What happens if I change the polarity of the battery?

Suppose if i charge a capacitor. Now if I reverse the polarity of the battery, how much is the heat produce during that process? My book says as change in energy = 0 (in capacitor) therefore heat produce 2cv^2. Can somebody actually explain me what will happen when the polarity is changed and provide me with some other way of explaining this answer?

• " Now i reverse the polarity of the battery .How much is the heat produce during that process . " How heavy is the battery, how much distance do you move it? :-) Serious: An ideal capacitor can not consume energy. What happens is that the energy is consumed in the battery, first when charging, then again when charging with the opposite voltage. I have to go now but I am sure somebody will give a long answer about this. Jul 20, 2019 at 5:22
• If there are no resistances involved there will be no heat generated. Show the circuit.
– Chu
Jul 20, 2019 at 6:13
• This sounds to me as theoretical nonsense, just like: electronics.stackexchange.com/questions/439582/… Jul 20, 2019 at 7:26
• @Huisman Yes, but ... . The ideal resistanceless examples fail to deal with the infinitely large current flows. Add those in and multiply and divide by a few infinities and it all works out :-). ie the ideal cases are unworkable because of infinities. The real world resistances make it calculable. [So. Yes. That's what you effectively were saying :-)]. Jul 20, 2019 at 13:57
• @Huisman Indeed, if you can ever get that zero Ohm circuit. What I'm saying (and I'd doubt that we disagree greatly) is that in practice you have resistance presence and the laws of Physics apply. If we have typical wire, typical capacitors, typical batteries, or even the very best, we understand "well enough" what will happen. As we approach "ideality" we probably need plasma physics, arc-flash shields and luck if we are to understand what happens (and not to die). Which is not what we get to in most cases. Jul 20, 2019 at 20:15

Ideal dielectrics = insulators such as capacitors and batteries have no effective series resistance (ESR).

Ideal dielectrics do not exist.

All dielectrics have ESR which follow the losses defined by Ohm's Law and thermal resistance that determines the temperature rise in the dielectric.

When reversing the charge on a polarized dielectric Capacitor ( Cap.), the insulation may break down and destroy the part, especially if the circuit impedance is as low as the ESR of the cap. If the reverse voltage is > 5% of the forward rating the risk of damage increases exponentially, although in current limited circuits some polarized caps can withstand -10% of Vmax for peak AC audio signals with some forward DC bias voltage.

When reversing a non-polarized capacitor the energy previously stored and not decayed by self-leakage resistance $$\E=0.5CV^2\$$ which is the same quantity of energy stored after apply the same voltage reverse.

However to invert these charges, the cap. will conduct $$\I_{peak}=\dfrac{2|V|}{ESR}\$$ in peak current with an exponential decay of $$\\tau= ESR*C\$$ [s] for current as the exponential reversal of voltage. The energy dissipated during thi discharge/charge process is from the ESR current*voltage power losses as explained in your textbook.

We say power dissipation=+ve is a positive value and generated power=-ve has a negative polarity, yet the energy product of the power P(t) and time integral is just an energy transfer ( as it is neither created or destroyed). So as the energy stored is the same quantity as before but just as if the leads were reversed yet power had to be generated by some supply and lost in the cap. during the process.

NPO/COG ceramix are low density ceramics are known to have low ESR as well as plastic film caps in the milliohm range thus will conduct very high current.

Batteries are like massive capacitors due to the high relative dielectric constant, , Dk ( or $$\\epsilon _r\$$), so a similar heat loss power process occurs going from one polarized voltage level to a higher state of charge (SOC) voltage to store more energy.

When the capacitor is first charged, it has received $$\\frac{1}{2}C(+V)^2\$$ joules of energy from the battery.

By reversing the battery, you create a potential difference which causes current to flow in the opposite direction, which will at first discharge the capacitor, until the voltage across it returns to zero. At that point the capacitor stores no energy at all.

But current continues to flow, since the battery and capacitor still have unequal potential differences, and this will begin charging the capacitor again, to the opposite polarity, until the voltage across the capacitor equals the battery voltage, this time −V.

At that point the capacitor has $$\\frac{1}{2}C(-V)^2\$$ joules of energy. Due to the squaring of potential difference V, energy stored is always positive, independent of the sign of the potential difference, and now it has the same amount of stored energy that it started with, when it was initially charged to +V potential difference. The only change is the sign of the potential difference.

So, during this process, the capacitor potential difference goes from maximum voltage +V, to zero, and finally to a potential difference of −V. It starts and ends with the same energy stored, a net change of zero joules.

As for how much heat energy is created, this relates to the fact that it is impossible to connect an ideal capacitor across an ideal battery, when the resistance around the loop is zero, because this would result in infinite current flow, absurd.

In reality the resistance around the loop cannot be zero, and the algebra shows that the energy transfer from battery to capacitor, and capacitor to battery is 50% efficient, due to energy lost in that resistance. That is, if a capacitor is charged to potential difference V, via a resistive path, so that it now has $$\\frac{1}{2}CV^2\$$ joules of energy, that same amount, $$\\frac{1}{2}CV^2\$$ joules of heat, must have been created in the loop's resistance.

Since these are the potential phases in your scenario:

1. Initial charging from 0V to +V potential difference, creating $$\\frac{1}{2}CV^2\$$ of heat,

2. Discharge of the capacitor, to 0V, creating $$\\frac{1}{2}CV^2\$$ of heat,

3. Charging from 0V to −V potential difference, creating $$\\frac{1}{2}CV^2\$$ of heat,

4. Perhaps another discharge from -V to 0V, again producing $$\\frac{1}{2}CV^2\$$ of heat.

All of those add up a total of $$\2CV^2\$$ of heat energy, but I think your question doesn't account for the last phase. Does the original question?