formula left hand side is one over rp. It's not rp equals one over r1 plus one over r2 Okay, and so if I substitute I get one over five Equals one over r1 r1 r2 r3. They're all 20 ohms. So it's one over 20 plus one over 20 And so on I just don't know how many are there and that's what I need to figure out, right? So let's say there are 20 or not 20. Let's say there are n number Okay, the sum number n and I need to figure out what that n is So what will this value equal to if there are n number? Well, let's say if there's one I would get one over 20 if there were two I would get two over 20 if there are five I would get five over 20 as the answer So if there are n number, what would this number be? This would be n over 20 And so now I just have to solve this for n. So n becomes Rearrangement 20 divided by five and that gives me four And so this means I need to connect four resistors of 20 ohm each in parallel to carry this out With this we can now jump to the next topic Circuits, so let's solve one problem on this So we are given a circuit With a battery and some resistors and we're asked to find the current and voltage across each resistor So can you give this a shot first? All right now whenever I should solve these problems earlier, I should think they are so easy I mean just by using ohm's law i equals v over r Right, and I know what v is v is 10 volt So for each one I substitute the value of r. So for this one v is 10 r is 2 I would be 5 amperes over here for this one again v is 10 r is 4 It will be 10 by 4 2.5 amperes for this one v is 10 r is 12 You get the point. So I got the current done, right? No, that's wrong mainly because What is given to us this voltage is across these two points. That's the voltage We don't know the voltage across each of the resistor And that's why we need to first calculate that Okay, and that's the reason when the current calculations are wrong So once we know what the voltage across each resistor is only then we can use ohm's law to figure out the current across that So the question now is how do we do this? Because we know the voltage across these two points It makes sense to calculate the resistance across these two points. In other words The first step we need to do is figure out what the equivalent resistance is between these two points Then we know the voltage across these two points Maybe from there we can calculate the current and then maybe we can figure out how to calculate The values across these resistors. So let's go ahead and do that So like we saw before we now know that these are in parallel These are in parallel with each other And then this entire resistance whatever that is will be in series with this one So let's redraw by replacing this with one resistor. So here is the Circuit and if I replace this by using the parallel formula, which I'm pretty sure you can do that yourself now You will get the value to be three ohms And now these two are in series. So if I add them I will get the final circuit with five ohm resistance And now I can say across these two points the voltage is 10 volt And I know what the resistance is so now I can use ohm's law and and from ohm's law now I can say ah the current over here this current has to be 10 by 5 two amperes But how do I find out the current and voltage across each resistor? How do I go from here to there? Well, we have to go backwards and this is why this is why while solving the circuit problems It's always nice if you can do it step by step So here's what I mean by going backwards when I go from this circuit back to this circuit I know that this equivalent resistance that came from the combination series combination of these two, right? So what does that do? Well, I know in series the current is the same So whatever is the current over here the same current must be flowing over here Oh, that means I know immediately the current through this resistor is two amperes and the current through this one should also be two amperes Did that make sense? Okay, I found the current through these resistors How do I find the voltage? Well, I know the current I know the resistance go back to ohm's law v equals ir this time And so two times two is four ir So this would be four volt This would be ir two times three six volt But that's not what I want. I want over here. So go back If you look carefully, we are already done for this resistor Now, how do I figure out what are the voltage and the currents across these two resistors? Again when I'm going back from here to here, I look at myself Which combination did this resistance come from? It came from parallel combination In parallel combination The voltage remains a constant. So I already know the voltage across these two points So the voltage across these two points, which means I know the voltage across this Is six volt whatever I have over here the voltage across this is also six volt So in parallel the voltage remains the same. So I found the voltage Now how I can use I can use ohm's law to find the current So again, if I use ohm's law v by r this time i equals v by r I will get six by four, which is 1.5 And over here six by 12 gives you or use me 0.5 amps And now look I have found current and voltage through each and every resistor And so the whole idea is first find the equivalent resistance Solve the circuit over there and keep going back If the equivalent resistance came from series combo, the current is the same We find voltage then if the equivalent resistance came from the parallel combo Then the voltage remains the same we find the current And also there could be different ways of asking the same question For example, one question they could ask is what's the current drawn by the battery? Okay, so the answer is two amperes again because two amperes is going through the battery Another another way to ask the question is they might put a voltmeter somewhere over here and ask What's the reading of the voltmeter that will be four volt? It's the same question of what is the voltage across this or they might put an ammeter over here and ask What's the reading of the ammeter? That's fine 1.5 amperes. Okay, so same thing With that we can now move on to the next topic power and heat So here's the first question. It's given a device is rated 100 watt 200 volt find its Resistance again, you know the drill Greater to pause and see if you can try Okay, so we are given power rating of that device and we are given its voltage And we need to somehow connect it to the resistance So how do you calculate power? What's the formula for power electric power? It's calculated as the product of voltage times the current. That's the formula for electric power But how do we connect it to resistance? I need to bring resistance into picture Well, we know the connection of voltage current and resistance ohms law So we can bring in ohms law and put resistance over here One way to do that is by substituting v equals ir We'll get ir times i which uses i square times r So that's one way to bring in resistance Another way it would be to substitute i equals v by r and so I get v square over r Now what's important is I think that there are three formulae to calculate electric power But no, there's only one formula This is just adding ohms law to this formula Okay, and that's important because ohms law does not work for all devices all cases something that you may learn in Later courses and so anyways if ohms law does not work v equals ir does not work Then you can't use this formula. So that's the reason why I stress that this is the formula to calculate electric power Anyways coming back to the question. We are given the power. So p is given We are given the voltage v is given. We're asked to find the resistance So I can go ahead and use this formula. Does that make sense? So I will rearrange that to give r And now I get all I have to do is plug in v is given to be 200 volts. So 200 square So I turned into 200 because usually things cancel out divide by power, which is 100 Like I said things are canceling out. And so you get 400 ohms. So that's the resistance Okay, let's do one more In this question, we're asked can a 50 watt 200 volt light bulb Run on a 10 ampere line Now at first you may be wondering what what is the meaning of 10 ampere line? Think of 10 ampere line is just a wire which can carry a maximum of 10 amperes So on such a wire if we were to run this bulb, will it work? That's the question You know what to do Okay, hopefully I've tried Over here. Yeah, I think it's better if I draw a diagram. It becomes much easier for me. So here's the line It can carry a maximum of 10 amperes of current It's not carrying any current right now But the maximum current it can carry is 10 amperes and on this line We are going to attach a bulb whose power rating and voltage ratings are given to us and the question is Will this bulb run or I think what is the question it's trying to ask is will this glow properly? Okay, how do we answer that question? What do you think? Well, what I am thinking over here is there's a limit on the current Right, so clearly if 10 amperes is not enough to make this bulb glow then the bulb won't go properly But if 10 amperes or less than that is enough to make this bulb glow then the bulb will go properly, right? So I just need to know how much current is needed to make this bulb glow. How do I figure that out? Hey, I know the power I know the voltage So from that I can calculate the current So I know what current from that I can I will know what current is needed for this bulb to glow And then I can answer this question. So again going back to power Power equation electric power is p equals v into i And so i equals p divided by v which is 50 divided by 200 Which becomes 5 divided by 20 that is 0.25 So I Becomes 0.25 amperes This means that this bulb will take 0.25 amperes or it will need 0.25 amperes Can this line handle 0.25 amperes? Of course it can handle up to 10 amperes So will it glow? Yes. The answer is yes Okay So again Important thing don't think that 10 ampere line mean it carries 10 ampere. No, this is the max that it can carry Okay, how much how much current it carries depends upon what what device you put across over there, right? Okay, anyways, here's a bonus question. Here's a follow-up bonus question How many such bulbs can be connected in parallel? Think about this and see if you can give this a shot All right, again, I'm going to draw a diagram since we are connecting bulbs in parallel Here's my diagram now a lot of bulbs will be connected in parallel. We don't know how much that's what we need to figure out But how do I calculate that? How do I figure out how many bulbs can be connected in parallel? Well, the way I'm thinking is now I know that each bulb Each bulb will take in 0.25 amperes So one bulb fine two bulbs will take in 0.25 and 0.25 total 0.5 That's also fine three bulbs are also fine But I'm pretty sure there'll be a limit after which the total current will exceed 10 ampere and it won't work So how do I calculate this looks like now a maths problem? Right. So how do I do this? Well, let's say there are n bulbs over here Then the total current that will be taken each one will take 0.25 total current taken by n bulbs will be n times 0.25 And we know that current At max since I want to find what's the maximum number of bulbs I can put over here at max that number has to be 10 ampere Does that make sense? So from this I can calculate what n is n is 10 divided by 0.25 which turns out to be 40 So I can put 40 bulbs in parallel maximum And if I put anything more than that it will start drawing more than 10 ampere's current and then Maybe the wire will melt or maybe there's a fuse somewhere that will blow up But whatever it is it's not good for us. So to answer your question maximum 40 bulbs can be connected in parallel All right another problem on power is this one Calculate the heat generated in 10 seconds when an electric ion of resistance 15 ohm takes a current of two amperes I'm not even going to tell anymore. You know what to do Okay, now at first it might seem wait a second. Where did this heat generated concept comes from? What formula do I use for this? But if you think carefully heat generated is basically energy This is basically energy Heat is a form of energy And what is power power is how much energy is dissipated or how much energy is consumed per second? Right. So in when it comes to resistive devices like electric ion or heat or things like that that power that energy is usually in the form of heat Descubated in the form of heat. Anyways power it tells me how much energy is consumed or dissipated per second So if I want to calculate how much energy is consumed or dissipated in 10 seconds I just multiply that number by 10. So you see this is also a problem on power electric power Okay, and so I'm going to go use the same formula P is equal to v into i I know what the I don't know what the voltage is. I know the resistance. So I can use v equals ir And get i square r over here and another value of i This is two amperes and r is 15. So if I plug in i square is two square, which is four 15 is just 15 15 times four is 60. So that gives me 60 watt This means every second 60 joules of energy heat energy is generated. That's the meaning of power So I know every one second 60 joules is generated in 10 seconds. How much heat is generated? Well, that's going to be 60 times 10 So the heat generated is just going to be 60 times 10 and that's going to be 600 Joules And so what's important is you don't need a new formula for heat, right? You might see in your textbooks that heat formulas given as i square r into t They call that the joules law now, that's great But if you think carefully, what exactly what are we doing? We're just calculating how much energy is generated by multiplying power with time because power is energy per second You multiply by whatever time you take and you get that energy. So I don't remember this as new Formula so this all comes under power. So if you know what power is how to calculate electric power We can calculate the amount of heat generated And that now brings us to the last topic the last question question on commercial unit Here we go so Find the cost of operating a five kilowatt device 20 hours a day for 30 days given per kilowatt hour is three rupees You know what to do Hope you've done that Let's see So we are given or what we are asked to do is find the cost how much money it takes to operate this device Okay, and what's given to us is cost per kilowatt hour What is this kilowatt hour? Well kilowatt is the unit of power. What is the unit of power? So this is power r is time. So kilowatt r is what is that power into time? We just saw it's energy Right now normally unit of energy is joule kilowatt r is the commercial unit of energy It's a big unit of energy. So this is also called unit by the way So but what's important is what's given to us is how much how much it costs to consume one commercial unit of energy So how do I calculate the cost over here? Well, we can figure out how much is the total energy consumed in the same units And then we can multiply that by three because I know each unit takes our cost three rupees So let's go ahead and figure out. What is the total energy consumed over here? How do I calculate energy consumed in general? We just saw it's power into time So energy consumed is power into time. We know what the power of the device is that's five kilowatt And we know what the time is. What's the time? The time given is 20 hours a day, but for 30 days So total how many hours? Well total we get 20 hours a day times 30 multiply them, right? So that this many hours and what's important is you have to keep the units in kilowatt and r So notice this will become r and that's important because our units is also commercial unit is kilowatt r So that's the one important thing. So if kilowatts are not given you have to convert that into kilowatt If minutes or seconds are given again, you have to convert that into r's. That's the important thing Anyways, once that is done, we can now just multiply five times two is five times 20 is 100 100 times 30 is 3000 So this is 3000 kilowatt This cancels r So that's the total energy consumed. We can also say 3000 units kilowatt r is also called unit, okay And we know for each unit we have to pay three rupees for these many. How much do we have to pay? So the total cost will be three times 3000 and that's going to be 9 thousand Rupees And there is our answer And that's pretty much it. Just one thing I need to tell before we end this is if you Struggled in any of the problems or any of the questions. Don't worry We have videos and Exercises explaining each of those topics in great detail. So you can go back to the lesson go back to the unit and practice more