# Charging of a battery

A battery has an Emf 6 Volts. It is completely discharged. It is charged by maintaining a potential difference of 9 Volts across it. If the internal resistance of the discharged battery is 10 ohms, find the current through the battery, just after the connections are made.

My textbook says that the net potential difference across the battery is 3 volts, but if the battery is discharged why would we subtract the emf of the battery while calculations? Wouldn't it act like a conductor when it is discharged completely?

• If a battery is fully discharged its voltage = 0
– user56903
Commented Apr 25, 2016 at 14:02
• Seems to me the textbook takes a very simplistic view on recharging a battery. When recharging the electro-chemical reactions that gives a full battery its EMF have to be reversed. In essence all the energy the battery delivered while discharging now has to be supplied by the recharging current. On top of that, internal resistance also has to be overcome.
– Gert
Commented Apr 25, 2016 at 14:02
• I'm curious about that textbook. What is the title and author? Commented Apr 25, 2016 at 14:07
• When the 9 volt battery is hooked up in a charging scenario, it is opposing the 6 volts of the discharged battery. Yes, this battery gives 6 volts when discharged, I am suggesting the problem assumes it is ordinarily a 9 volt battery. So, the charging voltage is 9 minus 6 or 3 volts so the current is 3 volts divided by 10 ohms. And, of course this is only at that first instance since the battery will be charged and it not always 6 volts, it is increasing until it matches the charging 9 volts. Commented May 7, 2018 at 3:13

The battery

How is the battery made up?

• The battery is made up of dissimilar materials at its cathode (-) and anode (+).
• Even in the discharged state, the emf of the battery is $$6V$$, or slightly above. It will not be possible to discharge the battery below this voltage without changing the chemistry of at least one of these: the anode or cathode.
• In the discharged state, i.e. the charge at which it becomes unable to, for example, turn a light on, the battery's emf is often not zero because the emf required by the load is significantly higher than zero, and also because the battery cannot go below its minimum charge voltage determined by chemistry as mentioned above. It can, however reach as low as 0.8V, as in alkaline batteries. The figure below shows the decrease in battery capacity of a lithium ion battery from terminal $$4.2V$$ to $$3V$$. Note also that the voltage settles at 3V (Source Ref).

Charging the battery

If we use a stable $$9V$$ charger, then, at the start of the charging, the voltage difference (emf) will be:

$$9 - 6 = 3V$$ ... at $$3V$$ emf, the current is:

$$3/10 = 0.3A$$.

Indeed, the resistance may change as the battery gets charged, as internal chemical activity changes.

• Totally wrong answer. If the battery is not damaged, and is capable of being charged, the chemical potential, in the case of the lithium battery, is 4.2V. In other words, Any voltage lower than say 4.2V will not charge a lithium battery at all. Check it experimentally if you dont believe it. Commented Nov 20, 2019 at 15:11
• @Kostas Please check your reasoning. The reference (based on experimental data) that I have quoted in the answer used specifically lithium ion battery. The voltage, and therefore the 'chemical potential' you mention, does vary with state of charge. If, for example, the 4.2V lithium ion battery is at 50% (capacity) charge, then its voltage is 3.8V according to the above graph, from the quoted reference. If you apply a voltage above the 3.8V, say 4.2V, the battery will charge. If you know of a phenomenon that allows such lithium battery to remain at 4.2V as it is being discharged, enlighten me. Commented Nov 22, 2019 at 0:59
• if you never heard of this mysterious "chemical potential", I will not be able to enlighten you. But here is a link, and it says, about a lead acid battery: Commented Nov 22, 2019 at 20:17
• To charge the battery, the alternator voltage output has to exceed a minimum charging voltage. This minimum charging voltage is 13.8 volts dc across the battery terminals, or at the output of the alternator. A single lead-acid cell starts to charge at anything over 2.25 volts. Since a 12 volt battery has six cells, any 12 volt lead-acid battery needs at least 13.8 volts to start to charge. This voltage will be enough to fully charge or maintain the battery on a trickle charge, but charging time will be very long at 13.8 volts. w8ji.com/battery_and_charging_system.htm Commented Nov 22, 2019 at 20:18
• @Kostas I acknowledge the minimum chemical potential must indeed be exceeded. The initial question had been misread to consider the second sentence as subsequent enactment of an action rather than a statement of state. The necessary edit has been made. Your point about lithium ion battery needing 4.2V for charging however needs reviewing. The graph above shows that this type of 4.2V lithium ion battery, when fully discharged, will charge at above 3V. Commented Nov 25, 2019 at 2:30

The simple answer is $$(9-6)/10=0.3 A$$ , but it is not easy to explain why this is the correct answer in reality and in theory. When they say that the battery EMF is 6V, I think they mean it is the voltage in the fully charged state. Now, the surprising and counterintuitive thing is that if such a battery were discharged to zero or near zero, and connected to a fixed 9V source, the current would not equal 0.9A for any amount of time, and most likely the current would go to 0.3A nearly instantly.

The battery usually 'supplies' electrical energy to the circuit element as charges flow through it. However in case of charging, the charging source provides energy required to reverse the spontaneous electrochemical reactions which make cell a cell, seemingly returning the cell or battery to its initial state (it's not the case though, with time through wear and tear, reversal process becomes inefficient and the battery is dead now) and actually it's the game of electrons and electron giving material.

In this question when the 9V battery is connected, the energy equivalent to 9 units(say) is supplied to the circuit and 6 units of those is utilized in 'charging process'. These 6 units are completely dumped to the discharged battery as soon as the connections are made. Think of it as filling an empty tank A with filled tank B of water. When A is completely empty, the whole of pipe is carrying water, but this is not the case towards the end. So, just when the connections are made, the supposed emf from the reference of the resistance is 9-6 = 3V. This explains everything.

Sorry for my poor english. My native language is french.

This exercise is based on a rudimentary model of a battery that would be equivalent to a pure voltage source and a series resistor (thevenin model). The "discharged" battery would have the same emf: simply, the useful products (PbO2, Pb and acid solution for a lead-acid battery) would be almost consumed and the recharging would aim to reconstitute these products.

This is obviously a very, very basic model of the battery. In reality, the charge characteristic is not linear. There are overvoltages linked to the kinetics of the reaction of charge to the electrodes, to the phenomena of diffusion within the electrolyte ..... Even at very low current, the voltage to be applied to start the charge is greater than the emf of the battery. Also, the resistance of the electrolyte changes as the concentrations of products increase.

Charging a battery is an electrolysis and one easily finds the characteristic of charging an hydrogen electrolyzer. The ideas are the same.

See for example: hydrogen electrolysis

Batteries are made up of one or more electrochemical cells, arranged in series.

In these cells an electrochemical Redox reaction takes place:

$R+ze^{-} \to R^{z-}$

$O-ze^{-} \to O^{z+}$

Where $R$ and $O$ resp. are a reducing agent and oxidising agent. This transfer of electrons provides the EMF of the cell.

As more current is drawn from the cell, the concentrations of $R$ and $O$ in the cell's electrolyte slowly drop and so does the potential across the electrodes, until it is basically zero.

By recharging the cell (by reversing the normal flow of current) the above reactions are reversed:

$R^{z-} - ze^- \to R$

$O^{z+} + ze^- \to O$

Thermodynamically the total amount of electrical energy the cell delivered during normal operation (discharge) must now be delivered by the charging current, to fully restore the concentrations of $R$ and $O$ to their original levels.

Cells are imperfect though: the electrolyte is conductive but does have some electrical resistance ('internal resistance'). For that reason both charging and discharging waste some energy as ohmic power.

• So what is your answer to the question? Commented Apr 27, 2016 at 13:11
• @sammygerbil: I think the question wants to be answered as $I=\frac{9V}{10Ohm}$ but that's only approx. correct at the start of the charging. Towards the end of the charging it would be more like $I=\frac{9V-6V}{10Ohm}$. It's a poorly formed question, IMO.
– Gert
Commented Apr 27, 2016 at 13:53
• Actually I meant the user's question, about why the pd across the cell isn't 3V. But I can see what you're saying is that emf builds up again slowly, from 0 at depletion. Commented Apr 29, 2016 at 1:12