| General > General Technical Chat |
| Veritasium "How Electricity Actually Works" |
| << < (159/185) > >> |
| PlainName:
--- Quote ---But the new question is what happens if after you charged the capacitor so no current but capacitor contains those 72mJ and you reverse the polarity on the source. --- End quote --- No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy. |
| electrodacus:
--- Quote from: dunkemhigh on May 21, 2022, 11:50:09 pm ---No, you're jumping ahead and going off on a diversion again. We have yet to determine exactly what is going through that resistor since you insist it can't be energy. --- End quote --- Of course energy goes through resistor. When have I ever mentioned that is not ? It is not going through capacitor but in to capacitor. The entire point I want to make is that energy travels through wires and wires and resistors are the same thing. If you reverse the polarity after capacitor was charged you will see that all energy now delivered by the supply ends up as heat on the resistor. All of it not just half as at initial connection when capacitor was discharged. |
| hamster_nz:
--- Quote from: electrodacus on May 20, 2022, 10:05:11 pm --- --- Quote from: hamster_nz on May 20, 2022, 09:50:24 pm ---Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)? Are we in Humpty Dumpty land again, where words only mean what you say the do? --- End quote --- If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor. In the particular example efficiency was just 50% so half of the energy ended as heat during transportation. You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device). What part exactly is not clear. --- End quote --- It is clear that you are completely wrong. When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be. But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron. But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron. For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron. In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2. And all work being performed by your source capacitor. The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor. |
| electrodacus:
--- Quote from: hamster_nz on May 22, 2022, 12:57:58 am ---It is clear that you are completely wrong. When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be. But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron. But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron. For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron. In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2. And all work being performed by your source capacitor. The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor. --- End quote --- You forget one super important thing. That is reversible so energy storage and not work. The work will be done when you discharge that capacitor. So all that energy you put in the capacitor 0.5 * C * V2 is still there in the capacitor ready to do work. In my most recent example with 12V ideal voltage source 1Ohm total circuit resistance 0.5Ohm for each wire and the 1000uF capacitor. Energy stored in capacitor was 0.5 * 0.001F * 12V2 = 72mWs But the ideal voltage source needed to supply 2x as much 144mWs in order to cover the losses in transportation of that energy that happens to also be 72mWs drop on that 1Ohm resistance as energy traveled through wires to get from voltage source to capacitor. Now try to calculate what happens if you after the capacitor is charged you disconnect the voltage source and connect it back but with reverse polarity. What you will notice is that now all the energy delivered by the voltage source ends up as heat in the wire (all of it not just half like when capacitor was discharged). |
| hamster_nz:
--- Quote from: electrodacus on May 22, 2022, 01:12:47 am --- --- Quote from: hamster_nz on May 22, 2022, 12:57:58 am ---It is clear that you are completely wrong. When there is 0V across a 1F capacitor it takes a very tiny bit of work to remove a charge from the - plate, and add on to the + one. It doesn't matter if it is the same electron or not - it's a little bit of work to pull the electron out of one plate, leaving it slightly positively charged, and a little bit of work to push it into the other plate, which was quite happily neutral and now won't be. But for the next time there is 1/(6.24 x 10^18) of a V between the plates, so that requires more work. If you do the math it will be 4x that of the first electron. But for the next time there is 2/(6.24 x 10^18) of a V between the plates, so that requires even more work than the previous one. Once again, if you do the math it will be 9x that of the first electron. For the (6.24 x 10^18)th electron there is now (6.24 x 10^18)/(6.24 x 10^18) = 1V between the plates. It is now taking a whopping (6.24 x 10^18)^2 times the energy it took to move the first electron. In fact if you sum up all the work done on each charge it will equal 0.5 * C * Vfinal^2. And all work being performed by your source capacitor. The inductor in the DC-DC configuration when the switch is closed it is taking lower current at the higher voltage to make establish a magnetic field, and then when the switch opens it the collapsing magnetic field moves higher current at a lower voltage, at just slightly higher than the voltage of the uncharged capacitor. --- End quote --- You forget one super important thing. That is reversible so energy storage and not work. The work will be done when you discharge that capacitor. --- End quote --- :-// Oh come on, you know the formula: work = force x distance, If you or I push down on a lever, lifting a weight, it is still work, even though the weight on the far end can move the lever back to the original position, undoing our efforts. |
| Navigation |
| Message Index |
| Next page |
| Previous page |