 The previous lecture we looked at control sources. Now so far we have looked at two terminal elements and also control sources and what we have done is to describe the relationship between the voltage and current of these elements. In this lecture what we will look at is the power and energy in these elements. Now before I get started are there any questions regarding the previous lecture or any of the previous lectures? Okay I think now you are able to see the journal as well as the webcam video. Is that correct? If someone is not able to see that please send a message in the chat window. You are able to see me. So if there are any questions regarding the topics we covered in the previous classes please ask questions. You can type into the chat window. Let us continue with the lecture elements. The independent sources, voltage and current source and also the resistive, capacitive and inductor and each of these is defined by some relationship between the voltage across the element and the current through the element. All these relations are. In this case I have shown all the elements with this passive sign convention okay that is if V is defined positive on top and negative on the bottom I goes into the upper terminal, the terminal on top okay. Now this product V I that is the product of the voltage across the device and the current through the device has some significance and that is nothing but the power that is going into the device okay. So with this sign convention with the passive sign convention V times Y is the power that is going into the device. Now this either can be constant or changing with time depending on V and I okay. If V or Y is time varying it is possible that this power is time varying and in that case it is known as instantaneous power. So this is what the instantaneous power is and this if it is positive the power is going into the device and if it is negative power is drawn from the device okay that is P of T is less than 0. So this is how the power is defined and it is extremely important to follow the passive sign conventions that is the appropriate signs of the voltage and current across the elements okay. Now what we will do is we will look at each of these elements and see what the power is okay because we know that the element enforces certain relationship between the voltage and current that means certain things for power and we look at that shortly okay. So let us take a resistor some V and some I V equals I times R where R is the resistance value okay. Now the power P of T is V times I now in general it can be dependent on time. Now the resistor does not care that for V and I are changing with time and each instant of time the voltage is proportional to the current and that instant of time and the proportionality constant is the resistance R okay. So if I substitute V equals I R from here I will get I square times R alternatively if I substitute I equals V divided by R I will get V square divided by R okay. Now given this type of relationship please tell me what is special about this okay we have V square by R or I square by R. So what does it signify? No answers but I think the probably the question was a little vague. My question is please concentrate on the fact that we have I square that is square of some quantity or V square so what does that signify? The answer V square and I square in this expression imply that positive okay or it could be 0 but it is never negative okay. This means that a resistor always absorbs power okay you can never get power out of a resistor. The resistor always absorbs power and typically we also say instead of absorbs dissipates. So in practice if you connect some voltage across a resistor some current will flow and the resistor absorbs the power of V square by R and it will go on heating up okay. So that is what usually happens in that is what happens in practice okay. The point is that the is always positive which means that the resistor absorbs power okay. So I think this must this is pretty clear. Now let me take some other elements let me take a voltage source V equals V naught that is the voltage of the voltage source is this quantity V naught and I have some current through the voltage source. Now I would like answers from the participants. What do you think this does? Will it absorb power or what is the condition or can we say anything about it at all? Can we make any statements about what the power is going to be in a voltage source okay. I see a number of answers that some of you say it absorbs power and some of you say it provides power or it delivers power and so on okay. Now the fact is that you cannot tell just by looking at this it can do either. See the point is P equals V times I. Now let us assume that the way I have defined it V naught is positive okay. So let us say V naught happens to be 5 volts okay. Now the property of the voltage source is that the current can be anything it constraints the voltage across it to be V naught but the current can be anything. Now if the current is positive a positive current is flowing in the direction shown P is V naught times I and I have already said V naught is positive and that will be more than zero okay. In this case it absorbs or it dissipates if I is less than zero that is a current is flowing out of the positive terminal when I is negative and P equals V naught times I will be less than zero at delivers. If you have a voltage source you cannot say that it is going to definitely absorb power or delivers power depending on the configuration of the circuit the current will be either going into the positive terminal or coming out of the positive terminal. So depending on these things it can either absorb or deliver power okay. Now we also know the I V characteristics of the voltage source equals V naught this means that the I V characteristics of a voltage source of value V naught and in this case I have assumed V naught to be positive okay. Now if the axis divides the plane into four quadrants now please tell me in which quadrant the voltage source would absorb power and where it will deliver power that is the value of I can be anything right you can see that the characteristic of the voltage source is in the first and fourth quadrants. So if my question is in which quadrant should the voltage source be operating in order to absorb power. Again there are some answers and this is really simple okay because in this quadrant in the first quadrant V is more than zero and I is more than zero. So obviously the product is positive okay in the first quadrant and in the fourth quadrant V is more than zero but I is negative. So obviously V is I V which is less than zero so this delivers power if it is operating here and this absorbs if the voltage source is operating in the first quadrant it absorbs power if it is operating in the fourth quadrant it delivers power and what is meant by operating in this quadrant in the circuit the current value happens to be such that it is in the first quadrant okay. Similarly when I say in the fourth quadrant current value happens to be such that it is in the fourth quadrant. So this is just another way of putting whatever I said earlier if I is less than zero it will be in the fourth quadrant okay. Now we can also take the case where this V naught happens to be negative okay it shows us in trend if V naught is negative then the characteristic of the voltage source would be over there. Now in which quadrant would it be absorbing power and in which quadrant would it be delivering power of answers but the principle is extremely simple you look at what the signs of V and I are and from that you determine the sign of V times I okay. In the second quadrant V is more than zero the voltage is negative and the current is greater than zero. So this is the same as this one okay as far as the IV product is concerned so in the third quadrant also it delivers power in the second quadrant it delivers power and in the third quadrant when V and I are both negative it absorbs power okay. We asked and you already asked let us say you have a circuit with number of sources and whether it is absorbing power or dissipating power in a particular condition you have to see whether this quadrant it is operating in okay finally you just have to calculate the product of V and I with the appropriate signs and decide whether it is absorbing power or dissipating power okay. In general whether it is a voltage source or whatever element it is in the IV plane I times V is positive in the first quadrant and I times V is positive in the third quadrant also so the element if it is operating in the first quadrant the third quadrant will be absorbing power and if it is operating in the second quadrant the fourth quadrant I times V will be less than zero and it will be delivering. So the question from Sadeq please go ahead now with this we can clearly see what happens with the resistor or a voltage source or a current source okay. First of all if we had a resistor what would be the characteristics it is a straight line passing through the origin okay. So the straight line the resistance characteristic would be something like that okay so you see that it is only going through regions which are absorbing power okay so the resistor always absorbs power exactly the same conclusion we reach from the power being either V square by R or I square times R okay. Now if you have a voltage source it is a positive voltage source with the chosen signs it can be either here or if it is a negative voltage source it will be over there okay it will be one of these two. Now you can see that it is either it could be either absorbing power or delivering power okay. Now similarly if you had a current source a positive current source with the chosen signs would be here and it can be either delivering power or absorbing power you would actually calculate the voltage across the current source to figure out what it is doing similarly not as negative if the current source has a negative value then also it can be either delivering it can be either delivering or absorbing power okay so again you have to calculate the uh calculate the voltage across the current source to be able to tell whether it is absorbing power or delivering power okay. I think now it must be pretty clear I did the voltage source in more detail but it must be pretty clear that even the current source can be either absorbing power or delivering power okay. Any questions about this any confusion some questions may be some confusion with the negative source a negative source simply means that the voltage value is negative that is all that is all you have to do okay. So just to get some practice let us do a couple of problems which are very simple okay what I will do is I will connect right plus minus like this and say 5 volts it means that in this polarity it is 5 volts okay that is the meaning of the symbol of the voltage source and similarly 2 amperes like this that means that 2 amperes is flowing from bottom to top please uh uh answer this now we have two sources in the circuit one is a voltage source and one is a current source now let us try this I will try to pull for the first time you can answer with the pull please mark the choices once I announce the pull voltage source is absorbing and the current source is delivering power the current source is absorbing and the voltage source is delivering the power okay. So these are the choices let me try the pull uh quickly okay you should be able to see the pull please answer if it is a b c or b setting in front of uh uh computers please try to answer the pull and those of you who are in an institution uh in front of a screen maybe you can announce your majority opinion to the mentor and then they can enter the pull okay please try that there is a feature of this adobe connect which we can try and use if it is useful we will do it more often okay and I think so many of you are seeing quiz shows and lots of things with sms voting so you should be very comfortable with this type of uh this mode of answering questions okay I think we can uh see the participants answers and in fact it is correct a majority opinion says it is c the voltage source is absorbing power and that is the current answer okay and in this case the answer is quite simple the uh what we do I mean always when you are not sure and especially in the beginning uh when you are just studying things just be systematic and calculate the uh calculate the power or anything else systematically okay now all the voltage source if I take v in this direction passive convention says i has to be in that direction okay clearly we see that v is plus 5 volts here by the way I have defined the voltage source and i is 2 amperes okay clearly v times i is plus 10 and the product of voltage and uh current is gives you watts okay this is something you would already be familiar with by its positive similarly if I take the passive sign convention for the current if I take v like this that is 5 volts this means that I have to take i in this direction and that is minus 2 amperes 2 amperes is flowing from bottom to top which also means it is flowing minus 2 amperes is flowing from top to bottom so v times i for this is uh minus 10 watts okay so that means that the current source is delivering 10 watts of power which is going into the voltage source okay now clearly uh this choices a and b will be observed I mean if you have a circuit with only two sources it cannot be that both are absorbing power or both are delivering power okay because that violates basic conservation of energy principles if both are absorbing power then you can ask where is it coming from similarly if everything is delivering power you can ask where is this going okay so it's only two elements it's the only possibility there is for one of them to be delivered delivering power and for the other one to be absorbing power in this case the voltage source absorbs power okay now just for a little more practice I will give you one more question I will make this 5 volts that is uh what is the meaning of this is that it's the same as saying it has 5 volts in this direction okay I have written this plus and minus inside the circle and say minus 5 volts that really means it is 5 volts with bottom uh the plus sign and top is the minus sign okay well I will introduce just one more complication okay and that is that I will connect the resistance here which is 1 ohm okay now uh voltage source please let me know whether it is delivering the question is in this circuit is the voltage source absorbing power or delivering power now and the majority opinion says that it is uh the voltage source is delivering power and that is the correct answer if you want I'll analyze it later and show you now let's make things a little more uh uh complicated okay my question to you is decided that the voltage source is delivering power so now we have decided that the voltage source is delivering power but of course we are doing engineering that means that we have to be able to quantitatively calculate things so now please mark how much power is it okay is it 5 watts or 10 watts or 7 watts or 3 watts what is it sorry the last choice is uh 25 watts not 3 watts okay so it is delivering power but how much is it is it 5 watts 10 watts 7 watts 35 watts something else I miscalculated everything I'll work for a couple of more responses before I close now if you look at the results of the pole maybe both b and b that is 10 watts and 35 watts have uh got more or less the same number of watts and I can see more rewards for uh 35 watts coming in of course I mean this is engineering although this is from we don't decide things per width there is only one correct answer here and let's calculate one of this okay many of you have said 10 watts because the answer to the previous question or there was some power of 10 watts in the previous question where we had only a voltage source and the current source remember in this case we also have an additional component the resistor so there is no reason that the answer should be exactly the same as before okay it could be coincidentally the same but you have to calculate it nonetheless okay now uh first of all basically to figure out how much power the voltage source is delivering what we need to do is to find out the current through the voltage source we already know the voltage across the voltage source so all they need to do is to find the current through the voltage source okay now this is a one ohm resistor and it's in parallel with the voltage source so obviously the voltage across this is 5 volts in this direction of minus 5 volts with the opposite polarity so what does it mean that means that the current in the resistor would be 5 volt divided by 1 ohm equals 5 ampere okay now you can clearly see that at this node we have 5 ampere through the resistor and 2 ampere through the current source and the remaining branch to satisfy case here must have 5 ampere plus 2 ampere equals 7 amperes okay now as far as the voltage source is concerned the voltage with this polarity is minus 5 volts and the current with that polarity following passive sign convention is 7 amperes okay so v times i is 7 amperes times minus 5 volts which is minus 35 watts this minus tells you that it is delivering power and the 35 watts is the amount of power that is delivered so the correct answer is b okay now it's very easy to generate even more problems of this type you can go on adding resistors and current sources and voltage sources and so on okay and you can get some practice yourself of course if you go to a basic book like Haydn, Kemmerle, there are a number of problems but even you can generate these problems yourselves okay the only thing I would say is be very systematic and simply calculate v times i okay based on the uh Kirchhoff's current log Kirchhoff voltage log and so on and for each element do that properly and you will get the right answer okay in this itself we can take as an additional exercise uh delivered or absorbed by the current source absorbed by the resistor okay you can calculate these things for yourself and maybe discuss it on the forum or whatever is convenient for you okay now I hope it's very clear that a resistor always delivering voltage source or a current source can absorb power or can deliver power depending on the circuit they are connected in okay now let me ask you a question I have a circuit with a single voltage source and a number of resistors okay I have only one voltage source and a number of resistors now uh can the voltage source be uh absorbing power is that possible I'm talking about a circuit with a single voltage source and number of resistors you know in January the voltage source could be absorbing or delivering power now in this case can the voltage source be absorbing power if so why or if not why not clearly I think all of you guessed it correctly if you have a single voltage source and all resistors it can it has to deliver power okay you cannot have all the elements in a circuit absorbing power then the question is where is the power coming from okay so you have to have at least one source in the circuit which is delivering power now depending on the circuit it could be more than one source but whatever the circuit you have you the at least one of the independent sources that you have must be delivering power okay so that part is correct now uh having discussed voltage sources current sources and resistors let's move on to what happens in inductors and capacitors okay relationship for the capacitor the capacitor current is c times dv by dt the instantaneous power v is v times i which is v times c dv by dt this itself looks a little uh uh complicated and we can't tell whether it is absorbing the power at some instant or delivering power okay either is possible depending on uh if v and dv by dt are both positive it's absorbing power similarly v and dv by dt are both negative it is absorbing power or if v and dv by dt have opposite sense it is uh it is delivering power okay so all of these things are possible now it is more interesting to look at uh slightly different quantity okay let me rearrange this first this is c times v times dv by dt now from some now from basic calculus you know that if you have if you try to differentiate v square what you will get is 2v times dv by dt okay time derivative of v square is 2v times v dv by dt and we have something like that over here okay so we have uh v times dv by dt basically so what we have is c times half of d by dt of v square okay i'll say c by 2 d by dt of v square okay so the instantaneous power that is going into a capacitor turns out to be directly proportional to the time derivative of v square okay and this itself can be either positive or negative right now uh a relative quantity to power is what is known as energy okay energy is nothing but the integral of power over time okay so let's say you would take a definite integral from t1 to t2 and you have some uh element with v across it and i throw it we know that power itself is v times y okay so the energy e that is absorbed by the element absorbed by the element is given by the integral of dt and when i put some limits here t1 to t2 okay this makes that energy e absorbed by the element over time from t1 to t2 okay so that is what we meant here so energy e and depending on how long you wait the different amounts of energy will be absorbed okay it's an integral quantity so you have to specify the time over which the energy is absorbed and that is equal to the integral of v times y dt okay now a relative definition we can look at what happens in a capacitor right we power is c by 2 d by dt of v square let me write the capacitor again vi and this is the capacitor c and the power which is v times y can be written as c times v times dv by dt which is c by 2 time derivative of v square okay now let me uh integrate this okay and i'll also choose to be zero and i'll assume that the capacitor has zero volts across it okay this is just a starting point i start from a capacitor that has zero volts and then vi and i are varied in some way okay they are related by i equals c dv by dt maybe i vary me or i vary i it doesn't matter but they vary in some way okay and finally i reach a certain time t2 okay so the energy absorbed by the capacitor from t1 to t2 there is nothing but integral c by 2 d by dt v square dt from t1 to t2 okay and let's say that at time t2 capacitor has certain vc volts now you can see that this time derivative operator and the integral cancel each other and you will be left with c by 2 times v square with the value of voltage voltage at t2 okay and this is given by c by 2 vc square so what this means is that a capacitor if you charge it up to a voltage vc will have energy it has absorbed all this energy right so it will have an energy c by 2 times vc square across it okay and also just like the power in the resistor was always positive the energy in the capacitor is always positive vc itself could be either positive or negative but the capacitor would have stored some energy okay is this clear any questions about this part of it what i did was try to find out what the power dissipated in a capacitor was and it comes out as some not power dissipated power absorbed by the capacitors and it comes out as some derivative of something now energy which is the integral of the power comes out to be half c vc square okay half of capacitance times the square of the voltage and that is always positive so if you charge a capacitor from its discharge state from 0 volts to a voltage vc it would have absorbed an energy or it stores an energy of half c vc square okay any questions about this how we delivered it or any sticky points so now i also started from t1 equal to 0 where the capacitor has 0 volts across it and nothing is stored in it okay and no charge no previous no energy and i took it all the way to voltage vc and it has an energy c by 2 vc square stored across it okay so from this we can say that but now the important thing is i didn't worry about how exactly the voltage changed okay it went from 0 to vc in whatever manner but finally the energy stored in the capacitor is half c v square okay so the energy in the capacitor is not dependent on how you reach that voltage but it's dependent only on the voltage across the capacitor okay now it doesn't matter let's say i plot the voltage across the capacitor v as a function of time t and let's say this is the voltage vc now i could have reached it as a straight line i could have reached it in some strange way like this okay i could do even wilder stuff and go there so as long as initially it is at 0 so that means its energy everything is 0 and finally i reach a voltage of vc so that means that once i reach here i would have totally absorbed an energy of half c vc square and that is stored in the capacitor okay now there is a question on what happens when you discharge the capacitor okay so let's say again so i called this time as t1 and i'll call this t2 where it is at vc and let's say it goes to 0 from here okay and i'll show some arbitrary way no it'll go to 0 at a time t3 what is happening during this time integral from t2 to t3 c by 2 of those whole thing okay this is the energy absorbed by the capacitor during this time okay now what do we get again the integral and the time derivative cancel each other and we'll end up with c by 2 v square with the voltage at t2 and the voltage at t3 okay now what is this going to be this will be c by 2 and voltage at t3 is 0 because at least 0 volts minus voltage at t2 is vc okay so we get minus half c vc square okay what it meant is it is negative so that means that the capacitor gave off the energy okay so if it is positive the capacitor absorbed energy and if it is negative the capacitor delivered energy okay so if you charge the capacitor to vc it would have stored an energy of half c v square and then now if you discharge it back to 0 it gives back the energy okay so the capacitor does not dissipate energy uh whereas in a resistor the energy you put into it you cannot get it back whereas a capacitor you charge it to some voltage it draws some energy from the source okay and then you discharge it back to 0 it gives back all the energy to the source okay so a capacitor does not dissipate any energy but it can store energy I hope that is clear so that means that the energy stored in a capacitor can be recovered okay you can use that and it is one of the important uses of a capacitor you store energy in it and then you can also get the stored energy okay now let us take the example of an inductor it is very similar to that of the capacitor proportional to the time derivative of the current so v times i which is l times i times d i by dt and this itself can be either positive or negative and we use the usual algebraic trick to write this as l by 2 d by dt of i square okay this itself is not an interesting quantity as the energy absorbed by the inductor equals and for the duration t1 to t2 equals the integral from t1 to t2 of l by 2 d by dt of i square with respect to time so again this will be half l and i square this will be the value of i at t1 and the value of i at t2 okay so if you start from an inductor that is that has zero current in it and go to a current i l okay the current of i l at t2 the energy absorbed would be i square from zero to i l which basically gives you half l i l square this means that stores an energy square that is half of inductance times the square of the current through the inductor and just by similar logic that we employed earlier okay we can show that by discharging from some current i l to zero current you can recover all of the energy okay so the inductor also does not dissipate any energy and it stores an energy of half l i l square it also doesn't define how you reach i l you could reach it gradually you could reach it abruptly and you could go back and forth between positive and negative values and reach that okay so as long as you are at the current of i l you will have an energy stored which is half of l i l square in the inductor okay any questions about this okay there is one question it says basically it asks what is the difference between an inductor and a capacitor now i and v relationships are different okay for a capacitor we have a voltage v across it on the current i through it i is c times dv by dt and in an inductor if you have a voltage across it on the current through it i is l di by dt okay so as far as circuits as far as the terminal characteristics are concerned it is like the roles of current and voltage have reversed okay in a capacitor the current has the time derivative of voltage and in an inductor the sorry i made a mistake here this should be v the voltage is the time derivative of current okay so that is all there is there and physically of course there is a difference a capacitor stores energy in the form of electric fields and an inductor stores energy in the form of magnetic fields okay and in other words another way of saying it is a capacitor stores voltage and an inductor stores flux linkage okay inductor stores current inductor stores are current interesting question which is that if a capacitor behaves as if a capacitor can absorb power and also deliver the power is it an active element or a passive element okay so this was the question again i would probably turn this back on the participants and ask what do you think it is is a capacitor an active element or a passive element and of course exactly the same question can apply to an inductor a capacitor can deliver power but deliver energy but the point is that you still consider it a passive element it can only deliver the energy energy which is stored in it okay so you start from a capacitor that is discharged now it cannot deliver any energy it can absorb some energy and then give it back okay that's a very useful mode of operation for a capacitor or an inductor you can use that to store energy sometimes what happens is there are many cases in which circuit needs a lot of energy at some particular instance so what you do is you arrange the circuits such that the capacitors is charged up and it stores energy and during those instance of very high energy demand it can give out energy okay but we still consider it a passive element because you first have to supply energy to it and then take it out okay i hope that is clear okay now before we go further let's be clear about a few things resistor V power is R V square by R always dissipates okay and a capacitor and inductor the energy is half CV square in a capacitor and inductor okay where i is the current at that instant you see that the instantaneous power in a resistor will be the instantaneous voltage square divided by R over the instantaneous current square times R now the instantaneous energy stored in a capacitor is the voltage at that instant square times half CV similarly the instantaneous energy stored in an inductor is the instantaneous current I times half L okay and we also said that the voltage in current source can either absorb or deliver power okay there is V times Y and you can also find out the energy you can integrate the power over some time to see how much energy has been delivered by the voltage source or the current source over a given period okay now we know that the voltage V is measured in volts and with the symbol positive V and the current I is measured in amplifiers the symbol is uppercase or capital A okay now the power which is V times I which is volts ampere is measured in is denoted by units of watts is measured in units of watts denoted by W okay so one watt equals one ampere times one volt and also equals one ampere square times one ohm from this popular this is just so that you get used to a different ways of arriving at the units of watts and it's also equal to one volt square divided by one okay the energy E is measured in joules denoted by J okay and we know that the energy is the time integral of so one joule equals one watt per second okay so I think these are mostly are familiar with these things but I am just refreshing your memory okay so we know that clearly the dimensions have to be consistent so half C V square is the energy in a capacitor so if you multiply part heads with volt square you will get units of joules similarly half L I L square is the energy stored in an inductor and if you multiply henry's the units for inductor with ampere square you get joules again okay so whenever you do any calculations please please do it complete with units okay there is not much point giving answers like the current is one right it doesn't make any sense it has some units so it has kind of one amp for one milliamp one so on okay similarly the energy could be joule or a picot joule or nano joule or whatever it is everything has to be specified with the everything has to be specified with proper units all right after that I made a mistake here the power is the rate of change of energy so one watt is one joule per one second or one joule is one watt times one second okay this means that if you deliver one joule of energy in one second the average power is one watt alternatively let's say you have a resistor and it is dissipated one watt and if you wait for one second it would have dissipated one joule of energy okay and now we can do a couple of calculations just to get you used to these calculations let's say I have one milliamp current so I think all of you know the pre-pixels nearly is 10 to the minus 3 and micro is 10 to the minus 6 nano is 10 to the minus 9 and picot is 10 to the minus 12 these are the ones that we most frequently use okay a 2 kilo ohm resistor connected across it okay now the question is what is the power dissipated in the resistor how much power is dissipated in the resistor please answer this how much power is dissipated in the resistor okay so now the current here milliamp and the power dissipated there is nothing but i square times r which is 2 times 10 to the minus 3 this is 2 milliamp right square times sorry 1 milliamp not 2 milliamp 1 milliamp times r which is 2 kilo ohms and kilo is nothing but by the way the average subscripts so the prefix is more than the unity kilo 10 to the 3 and mega of 10 to the 6 okay and giga of 10 to the 9 so this will give you two times 10 to the 3 ohms the entire result will come out to be 2 times 10 to the minus 3 watts or 2 milliwatts okay so the power dissipated is 2 milliwatts some of you said it is 2 watts but please mind these units properly okay now next is let us say a 5 volt voltage source across a 10 nano farad capacitor okay how much energy is stored in the capacitor how much is that please calculate and similarly you can also calculate a 1 milliamp is flowing through a 100 milli henry inductor how much energy in the okay i mean by and large you have got this it is just simple arithmetic here it is half and 10 nano farads which is 10 times 10 to the minus 9 farads times v square which is 5 square which gives you this is 25 25 times 10 is 250 divided by 2 is 125 125 times 10 to the minus 9 joules or 125 nano joules and in this case the inductor energy is half li square the current is 1 milliampere that is 10 to the minus 3 amperes times sorry the inductor is 100 milli henrys which is 100 times 10 to the minus 3 henrys times 1 milliampere which is 10 to the minus 3 square okay and this gives you 50 times 10 to the minus 9 joules or 50 nano joules okay do any of these calculations by yourselves from here onwards okay any elements or any simple circuit if you should be able to calculate the voltage and current in each element and then calculate how much power of energy is dissipated or delivered in each element okay any questions about anything we have done so far so what we have learned in this energy is the definition of power and energy in electrical elements an inductor and capacitor store positive energy and they can also be recovered they do not really dissipate or lose any energy so that is an important feature that distinguishes them from resistors which constantly lose energy okay they all continuously dissipate power as long as some current is flowing through them and voltage source and current source are the sources of energy but it is possible for them to also dissipate energy or power okay so if there are no further questions what we will do is we will stop here and continue with the next lecture where we will discuss another element which is known as the mutual inductor and then move on to circuit analysis okay any questions okay then I will see you in the next lecture