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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?

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  • $\begingroup$ If a battery is fully discharged its voltage = 0 $\endgroup$ – user56903 Apr 25 '16 at 14:02
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    $\begingroup$ 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. $\endgroup$ – Gert Apr 25 '16 at 14:02
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    $\begingroup$ I'm curious about that textbook. What is the title and author? $\endgroup$ – garyp Apr 25 '16 at 14:07
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    $\begingroup$ 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. $\endgroup$ – K7PEH May 7 '18 at 3:13
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The battery

The main issue to consider is the behaviour of the battery as it discharges and charges:

  • In the charged state, the emf of a $6V$ battery is $6V$, or slightly above.
  • 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. It can, however reach zero, as in alkaline batteries. The figure below shows the charge profile whereby the charger is $3V$ above the battery 'full charge' voltage. Note also that the charge vs capacity graph is not linear (Source Ref).

Therefore, The battery voltage when a $6V$ battery is 'discharged' will be anything from approximately $3V$ to $0V$. Let's say it is at $1V$.

Charging the battery

  • A charger must cause current to flow in the opposite direction to the no-charging, load-connected state.
  • The battery being charged must have a lower voltage difference than the charger, otherwise no charging will occur.

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

$9-1 = 8V$ and therefore current will be:

$8/10 = 0.8A$

As the battery reaches full charge, the emf will be:

$9 - 6 = 3V$ or less if overcharging is done. at $3V$ emf, the current is:

$3/10 = 0.3A$, i.e. the current reduces as the battery gets charged up.

Indeed, the resistance may change as the battery gets charged, as internal chemical activity changes, but we ignore this here.

enter image description here

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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.

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enter image description here

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.

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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.

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    $\begingroup$ So what is your answer to the question? $\endgroup$ – sammy gerbil Apr 27 '16 at 13:11
  • $\begingroup$ @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. $\endgroup$ – Gert Apr 27 '16 at 13:53
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    $\begingroup$ 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. $\endgroup$ – sammy gerbil Apr 29 '16 at 1:12

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