Energy from each battery in a double-battery circuit

  • Thread starter ezadam
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  • #1
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Homework Statement



Find the energy delivered by the 4.00 V battery in the circuit.
R1=8.8 ohm, R2=2.5 ohm, V=15 volts, [tex]\Delta[/tex]t=3.30 minutes

http://www3.zippyshare.com/scaled/51639486/file.html

I have already figured out the correct values and directions (see image) of the currents:

I1=0.84 A, I2=1.39 A, I3=0.56 A

Homework Equations



U=P[tex]\Delta[/tex]t = VI[tex]\Delta[/tex]t

The Attempt at a Solution



I know the correct answer which is: P=VI3[tex]\Delta[/tex]t but I am not convinced with it. What bothers me in it is the current we chose: Isn't I3 the current contributed from both batteries and not the 4.00 V battery's own produced current ? Thus, shouldn't we try to find the current exclusively from the 4 V battery and use that value in our calculations ?
 

Answers and Replies

  • #2
gneill
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In point of fact, your own calculations show that current is flowing into the 4.00 battery and not out of it. So that battery is absorbing energy (it is being charged), not delivering it!
 
  • #3
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Yes you are totally right, but that absorbed energy is a net energy.

In other words, the 15 V battery provides a large amount energy, part of it gets dissipated in the resistors, but the remainder of this energy is still larger than the energy in the 4-volt energy in magnitude. The fact that currents from both batteries are in opposite direction says that the low-energy battery (the 4-volt one) will be absorbing the net energy.

We want the power from the 4 volt battery only, I mean we want its contribution to the circuit in terms of power without the contribution of the 15 volt battery

I don't really know where the flaw of my reasoning is ...
 
  • #4
gneill
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The 4V battery is sinking current (current is flowing into its + terminal). So rather than supplying energy to the circuit, it is absorbing energy and storing it. The magnitude of this energy consumption (looking at the 4V battery alone) is I3*4.00V.
 
  • #5
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But isn't the current that's flowing into its positive terminal the difference between part of I2 that went through the branch and the 4 V's battery's current ?

If this is correct, then calculating the 4 V battery's power means that we need to use the battery's own current

I'm really sorry for my loads of questions and my apparent stubbornness, but I really need to clear this point out.
 
  • #6
gneill
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After all the calculations are done, when you finally arrive at the total current I3 in the branch containing the 4V supply (and note that in this case I3 is the actual current that will flow in that branch, it's not a "partial" current, or abstract loop contribution), that is in fact the current that will be either flowing through the battery. If you put an ammeter in series with the battery, you will read the value of I3.

If I3 flows into the battery, then the battery is absorbing energy, not delivering energy to the circuit. If I3 flows into the battery, then there is no sneaky other bit that's flowing out while some other current flows on top of it to hide it. I3 is I3.

Don't be concerned that batteries can act as current sinks as well as sources; A battery will source or sink any amount of current required in order to maintain its designated potential difference. If that means absorbing current via its + terminal instead of sending current out via its + terminal, so be it.

If you consider a battery charger, for example, it deliberately forces a controlled amount of "reverse" current into a rechargeable battery, which then absorbs and stores the energy for release later when it powers a circuit.
 

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