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## Class 10 Physics (India)

# Electricity class 10 numerical: CBSE board practice

Prepare for class 10 CBSE exam for the chapter Electricity by revising difficult board-type numerical questions. Created by Mahesh Shenoy.

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- what is electricity(0 votes)
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Electricity is a form of energy caused due to flow of charges.*Hope it helps. Feel free to comment if your doubt persists...Have fun learning :)*(1 vote)

## Video transcript

to solve various problems from electricity chapter we need to understand and remember for formula one is Ohm's law a formula on resistivity series and parallel resistance formula and finally the power formula all the other formulas 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 are 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 are 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 huffs 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 divided 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 4 volts we know the value of the current so we'll take that set and so if we plug in when V is for world we get the current to be point 2 volt sorry 0.2 amps and that gives us 4 divided by point 2 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 them are 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 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 Ohm's 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 we 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 material 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 of 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 have 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 2 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 or 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 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 don't I know now I have to figure out what is a the cross-sectional area that's them 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 lengths to be the same and even the resistance to be the same so if I were to calculate the area of goal 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 grade you 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 are 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 gonna 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 sum up they 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 these are also not in series this and these are also no 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 the connect 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're 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 panel 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 1 over R P let me do that over here I'll do 1 over R P equals 1 over 10 plus 1 over 40 and then I take the common denominator of 40 so I'll get 4 plus 5 I'll get what 5 or 40 and I might think hey that's the answer Phi over 40 or maybe 1/8 but remember this is 1 over R P so do the reciprocal so the R P that is the actual value becomes 8 ohms so the equivalent resistance over here becomes 8 ohms and so I replace that two resistors with one resistor of domes and now I have two resistors again in series so I can go ahead and add them seven tha's 8 is 15 so that gives me my final answer 15 ohms I have reduced this entire circuit to one equal and 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 will present the entire solution okay so here is 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 1 ampere across 5 volt again grid 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 ok I guess for me our diagram is necessary to make sense of this so we are asked to find how many 20 of resistors are necessary to carry 1 ampere across 5 volt so here's how I'm thinking so however you i have wire across which 5 volt of voltage is put and it needs to carry 1 ampere of current all right I'll 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 1 ampere now since there is current and voltage and again resistors I'm guessing I need to use Ohm's law over here so what I do is I look at this and I say hey there is voltage U and this current given so from that I can immediately use Ohm's law and figure out what this resistance must be right so the resistance according to Ohm's law must be your eye that has to be fiims so this means look at how the question changes so this means we have to figure out how many 20 ohm resistors are necessary in parallel to get fiims 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 firearms in other words this is Rp in other words is given that equivalent resistance or we figure out the equivalent resistance needed is 5 ohms and for that equivalent resistance of firearms 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 1 over R 1 plus 1 over R 2 and so on I can remember for the para formula left hand side is 1 over RP it's not our P equals 1 over R 1 plus 1 over R ok and so if I substitute I get 1 over v equals 1 over R 1 R 1 R 2 R 2 they are all 20 ohms so it's 1 or 20 plus 1 or 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 also not 20 that's it there are n number ok there's some 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 see if there's one I would get 1 over 20 if there were 2 I would get 2 or 20 if there are 5 I would get 5 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 a rearrangement 20 divided by 5 and that gives me 4 and so this means I need to connect four resistors of 2000 me 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 are 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 actually think they are so easy I mean just by using Ohm's law I equals V or 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's 10 R is 2 I would be fine amperes over here for this one again V is 10 R is 4 it'll be 10 by four two point five amperes for this one V's ten 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 the equivalent resistances 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 world 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 3 ohms and now these two are in series so if I add them I will get the final circuit with 5 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 on and from Ohm's law now I can say ah the current over here this current has to be ten by five 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 well this is why while solving this circuit problems it's always nice if we 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 Karamazov flowing over here whoo that means I know immediately the current through this resistor is two amperes and the current of 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 homes law V equals IR this time and so two times two is four IR so this would be four world this would be ir 2 times 3 6 volt but that's all 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 current across these two resistors again when i'm going back from here to here i look ask myself which combination did this resistance come from it came from parallel combination in parallel combination the world age 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 6 volt whatever I have over here the voltage across this is also 6 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 PI R this time I equals v by r I will get 6 by 4 which is one point five and we're here six by 12 gives you or use me point five hams 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 equivalents came from series combo the current is the same you 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 their reading of the voltmeter the relief or 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 one point five 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 graded it a pause and see if you can try okay so we are given power rating of that device and we're 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 is 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 in the picture well we know the connection of voltage current and resistance Ohm's law so we can bring in Ohm's law and put resistance over here one way to do that is by substituting V equals IR you will 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 equal to V by r and so I get V squared 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 Ohm's law to this formula okay and that's important because Ohm's law does not work for all devices all cases something that you may learn in later courses and so anyways if Homestar 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're given the voltage V is given we're asked to find the resistance so I can go ahead and use this formula so make sense so I will rearrange that to give R and now I can all have to do is plug in V is given to be 200 volts a 200 square so that went into 200 because usually things cancel out divided by power which is 100 like I said things like cancelling out and so you get 400 ohms so that's the resistance okay let's do one more in this question were asked can of 50 watt 200 volt light bulb run on a 10 ampere line now at first you may be wondering 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 you have tried over here I think it's better if I draw it 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 is trying to ask is will this go properly okay now how do we answer that question what do you think well what I'm 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 bug won't grow properly but if 10 amperes are less than that is enough to make this bulb cool then the Bell will grow properly right so I just need to know how much current is needed to make this bug grow 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 belt to glow and then I can answer this question so again going back to power power equation electric power is P equals re 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 point two five amperes this means that this bulb will take point two five amperes I will need point two five amperes can this line handle point of amperes of course it can handle up to 10 amperes so will it go yes the answer is yes okay so again important thing don't think that 10 ampere line means it carries 10 ampere no this is the max that is can carry okay how much how much current carries depends upon what what device you put a cross over there right okay anyways then here's a bonus question he's a follow-up a bonus question how many such bulbs can be connected in parallel think about this and see if you in here this a short all right again I'm gonna draw a diagram since you are connecting bulbs in parallel here's my diagram now a lot of Bulls 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 books can we get 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 too we'll take in 0.25 and 0.25 total point five that's also fine three balls are also fine but I'm pretty sure there will be a limit after which the total current will exceed in ambient won't work so how do I calculate this looks like now a math 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 point to five total current taken by n balls will be n times 0.25 and we know that current at max since I want to find what's the maximum number of balls I can put over here at max that number has to be 10 up here does that make sense it's from this hanger where n is and 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'll start drawing more than 10 amperes current and then maybe the wire will melt or maybe there's a few Sun whether to blow up but whatever it is it's not good for us so to answer a 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 iron of resistance 15 ohm takes a current of 2 amperes I'm not even going to tell any more you know what to do okay now at first it might same 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 elect resistive devices like electric iron or heaters things like that that power that energy is usually in the form of heat dissipated 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 ten seconds I just multiply that number by ten so you see this is also a problem on power electric power okay and so I'm gonna go use the same formula P is equal to V into I I know what the whoa 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 rs.15 so if I plug in I square is two square which is four fifteen is just fifteen fifteen times four is 60 so that gives me 60 what 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 generated in ten seconds how much heat is generated well that's going to be 60 times ten so the heat generated is just going to be sixty times ten and that's going to be 600 Jews and so what's important is you don't need a new formula for heat right you might see in your textbooks that heat formula is given as I square R into T they call it the Jules law and 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 twenty hours a day for 30 days given cost per kilowatt hour is three rupees you know what to do how have you 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 his cost per kilowatt are what is this kilowatt are 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 its energy right now normally unit of energy is Joule kilowatt R is the commercial unit of energy the 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 or costing 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 its power in the 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 or the time the time given is twenty hours a day but for 30 days so total how many hours well total we get twenty hours a day times thirty you multiply them right so that this many R's 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 in it 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 ours that's the important thing anyways once that is done we can now just multiply 5 times 2 is 5 times 20 is 100 100 times 30 is 3000 so this is three thousand kilo watt this cancels R so that's the total energy consumed we can also say three thousand units you know what 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 three thousand and that's going to be nine thousand rupees and there is our answer and that's pretty much it just one thing I need you 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