In school, I have to memorise some equations, so I made it into a course. Check it out: https://www.memrise.com/course/1512347/circuit-equations-basic/
P.S. If I made any mistakes, please tell me!
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Hi, interesting course.
In level 3, I think your first Energy Formula should probably read:
E = (1/2) Q V
- and certainly, if you’re thinking about a capacitor.
A better known version of this formula would be:
E = (1/2) C V^2
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html
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From what I learnt, the energy formula is E = Q x V
@Mahirislam You’re correct that E = Q x V. This is simple to prove from the fact that the definition of the Volt is Joules/Coulomb, or V = E / Q; rearranging that definition yields, Q x V = E.
Hi @xvg11,
I don’t think you can depend on the definitions of units to give you the correct formula.
I’m certain that you need to include the factor of (1/2) in any real situation. See the explanation at 1’30’’ in this video:
@ian_mn You are extremely mistaken. You cannot ignore the fundamental definitions of the physical quantities. They are definitions, or mathematical identities, and are not susceptible to differing opinions.
I don’t need to look at the video to know the derivation of the formula for the energy stored in a capacitor. It is indeed 1/2 CV^2. This is a result of integrating the work done on each unit of charge as the capacitor is charged, and the voltage rises from its initial value of 0 to its final value of V.
The equation for work done to charge a capacitor depends on the definition of work being E = Q*V.
I know this stuff inside out, and could work out the equations in my sleep. If you review any physics 101 text, you can easily confirm that what Mahirislam and I am saying is correct.
Hi @xvg11, I think you might need to watch the video. Anyone directly applying this formula [without including the factor (1/2)] for calculating total stored electrical energy will get the wrong answer every time, unfortunately. 
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@ian_mn I truly couldn’t care less what you think.
My only interest in posting in this thread is to prevent you from confusing those who do not know. You are wrong. The facts are as I stated. You are confusing yourself by applying the “energy stored in a capacitor formula” where it does not apply.
E=QV. Definition. Identity. Accepted reality by every electrical engineer and everyone who has passed physics 101.
Link: “E=QxV” source, BBC GCSE Physics review. http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/electricity/chargepowerrev4.shtml
Link: “E=QxV” source, GCSE Science, http://www.gcsescience.com/pe5.htm
Now please stop trying to confuse the uninitiated.
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Hi xvg11,
I still think it would be helpful to you to watch the khanacademy video. It’s very clear and easy to follow - why not give it a try.
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@ian_mn I don’t know if you are trolling, or if you are just in over your head, because I looked at your video, and the first thing that he says is:
“In order to calculate the energy stored on a capacitor, we first need to remind ourselves what the formula is for electrical potential energy. If a charge Q moves through a voltage of V, the formula for the electrical energy, E, is just Q times V.” He then writes the equation E=QxV!
You are really making yourself look ridiculous at this point.
Hi @xvg11
Sorry you feel that way. But I’ll give you a simple example that may help clear things up quite quickly:
You have a simple insulated structure that’s been charged up to 5 volts. And the stored charge happens to be 1 micro-coulomb. Using the correct formula, the stored electrical energy will then be:
E = (1/2) Q V = 0.5 x (1e-6) x 5 = 2.5 micro-joules
Using the formula without the factor of (1/2), you would get 5 micro-joules, which is wrong. You could watch the Khan Academy video at 1’30" for more detail on the theory behind this - it’s very easy to follow.
This link may help clarify the situation further:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html
@ian_mn I’m not confused about this subject matter whatsoever.
You are the one confusing yourself by bringing in the extraneous subject of charge stored on a capacitor which is charged to a particular voltage. But even so, your link further confirms that what I have said is correct. You are misunderstanding your own link.
I know for a fact that the energy stored on a capacitor is 1/2 QV. That’s because the voltage on a capacitor varies from V to 0 as it is discharged, so instead of using the fully charged voltage, (V), in the equation E= QV, you have to take the average voltage across its plates instead, which is (1/2) x V, giving you E=(1/2) Q V.
But the equation E=QV holds true for all cases.
You just have to understand that in the case of a capacitor, the effective V is 1/2 of the initial V. This is where you are confusing yourself.
If you had looked at my links, you would see that this is correct. The BBC which says, “E=QV”, is not wrong. Just read what they said. If you are too lazy to look at the link, here is the text:
E = VQ
For a given amount of electrical charge that moves, the amount of energy transferred increases as the potential difference (voltage) increases.
Calculating energy transferred
You can work out the energy transferred using this equation:
E = VĂ—Q
E is the energy transferred in joules, J V is the potential differences in volts, V Q is the charge in coulombs, C
For example, much energy is transferred when the potential difference is 120 V and the charge is 2 C? Energy transferred = 120 Ă— 2 = 240 J
You don’t have to take my word for it. You can see that the University of Illinois explains this as well, and shows you exactly where you are going wrong, and confusing yourself terribly.
https://van.physics.illinois.edu/qa/listing.php?id=1769
“Electrical E = QV”
This formula as it stands refers to the energy gained or lost by transporting a charge Q from one place to another where the difference in the voltages is V.
For example, if a charge Q flows from one terminal of a battery to another, then QV is the energy delivered (say, to a light bulb) and taken away from the battery when this happens. You can add energy to rechargeable batteries (don’t try this with non-rechargeable batteries) by forcing charge to flow in the other direction. This formula assumes that the voltage difference remains constant.
The energy stored in a capacitor has a similar form, E=QV/2. The difference here is that the voltage of a capacitor changes as you add charge to it, and you have to add up all the contributions V*dQ for each little bit of charge dQ you put on the capacitor, where Q=CV is the expression which gives the voltage as a function of how much charge is on the capacitor. The capacitor relationship also appears as E=CV^2/2.
Ian, that’s the physics department at a renowned research institution, in addition to all of the other links that I have cited. All confirm that E=QV.
You owe it to yourself to ask yourself if the whole world is wrong, or maybe you are just misunderstanding something.
Hi @xvg11
From your own quote above:
This is what I’ve been saying throughout these posts, but apparently I’ve still managed to confuse you. Anyway, good luck with your GCSE exams over the next few weeks!
@ian_mn You may think that you are talking to a child such as yourself, but you are mistaken. You may think you sound cute, but you sound like a jerk.
The quotes and references that I’ve posted confirm everything that I’ve said to be correct. You, on the other hand, have amply demonstrated that you are trolling. I’m through with you.
There’s no reason to guess that when Mahirislam writes a formula for energy he (or she?) would be talking about the energy stored on a capacitor. After all, many circuits don’t even have capacitors, but they still have energy. By convention, when we use a letter like E we are talking about the general energy supplied by the source.
E=QV. And yes, it is actually quite useful to use the units to make sense of the terms, because the units have been defined by this basic relationship!
Now if that energy happens to be used to charge a capacitor, exactly 1/2 of the energy will be lost as heat, and 1/2 of the energy will be stored on the capacitor. That’s a rule that can be quite puzzling. Hence, it becomes quite memorable. But you shouldn’t think of it as a formula to memorize unless one actually includes more words, such as “When a cell provides energy VQ to charge a capacitor, the resulting energy stored on the capacitor will be (1/2)VQ.” You wouldn’t just write E=(1/2)VQ, because that is just 1/2 the energy. The other 1/2 has dissipated as heat.
Here is another explanation. http://www.ee.sc.edu/personal/faculty/simin/ELCT102/16%20Capacitor%20energy,%20connections,%20RC%20transients.pdf
Even though the pdf is all about the energy stored on a capacitor, and they do get to your equation, note that at the top of the second page the authors state clearly, “Electric Energy = Charge x Voltage: W = Q × V” .
Hope this helps!
@ Mahirislam, what a sweet little course. I especially like the mem you made for ohms.
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Thanks! Also, in the previous post, I’m a he.
The most widely accepted answer is E = Q x V. I got taught the formula this way many times in school.
Hi @xvg11
I certainly don’t believe you’re a young child, but based on your previous posts you appear to exhibit the traits of a confused (and possibly angry) teenager - and I feel sorry for you. As I said before, good luck with your GCSE exams - I would strongly advise you to stop distracting yourself with online activities, and concentrate on studying instead!
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What an ugly post and exchange @ian_mn. I hope you use it as a reminder in the future of what not to do. It’s assumption after assumption after assumption and instead of letting the assumptions go after they’ve been proven not relevant or incorrect you double down on them. That shows a rigidity of mind that is the complete opposite of what you want if you’re going to try to educate people or engage them in conversation. The right thing to do would be to stop and reflect on your assumptions, either about the physics or the people in this conversation, and continue with more humility about what you really know and how you address people.
It’s a beginner mistake in (online) conversation that you see a lot.
We have two ears and one mouth so that we can listen twice as much as we speak.
– Epictetus
It didn’t help here that @xvg11 was so abrasive.
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