# 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. – Oldfart Jul 20 at 5:22
• If there are no resistances involved there will be no heat generated. Show the circuit. – Chu Jul 20 at 6:13
• This sounds to me as theoretical nonsense, just like: electronics.stackexchange.com/questions/439582/… – Huisman Jul 20 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 :-)]. – Russell McMahon Jul 20 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. – Russell McMahon Jul 20 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.