7th May, 2022

University of Bristol

Question

Asked 23rd Apr, 2016

Capacitors and batteries are similar and different. One stores energy as electric field, the other one as a chemical reaction. However when charging a capacitor (RC circuit), 0.5CV^{2} [J] of energy is charged and 0.5CV^{2} [J] of energy is lost as heat in the resistor.

I'm not familiar with the way chemical reactions in batteries work so i wonder does this rule apply to batteries as well? If there is an electron flow towards the battery, i would assume that it does.

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A practical example about the efficiency of battery storage in the home. I lose about 30% of the stored electricity, just comparing what goes into the battery with what I get back.

Every battery has an internal resistance. You can think of an actual battery as a perfect battery in series with a resistor. As you charge the resistance converts some of your charging energy to heat. (note that watts = power = energy/time = Joules/sec). Say you're charging a 10V battery at a rate of 100 watts (stored energy per time). For a perfect battery that would require you apply 10V at 10 Amps. But if the battery has a 1 ohm resistance you'd actually need to apply 11V meaning you'll need to supply 110watts of power to get 100watts to the battery. The remaining 10 watts goes as waste heat. But that's not all! You again loose energy discharging the battery. But that's is a variable amount. If you drain the battery at low power for a long time you get better efficiency than if you drain it quickly at high power. That's why EV drivers with a lead foot dramatically shorten their vehicle range. Finally the chemical process releasing energy in the battery also occurs slowly when not in use and you loose energy over time with the battery just sitting. In effect there's also a very high short circuit resistance.

Answer given by Jambaugh

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Every battery has an internal resistance. You can think of an actual battery as a perfect battery in series with a resistor. As you charge the resistance converts some of your charging energy to heat. (note that watts = power = energy/time = Joules/sec). Say you're charging a 10V battery at a rate of 100 watts (stored energy per time). For a perfect battery that would require you apply 10V at 10 Amps. But if the battery has a 1 ohm resistance you'd actually need to apply 11V meaning you'll need to supply 110watts of power to get 100watts to the battery. The remaining 10 watts goes as waste heat. But that's not all! You again loose energy discharging the battery. But that's is a variable amount. If you drain the battery at low power for a long time you get better efficiency than if you drain it quickly at high power. That's why EV drivers with a lead foot dramatically shorten their vehicle range. Finally the chemical process releasing energy in the battery also occurs slowly when not in use and you loose energy over time with the battery just sitting. In effect there's also a very high short circuit resistance.

Answer given by Jambaugh

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Thank you Mushtaq Ahmad, but i would like to receive a more theoretical answer. If less than 50% of the energy is lost, why does that happen? What is the difference between a "Resistor-Battery" circuit and an RC circuit?

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I am sure you are corrent but you pretty much say "it's less than 50%, believe me" and i want to know why. Here are my thoughts on this matter.

What happens in an RC circuit during charging. The voltage drop on the capacitor increases exponentially and the current decreases exponentially with the same time constant. What about the resistor? Both the voltage drop and the current drop decrease exponentially.

So what could be the difference between battery and capacitor then? The only thing that comes into mind is that the voltage and current of the battery being charged do not change exponentially. But then the power loses would still depend on the instantaneous current. High current means more losses (it's P~I^{2}). And if there are few losses this means there is a lower current. But then again this means that the charging process will lose less energy if the battery charged slower. And this means that charging a battery faster will create significantly more losses.

I'm currently not 'fit enough' do do the math without errors, but I feel that if you charge the capacitor with a lower constant current (aka a voltage ramp), the losses in the resistor go down.

(due to I^{2}R).

This is what's typically happening during battery charging: until a certain point (depending on battery chemistry), batteries are charged in constant current mode. The "topping" is delivered in constant voltage mode, but this is where currents are already comparably low. What's adding to that: batteries are never discharged to zero. Thus the extreme inrush current when charging the RC with constant voltage does not occur during battery charging.

Anyway, battery charging can waste a lot of energy (during the so-called "fast charging"). Losses of 25+ % are not unknown during fast charging.

Regards

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Thank you U. Dreher. This is an excellent explanation. Charging at constant current reduces the losses.

Also you are right about the capacitor. I was able to prove that the energy released by the voltage source which charges the capacitor at two steps (V1 and V2>V1) is less than the energy released by a source which charges at a single voltage level (V2).

So the energy losses of the RC circuit would be lower if the capacitor is charged with a voltage ramp/constant current.

I would like to see a better answer. As of the original question excellently points out we loose 1/2 the energy when charging C2 from C1. Notice that if you put in an appropriate inductor L in series you can (teoretically) transfer all energy without loss. Since a battery is a non-linear capacitor it must be possible to provide accurate answer.

Since I'm interested for a practical purpose I consider a modelling approach. As a start I found this model:http://ltwiki.org/files/LTspiceIV/lib/sub/NiMH.lib and hope to find time to proceed.

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No clear answer on the energy efficiency of a rechargeable battery. Can we assume it is between 30-40%. It will be helpful if somebody answers.

@ Venkataraman Sekkar

A "clear" answer cannot be given as charging losses depend on charging mode (slow vs. fast), temperature, battery chemist, form factor and alike.

Current traction batteriy systems of EVs show a minimum of some 20 - 25 % charging losses during "normal charging" - more during fast-charging.

Resulting in efficiency figures of 75 - 80 % for normal charging.

For fast charging I unfortunately have no figures. I would not be surprised to see efficiency go down significantly - even to no more than 30 or 40 %. But the additional losses are only partly due to bettery charging efficiency: a good deal of energy is then required to provide battery cooling - preventing battery overheating.

Might well be the additional losses are to be attributed 1/3 to increased charging losses and 2/3 to cooling.

If this is the reality, is solar power or EVs economical alternative conventional energy systems as they are made out to be.

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This IS reality!

Fast-charging may be convenient, in terms of efficiency it is an awful idea.

Re "comparing to conventional energy systems": forget the fancy efficiency figures of e.g. Internal Combustion Engines (ICEs): these may be high (36+ % for Diesels, IIRC), but only for the maximum efficiency point (somewhat like the MPPT for PV). Practical ICEs may be at some 10 % overall. This is why hybrid cars have a significant lower fuel consumption - although they are losing a considerable amount of energy when charging/discharging the battery. But it helps to keep the ICE longer at the point of maximum efficiency.

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The answers are complicated. The question is simple. What is the portion of energy available from a battery after fully charged. Can anybody give an answer. Say it could be 20 or 30%. But what is the fraction of energy that can be reaped from a battery.

No direct answer could be given as it depends on the charging mode. If the battery is charged with constant voltage, the losses would be one (probably close to those of a capacitor). If charged with constant current, it would be another and would depend on the magnitude of the current.

Anyway, i haven't seen a scientific paper which investigates this. It's worth investigating.

Fine, Assume under domestic conditions, of 220 V, AC, is fully charged in about 3 hrs. We can easily find out how much energy is spent for charging. Now if the battery is connected to an utility, say a vehicle. Now energy is drawn from the battery to power the vehicle. Under such conditions what could be the ratio between input and output energies.

I'm sorry to joint this thread so late. As was pointed in the initial question, charging an ideal capacitor in constant voltage mode is always 50% efficient regardless of the effective R. It is more interesting to look at this theoretical problem in constant current mode. The charging efficiency in this mode is t/(t+2RC) where t is time and R is the effective charger series resistance. Obviously, if the final charging time t<<2RC (t=final voltage/constant current), then the charging efficiency approaches 1. For rapid charging, the efficiency could be <<0.5. A similar issue must apply for the nonlinear case of charging batteries, but it is difficult to find a clear discussion based on circuit theory.

Typo in my response: you want t>>(2RC) to have high efficiency. The key is having the charging time much longer than the RC time constant.

Besides being theoetically interresting, the question is becoming increasingly of practical importance as we slowly gets dogmatized with an axiom that electric cars etc are environmentally friendly. However if overall global eficiency of battery use is low, then we are just fooling ourselves. Therefore id like to know how much energy is really lost in using batteries for energystorage. (Maybe also including the manufacuring and disposal energy costs). It's food for thoughts that nobody has the answer to attery efficiency ...

If you avoid quick-charging and don't drive too fast, you might be able to achieve some 70 % wall (outlet) to wheel efficiency. (This "some 70 %" figure is from the "marketing materials" for a current vehicle. An efficiency at this level is realistic, given some preconditions are met.)

I don't expect well to wheel efficiencies for current ICEs beyond some 10 % overall. (Full hybrids are capable to achieve 20 .. 30 % well to wheel - depending on the regime, the driving habits, weather etc.)

To give efficiency figures for vehicles is quite difficult (see recent and historical discussions about test cycles), and test cycles have limited relevance for the actual energy consumption.

"Synthetic fuel" synthesis (widely discussed these days) will certainly have higher losses.

Regarding manufacturing/disposal: cancel "disposal", set "recycling".

While these processes will certainly consume more energy for battery-operated vehicles, look at the 10 % efficiency of ICEs: there's a lot of headroom. And you may not take hybrids into account: these require the same batteries as the pure EVs - they just have lower battery capacities.

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