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Energy of a capacitor _ Circuits _ Physics _ Khan Academy

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Language: en

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SPEAKER: Look at this capacitor.
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Look what happens if I connect this to this bulb.
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CHILDREN: Boo!
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SPEAKER: Yes, nothing happened,
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because this capacitor is not charged.
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But if we connect it to a battery first,
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to charge the capacitor and then connect it to the bulb,
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the bulb lights up.
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CHILDREN: OO!
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SPEAKER: The reason for this is because
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because when a capacitor is charged, it not only stores charge,
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but it also stores energy.
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When we connect the capacitor to the battery,
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the charges were split.
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These split charges want to connect when they are allowed to,
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because the opposites attract.
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If you finish the chain with a few wires and a light bulb,
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electricity will flow.
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And the energy stored in the capacitor is converted to light and heat,
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that come out of the bulb.
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After the capacitor is released from its charge
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and there are no more charges left to transfer,
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the process stops and the light goes out.
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The kind of energy stored in capacitors,
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is the electrostatic potential energy.
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If we want to find how much energy
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stored in one capacitor,
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we have to remember what the formula is
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for electrostatic potential energy.
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If a charge, Q, moves through a voltage, V,
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the change in the electrostatic potential energy of this charge
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is just Q over V.
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Looking at this formula, what do you think the energy of a capacitor will be,
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which has been charged up to charge Q and voltage V?
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CHILDREN: Q by V.
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SPEAKER: Yes, I thought so.
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But it turns out that the energy of a capacitor is 1/2 QV.
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SPEAKER 1: Where did that 1/2 come from?
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How is energy not just Q over V?
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Well, the capacitor energy would be Q over V,
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if while he is discharging his charge,
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the entire charge drops through the total initial voltage V.
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But while he is being released from the charge,
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all charges will not fall through the total voltage V.
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In fact, only the first charge to be carried,
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will fall through the total initial voltage V.
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Any charges that are then transferred,
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will fall through less and less tension.
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The reason for this is that every time a charge is transferred,
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this reduces the total charge,
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stored in the capacitor.
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And while the capacitor charge continues to decrease,
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the capacitor voltage continues to decrease.
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Remember that capacity is defined as
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the charge stored in the capacitor,
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divided by the voltage across this capacitor.
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As the charge decreases, the voltage decreases.
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As more and more charge is transferred,
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there will be a point where only one charge falls
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in 3/4 of the initial charge.
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You wait a little longer and the moment will come
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in which the charge is transferred through
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only half of the initial voltage.
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You wait a little longer and there will be one charge
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transferred only 1/4 of the starting voltage.
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And the last charge to be transferred,
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falls through almost no tension,
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because there is no charge left in the capacitor,
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to be stored.
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If you sum up all these downs
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in electrostatic potential energy,
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you will find that the total energy drop of this capacitor is just Q.
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The total charge, which was initially in the capacitor,
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1/2 of the starting voltage of the capacitor.
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That's 1/2 here because not all the charge
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dropped through the total initial voltage V.
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On average, the charges dropped in only half of the initial voltage.
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If you take the charge stored in the capacitor at any time,
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and multiply the voltage across the capacitor at that same moment,
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divide by 2
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and you will have the energy stored in the capacitor
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at that particular moment.
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There is another type of this equation that may be useful.
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Since capacity is defined as the charge on the voltage,
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we can write that this charge is equal to
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voltage capacity.
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If we replace the charge with the voltage capacity,
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we see that the energy of a capacitor
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can be saved as 1/2 in capacity
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on the voltage in the capacitor squared.
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But now we have a problem.
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In one of these formulas V is squared,
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and in one of these formulas V is not squared.
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Before, it was hard for me to remember what was.
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But here's how I remember it now.
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If you use the formula that has C,
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then you can "C" lie V ^ 2.
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And if you use a formula that doesn't have C,
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then you cannot "C" lie V ^ 2.
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These are the two formulas for energy stored in a capacitor.
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But you have to be careful.
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The voltage V in these formulas
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refers to the voltage across the capacitor.
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The battery voltage in the task is not necessary.
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If you're just looking at the simplest case with one battery,
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which has fully charged a capacitor,
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then in this case
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the voltage across the capacitor
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will be the same as the battery voltage.
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If a 9 volt battery has one capacitor charged
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up to a maximum charge of 4 pcs,
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then the energy stored by the capacitor,
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will be 18 joules.
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Because the voltage across the capacitor
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But if you have a case where multiple batteries
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are connected to multiple capacitors,
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then to find the energy of a capacitor,
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you have to use the voltage across that particular capacitor.
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In other words, if they give you this chain with these values,
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you can determine the energy stored in the middle capacitor,
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using 1/2 QV.
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You just have to be careful to use it
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the voltage of this capacitor,
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not the battery voltage.
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Introduction of 5 charge pillars
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lets you find that the energy is 7.5 joules.
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CHILDREN: Oooh!
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[Music]

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