Charging a capacitor in sereis and discharge in parallel?

[infamous_Q]

Ok. So putting capacitors in series decreases the capacitance of the circuit (and correct me if I'm wrong any where in this, and I'm likely to be). So this would mean it would take less time to charge, I'm assuming. What if you charged a set of, say 3, capacitors in series, then hooked them up in parallel to discharge (assuming that in parallel the capacitance is increased).
Thanks for any help guys.

 

[chroot]

If you wire up three capacitors in series, they do not store as much energy as they would in parallel.

- Warren

 

[mugsby]

but in series you can increase the voltage, so in theory you should have the same amount of power.

 

[infamous_Q]

so then...charging in series would take less time. And then discharging in parallel would provide the same amount of power as was put in, just over a longer period? ...theoretically..

 

[mugsby]

my best guess is that it would depend on the caps ESR.

 

[Cliff_J]

infamous - no free lunch here. Power implies the rate at which work is done (or energy is used, same thing) so you gain nothing with regards to power or energy. As already pointed out, if you store a charge in a capacitor at a lower voltage, the energy is much lower.

energy in joules = .5 * capacitance * voltage^2

Its like 3 water buckets and some hoses, you can only store so much water in the buckets and how high the buckets are off the floor determines the pressure on the hose (which you will also need to fill the buckets with the hose).

One application of the opposite to what you descrive, is when constructing something like a rail gun where the very high voltages are impractical to build. So many capacitors are charged in parallel, then connected in series to get the high voltage desired.

 

[infamous_Q]

yes. ok. i get the no free lunch part. But assuming you charged to full capacitance, wouldn't the energy inside the capacitor be the same? Regardless of whether it was in series or parallel?

 

[Cliff_J]

Yes, the energy is dictacted by the formula above - so regardless of how they are hooked up, if you charge the caps to the same level they will hold the same amount of energy.

Even the physical size indirectly effects the amount of energy. Because the plates can only be manufactured so thin and the dielectric so thin before it reaches its breakdown voltage that the typical electroylic cap size is going to be roughly the same for the energy stored. Yes a high voltage allows a much higher energy storage, but the dieletric would need to be thicker do either less material will fit in the same size or it gets bigger. Newer technologies like carbon caps are changing this and so on, but if compared within similar families its surprising how true it is.

 

[vanesch]

yes. ok. i get the no free lunch part. But assuming you charged to full capacitance, wouldn't the energy inside the capacitor be the same? Regardless of whether it was in series or parallel?

Of course. Imagine each capacitor has capacity C, and you charge each of them to a voltage V. That means, each of them has taken up a charge Q = C x V.

Now, to do that in series, you need to apply a voltage of 3 x V and deliver a charge Q. If you do it in parallel, you will have a voltage of 1 x V, but a charge of 3 x Q.

cheers,
Patrick.

 

[infamous_Q]

so in series you can charge at a higher voltage, and then discharge at a higher current? (although this would mean that the capacitor would discharge much faster than it charged). And if charged in parallel it can be at a high current, and discharged at a high voltage (over a longer period of time). correct?

 

[chroot]

You don't seem to understand anything we're saying, infamous_Q.

If you wire three capacitors up in series, the total capacitance is 1/3 of the individual capacitors.

If you wire up three capacitors in parallel, the total capacitance is 3 times that of the individual capacitors.

That's all there is to it. There's no reason to think about "charging times" or "higher currents." The entire behavior of a capacitor is modeled by a single equation:

$I = C\frac{{dV}}<br /> {{dt}}$

When you wire your capacitors in different configurations, all that's changing in this formula is C.

- Warren