E4 experiment solution on electricity.To determine pd of cell.

Added: 2026-05-26

[00:00:01]All right, good afternoon students. In this video, we are going to be solving everything on this particular manual E4 experiment.

[00:00:11]Now that we have gotten this X the E4 experiment, the aim is to determine the potential difference across a resistor. And we have our readings here. Now these are the readings. These are the readings. Remember that this my reading I use remember that in our previous video we use that E is equal to 2.7 volt then I have taught you how to find all these readings okay we did the manipulations to get all these readings so from this your experiment the examiner says you should plot a graph test question that we should plot a graph.

[00:01:02]He says we should plot a graph.

[00:01:05]The examiner says we should plot a graph that we should plot a graph of R against I inverse starting asex from the origin that is starting both asex Dixon is supposed to be both aix from the origin.

[00:01:25]So if you want to plot this graph, this is how it will look like. Okay, this is how the graph will look like.

[00:01:33]The meaning of starting both axis from origin means that this is 0.0.

[00:01:38]So next point is R. You put your arrow here and label it as ohms. And then this is your what? Your I inverse. Make sure you put the unit of this.

[00:01:51]Okay, this is how the graph will look like. And when you plot this graph, this graph will look like this. This person in here is the internal resistance of the cell. And you can find your slope from here. Okay, this is your slope. This is how the graph of this experiment will look like. But then in this video, I have plotted my graph. So I want to show you my own graph. And then you see my readings so that from there you will know what to do.

[00:02:26]Now look at my graph.

[00:02:29]Look at my graph.

[00:02:31]I plotted this graph with this. Now look at my graph. Look at my graph. You can see the graph. Now this is how the graph will look like.

[00:02:42]I've done my own graph. So when you plot your graph as from Monday, make sure your graph will look like this. Can you see it?

[00:02:51]You see? Can you see the graph? You see how it takes? And then I will show you the scale my scale. I plotted this graph using this my reading. This is the reading I used in plotting the graph. So look at this first. If you look at this graph, you find out that on my arrow ais, which is this point, I will come here and write and make sure you put ohms.

[00:03:18]You put this ohms here. Then this one is my I inverse. Say what I did here. This is my I. Let me bring it so that you can see it very well. Can you see it? Now this is my I inverse and I put a inverse there. I put it in bracket. Now look at the graph very well. You see that on this aix you can see that on this aix big six two cm I have one another 2 cm two. So this is 1 2 3 4 5 6. Okay. So this the scale is going to be 2 cm represent 1 ohm on arrow ais.

[00:04:07]So what I have here is 2 cm represent okay what I did is 2 cm represent one unit on arrow ais yes you can write one unit on arrow ais you can also say 2 cm you can also say 2 cm represent represent what 1 ohm on arrow ais it is even better you use this particular one two cm represent 1 ohm on arrow ais. Okay, are you seeing it? 2 cm represent 1 ohm on arrow ais. I have talked to this before. You can go through my previous video and see how to get your scale. This is correct also. 2 cm represent. Okay, let's get for the other one. Look at this.

[00:05:04]This is 0.5.

[00:05:06]Dex is one Dex is 1.5 D is 2.0 zero. So which means that on this AIX what I have is what on this AIX what is it equal to D6 2 cm represent 0.5 per look at it I have 2 cm represent 0.5 per on what on I inverse what aix You see what I did? So you can write this as your scale on I as you are correct. You can also use the second one. Can also use this particular one. Okay. It all depends on what your lecturer has been marking for you. So you can use this one. You can use this particular one. But for me I prefer this. I prefer this. For me I prefer this. Okay. So that's how you can do this. Now look at this my graph.

[00:06:17]I want you to look at this my graph. So this is how your graph will look like.

[00:06:21]It must not pass from the origin. It must pass from this side. While in exam this intercept represent the internal resistance of the cell. So I am not here to teach you how to plot a graph. So the only thing I will tell you is just for you to look at this reading. Look at this reading very well.

[00:06:43]You can look at this reading. You can pause the video and copy the readings.

[00:06:48]Use this reading and practice the graph.

[00:06:51]Plot your own graph. Try it and see how it will look like. It will be very simple. Okay. So if this is what you have, let's now go straight.

[00:07:04]Let's go and solve the question or the problems. Let's solve some of the problems. The first question here says we should plot a graph of R against I.

[00:07:17]It says we should plot a graph of what?

[00:07:21]Plot a graph of R against I inverse which I've shown you. This is how it will look like. And this is the original what graph. Okay. Now let's solve. Next question says we should determine. So the next question says determine the slope. It says determine the slope x and the intercept on the vertical axis. So I come let's go there. I'll just write that my slope my slope is equal to what?

[00:07:54]Change in R you put the unit. Okay? But ohms all over change in I inverse you put a what? You put a unit. Are you getting it? You're going to put a unit.

[00:08:10]So that my slope here is now equal to let's get this. Let's get it. If you check the X, you have this to be 6.2.

[00:08:23]two we're going to have this this point is four.

[00:08:30]Okay.

[00:08:34]So if you look at it you notice that from this my graph you notice that from this my graph let's this is my slope right this point. Now if you try it here since here is equal to four. Okay, let's count from here. 1 1.5 2 2.5 3 3.5 4 4.5 So this point this is 4 what? Five. This point is 4.5.

[00:09:10]This is 4.5. Okay, it's 4.5. fix point.

[00:09:14]So you write 4.5 minus minus you locate this point.

[00:09:27]Okay, which is 1, which is 1.5. This point is 1.5. This point here is 1.5. So you say -1.5 divided by you come this way if you th if here is two okay let's start from here now and th here is two here is 2.1 2.2 so this point is 2.2 what five 2.25 25. Plot it and you verify [clears throat] this is 2.25 minus if you press this point from your own you have it to be 0.85 because here is 0.6 0.7 0.8 and the middle is 0.85. So minus 0.85 which will now give me what? 3 all over because 4.5 - 1.5 is equal to 3. 3 / what? This one will give me 1.4.

[00:10:43]So that the slope will now be equal to what?

[00:10:48]2.1.

[00:10:49]This is 2.1.

[00:10:51]And this slope must have a unit. Look at the unit. This is e.

[00:10:58]You can say ohm. If this one is crossing over, it becomes like this. That is the unit of this. This is a amp. So that is just it. Or or you can put it this way that your slope is equals to 2.1 amp ohm anyone. But I think I prefer this particular one. So this is how to calculate the slope of the graph.

[00:11:27]Are you seeing it? This how you can calculate the slope of the graph.

[00:11:33]This is how you can calculate the slope of the graph. Okay, this how you can calculate the slope of the graph. And then we have gotten this. So that is correct.

[00:11:44]The next says state two precautions.

[00:11:48]State two precautions. I've talked to you about two precautions here. The first one is that you can write number one I ensure tight connections we are made across the terminals. Very simple.

[00:12:02]I ensured tight connection we are made across the terminal. So you can write this one. It is correct across the terminals. Okay.

[00:12:25]I ensure tight connection. We are made across the terminals. So when you write this one, you are correct. So number two is that I will allow you to use any other precaution of your choice so that it will not be the same thing. It will not look as if everybody is writing the same thing. So use any precaution of your choice and fill in this number two possibly the one that you have been taught. Okay, use this. While you are doing this, I advise you to join the white father day tutorial classics in unis. If you have not joined us, join us now. Join us now. Now more especially now that I started physics 101 102 sorry physics 102 from the beginning. Okay.

[00:13:16]So let us look at this second question.

[00:13:18]Next one says question number four. He says a parallel combination of 3 ohms and 6 ohm resistors is connected in a series with a resistor of 5 ohm and a battery of internal resistance 1 ohm.

[00:13:33]Calculate the effective resistance in the circuit. So class this is the diagram first. You need to learn this before you go there on Monday. This is the diagram. This question is on your manual. So this is the diagram they said a par combination of this is 3 ohms and 6 ohms they are connected in par you can see it resistor is connected in series excuse me it's like that is a mistake all right it's not a mistake the question is correct the question is perfectly correct it says a par combination of 3 ohms these are the par combination of 3 ohms and 6 ohm is now connected in series with a 5 ohm resistor. Look at it. And then this is connected across a set of internal resistance R small R which is equal to 1 ohm. So the examiner says we should calculate the effective resistance of the circuit. So how can I solve this? I will just write my answer.

[00:14:42]I will write answer or solution.

[00:14:45]So I'm going to say let this one be my R1. Let this resistor be my R1. I make it to my R1 which is 3 ohms. Come on.

[00:14:55]Let this one be my R2. I will make it to be R2 which is equal to 6 ohm. Are you getting it? And then let this particular one be my R3 which is equal to 5 ohm resistor. So the first thing I will do is to say let let the effective resistance of the par connection. I can say let the effective let the effective resistance let the effective resistance that is this two of the par connection let the effective resistance of the par connection be what? Let it be equal to R4.

[00:15:48]Let it be equal to R4. That means since these two resistors are in par, I will label it to be equal to R4 and get their effective what? To get their effective resistance. So now there is this formula that if I now want to get this one in par, remember the one you are taught in secondary school is just that small 1 / R1 + 1 / R2. So I'm not going to use that formula. I will use this formula to combine this one in parallel. You use that R4 now is equal to what?

[00:16:22]R1 * R2. That's * divided by what? R1 + R2.

[00:16:33]So this is when two resistors are connected in par and then you want to get their effective resistance.

[00:16:41]This is the formula you will use. So it's supposed to be D resistor mult* by D1 divided by divided by D and D. So what will be the effective resistor? That's the simplest shortcut. D will now be 3 * 6 all over 3 + 6 which is 18 / what? 3 + 6 is what? 9. So 18 / 9 18 / 9 is going to give me what? 2 ohms.

[00:17:18]Is going to give me 2 ohms. So 18 / 9 will give me 2 ohms. So you see that the par resistors is equal to 2 ohm. This is what I have. Even if you use that your 1 / R1 plus 1 / R2, you still get it to be equal to 2 ohms. is the same thing but this is the shortest method of solving this. Okay. So now that we have gotten that now that we have gotten that effective resistance between these two person that are in par is equals to true. So we need to connect it in series with this now to get the the overall the effective what resistance. So I can now say or I can put this one say therefore I can say therefore I can say there are four I can say therefore arrow four and arrow what arrow 3 are now connected in what in series.

[00:18:26]So when you connect this thing in series what will be the effective resistance?

[00:18:30]Therefore the effective resistance EQ become what?

[00:18:35]R4 + R3 which will give me what?

[00:18:40]2 which is equal to 5. Okay which is equal to 2 ohm plus what? 5 ohm which will now give me what? Seven. Which will now give me seven. But remember we are asked to calculate the effective resistance of the circuit where you were given that the internal resistance arrow is equal to 1.0.

[00:19:05]So what I'm saying is that if you write this as 7 ohms for them some lecturers will mark this but it will even be better you just put this as 2 + 5 then plus that 1 ohm which will now give us 8 ohm w resistor.

[00:19:24]which is equal to 8 ohm resistor. Okay, which is now equal to 8 ohm resistor because the examiner says we should calculate the effective resistance in the circuit.

[00:19:38]Okay, but I'm saying that even if you stop here, the examiner will still give you your mark. Are you getting it? So that's exactly how we can solve this.

[00:19:51]So let's take the last question which says that an electric bell takes a current of 0.2 amps from a battery of two dry cell connected in series. Each cell has an emo of 1.5 volt and an internal resistance of 1.0 ohms.

[00:20:12]Examiner said from your question there calculate the resistance of the cell.

[00:20:21]I I says what amount of current will be delivered to the bed? Says take if the cell we are arranged in par.

[00:20:32]So this is very simple and if you are looking at your manual you see that this is the diagram of the circuit. try to learn it on time while you are watching this video. Use your pen and then you write all these things down. So this is the the the secure diagram. This is a secure diagram. This is the two sets that we are connected in series each emf of 1.5 volt. EMF of 1.5 volt. The internal resistance is one. The internal resistance is one here. So I will now say solution or I can say answer solution.

[00:21:14]I'll say solution using the formula using the what the formula because I want to calculate the resistance of the cell. The formula we use is that I is equal to E all / R + R. This is the formula I'm going to apply.

[00:21:36]Are you getting it? So, how can we get this our E? This is the formula we apply to make this subject formula. That's what I'm looking at. Okay, I want to make this our subject formula and get it because that is what they say we should calculate this out. Now, how can I get this E? Remember that these two sets we are connected in series.

[00:21:58]So I will say E that is the total em of the set is now equal to E1 + E2.

[00:22:09]The reason is because I can say since they are connected in what? thinks they are connected in in series because when two sets are connected in series to get their effective emo you add but when they are connected in par their effective effective emf is the individual emf of the set. So my E is going to be E1 plus E2. Okay.

[00:22:43]So that my E now will now be equal to 1.5 + 1.5 which will give me 3.0 volt. So I've gotten my E to be 3.0 volt. Remember that from this formula from this formula I've gotten E. So we need to also find this. So how do I get my R? To get the internal resistance.

[00:23:14]So to get R, we say that R is equal to R1 + what? R2.

[00:23:24]Since they are also connected in series.

[00:23:27]Since they say are in series, look at it. Since these two sex are in what?

[00:23:34]Since they are in series, D and D are in series. Okay. So, it's going to be 1 + 1, which will give me 1.0 + 1.0, which is equals to 2 ohms.

[00:23:50]So, I've gotten my R to be equal to 2 ohms.

[00:23:54]So now using the formula I is equal to E / R + we are given that I is equal to what 0.2 amp from the question arrow is this and my E is equals to 3.0 0 volt and what is my R 2 ohms. So from here I have this as 2 ohms I have this as unknown which is what I'm looking for.

[00:24:28]I have E as 3.0 and I is this. So what I will do is just to substitute all these things and then I'll get my final answer. Now look at it. The I was given as 0.2 amps.

[00:24:42]So we now move by saying 0.2 is = 3 all over arrow plus what?

[00:24:52]+ two. So next thing I will do is to cross multiply.

[00:24:59]So when you cross multiply when you cross multiply you have that this is 0.2 2 bracket r + 2 which is equal to three. Right? So when you use this and multiply this you have 0.2 r + 0.2 * 2 will give me 0.4 which is equal to 3.

[00:25:24]So 0.2 r will give me 3us 0.4.

[00:25:30]This is because this one will go over to this side and 0.2 2 R will give us what?

[00:25:37]I'm having it to be 2.6.

[00:25:39]So only my R will give me 2.6 / 0.2 which will now give me 13 ohms. So therefore the R of the B is equal to 13 ohm which is the final answer. So this is going to be my final what answer.

[00:26:00]Okay. So I said use your pen and copy all these things then you thank me later. So join the white father day classes now before it will be late. That is the answer. Now the last question says I I of that six look at it says it says we should calculate the what? It says what amount of the cell sorry it says what amount of current will be delivered to the bell take if the sex were arranged in par. So if the sex we are arranged in par for I I this is how it will look like when the sex are arranged in par you have this okay this is what we're going to have. So this is my E2 and this is my sorry E1 and E2. So I will also use the formula using the formula the formula I will use is this that I is equal to E all over R + R that's the formula I apply so since they are in par the effective EMO E is equal to E1 which is also equal to E2 okay I will explain more in electricity in physics 102. So that is what I'm going to be having which is now equal to 1 what 5 volts that's 1.5 volt. So we have gotten in from this formula. Okay we have gotten in from this formula and then our R is equal to what? 13 ohms.

[00:27:49]Our array is 13 ohms.

[00:27:52]from here from this point this is the R which is 13 ohms because you have gotten the value of this R as 13 ohms. So this is what I will have that my R is 13 ohms and now what is my R? What is my R?

[00:28:10]Since my R what is my R? So since these two sets are connected in remember that this is one and this is also what? This is also one. So since the two sets are connected in par since they are connected in par one can say that our r so I can use that your normal secondary score 1 / r1 is equal to 1 / r + 1 / r2. So we can use this to find the effective internal resistance of the cell. So 1 / R will now give me 1 / 1 + 1 / 1. So 1 / R will give us what? 1 / 2. So when you make your R subject formula arrow will give us what?

[00:29:09]No, sorry.

[00:29:10]Sorry.

[00:29:12]This is what I will have. Sorry. This will give us that. Sorry. We're going to say that 1 / r is equal to the lucm here is 1. Okay. The lucm here is going to be one. One go one. So this is 1 + 1 which will give me 2 over 1. Right? So we now have that 1 / r= 2 / 1. Then when you cross multiply this because you want to make r subject formula you have that 2 r is equals to 1. What is r? 1 / 2. So r is 0 what? 5. So our r will be 0.5.

[00:29:57]Our r will be 0.5.

[00:30:00]Right?

[00:30:02]So my R will be 0.5.

[00:30:04]So if my R is 0.5, we can now say that therefore I is equals to E all over R + R. So the current will now be what?

[00:30:19]1.5 / 13 + 0.5.

[00:30:27]So the current will give me 1.5 all divided by 13.5 which the final answer will give me 0.11 amp. So the current for this I I is 0 what 111 amp. So when you solve it you are going to score 30 over 30. This is my prayer for you. So for those of you that wants to join the white far day, join us now that we just started physics 102 and mass 102. Join now. Rush us. So this is my phone number 070 36221432.

[00:31:12]Thank you and God bless you. I love you guys. Bye-bye.