Math problem: Energy stored in Cap... - Page 1
Forum section: electronics / beginners

author: Axel1973
author_rank: Regular Contributor
timestamp: May 27, 2012, 05:42:01 pm
content: Hi Experts,today im puzzeling about how to calculate the amount of energy stored in a capacitor at a specific point in time. therefor i searched google for quiet a while and i found several equations to calculate it. but for some reason i just dont understand it very well.for example i found this simple one:E == Energy (Joule)U == Voltage (Volts)I == current (Ampere)t == time (seconds)C == capacity (farad)E = U*I*tReading about Coulombs i know that 1 coulomb == 1As (ampere second). So i though i could substitute the "As" with Q and do aE=U*Q right?Now i asked myself, how to determine Q of a capacitor. I googled and found this:C (farad) = Q/USince i know the Voltage and the Capacity of a given Cap i can transform this equation toQ = C*U , right?So i got Q now and can substitute it in the first equation...E=U*Q ==> E=U*C*U ==>E=U²*CBUT...then i found another equation thats supposed to determine the amount of energy stored in a CAP. It looks like this:E = 0.5 * U² * CNow im puzzeling.. WHERE are the "times 0,5" coming from? I mean, i see where the U²*C is coming from and how it is formed. But 0.5*X?? Where is my mistake? What am i missing here?Thanks in advance for any good afford in explaining me.greetsAxel

author: 0xdeadbeef
author_rank: Super Contributor
timestamp: May 27, 2012, 06:01:38 pm
content: The antiderivative of x is 0.5x^2

author: jahonen
author_rank: Super Contributor
timestamp: May 27, 2012, 06:02:56 pm
content: Problem is that Q is not constant, but a linear ramp, when going from zero to a specific voltage. So you must do an integration, and as we know, int(x,x) is 1/2*x^2. 0.5 comes from an integration rule.Regards,Janne

author: Omicron
author_rank: Regular Contributor
timestamp: May 27, 2012, 06:36:15 pm
content: Quote from: Axel1973 on May 27, 2012, 05:42:01 pmYour starting equation is wrong. This equation assumes that I and U are not changing over time. When you charge a capacitor with a constant current then the voltage will be changing over time. The correct way to calculate the energy is this:- draw a graph with on the vertical axis U * I and on the horizontal axis time.- the energy at any time T is the area underneath the graph between time t = 0 and time t = TThe only assumption we make is that at t = 0 there was no charge on the capacitor.If you do this you will find the correct equation.

author: IanB
author_rank: Super Contributor
timestamp: May 27, 2012, 06:51:41 pm
content: Do you know about calculus?The correct formula for the energy stored in a capacitor is the formula that says:E = 0.5 * C * U²But deriving this formula in the most natural way requires the use of calculus.Here is a simple explanation without calculus.Firstly, energy is defined as "the potential to do work". That means that if energy is stored in a device you may use that energy to do work, and conversely you can determine the stored energy by calculating the amount of work that was done to store that energy.Now energy is stored in a capacitor by moving charge into the capacitor against a potential difference (the voltage on the capacitor). The more voltage the capacitor has, the harder it is to move additional charge into the capacitor and the more work you have to do to achieve it.Suppose the capacitor is 1 farad in size and you want to add 1 coulomb of charge to it. The capacitor starts out discharged at 0 volts. If you add 1 coulomb to it the voltage will increase by 1 volt. Let's consider the work required to do this.If we add the first 0.1 coulombs to the capacitor the voltage is starting at zero, so it requires no work to start with, but when we are finished the voltage has increased to 0.1 volts.If we try to add the second 0.1 coulombs of charge, the voltage is already 0.1 volts. So now we must overcome that opposition. The work we have to do is charge x voltage (amps x seconds x voltage = watts x seconds = joules). So the amount of energy stored was approximately 0.1 x 0.1 J.Let's add another 0.1 coulombs. Now the existing voltage has gone up to 0.2 volts, so the work done and the incremental energy stored is 0.1 x 0.2 J. Notice how the work is increasing as the potential on the capacitor is getting higher?For the next 0.1 coulombs of charge the voltage starts at 0.3 volts and so the work is 0.1 x 0.3 J.We can repeat this all the way up to 1 coulomb to find the total work done and the total energy stored. To make it more accurate we can use smaller and smaller charge increments and do more and more summations. Mathematically this is called integration.However, we can look at this in simpler terms.The total charge stored on a capacitor is given asQ = C * UWhen this charge was moved into the capacitor the voltage started out at zero and ended up at U. So the average voltage that the charge had to overcome was 0.5 * U.To find the energy we can multiply the charge by the average voltage, so thatE = (C * U) * (0.5 * U) = 0.5 * C * U²This is where the energy formula comes from.