Energy in DC Circuits: 12V, 2ohm Resistor(s)

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The discussion revolves around two ideal DC circuits powered by a 12V battery, one with a single 2ohm resistor and the other with two 2ohm resistors. Participants explore the implications of current flow and power dissipation in these circuits, particularly focusing on the relationship between energy supplied by the battery and the time each circuit can operate before the battery is depleted.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants examine the calculations of current and power dissipation in both circuits, questioning whether the conclusions drawn about the operational time of the circuits are correct. They also discuss the implications of treating the battery as an ideal voltage source versus a real one.

Discussion Status

Some participants have offered insights regarding the differences in energy supplied by the battery in each circuit and the resulting implications for battery life. There is an ongoing exploration of the assumptions made about the battery's behavior and the electrochemical processes involved.

Contextual Notes

Participants note the importance of considering the battery as an ideal source and the potential complications that arise if this assumption is not held. The discussion reflects uncertainty about the implications of the differing rates of energy supply in the two circuits.

pivoxa15
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Consider two ideal DC circuits both with a 12V battery. The first has one 2ohm resistor and in the second circuit contains two 2ohm resistors.

After some calculations, it is evident that a 6A current flows in the first circuit and a 3A current flows in the second. Assume the only power dissipated is via the resistors. The total power dissipated in the first circuit is 72W and in the second is 36W. The power supplied by the emf must also equal these rates. However, the two emfs are identical so must have supplied the same total amount of energy at equilibrium when the batteries are dead. Hence from the power of dissipation of each circuit it means circuit two will carry a current for twice the amount of time as the first circuit. Is this correct?
 
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You know the rules around here.

What do you think?
 
I made this problem up and have suggested what I think the answer may be in the second last sentence in the OP.
 
No one has answered yet. Is it too straight forward? I am a bit fuzzy on electronics. But the solution I suggeseted seem to go with general physical principles.
 
pivoxa15 said:
Consider two ideal DC circuits both with a 12V battery. The first has one 2ohm resistor and in the second circuit contains two 2ohm resistors.

After some calculations, it is evident that a 6A current flows in the first circuit and a 3A current flows in the second. Assume the only power dissipated is via the resistors. The total power dissipated in the first circuit is 72W and in the second is 36W. The power supplied by the emf must also equal these rates. However, the two emfs are identical so must have supplied the same total amount of energy at equilibrium when the batteries are dead. Hence from the power of dissipation of each circuit it means circuit two will carry a current for twice the amount of time as the first circuit. Is this correct?

Yes, if you consider the battery to be an ideal voltage source.
 
antonantal said:
Yes, if you consider the battery to be an ideal voltage source.

If we don't than all sorts of things could happen? i.e. too many things could happen and in the end we wouldn't know the net result.
 
In these two cases the rate of energy supplied by the battery differs for the two circuits (consider first connecting one circuit and then the other to the same battery). It just means that the electrochemical processes need to progress at a quicker rate in the first case than in the second. That is as the charge are removed quicker from the poles more chemical processes runs to replenish the charge at the poles.
 
andrevdh said:
In these two cases the rate of energy supplied by the battery differs for the two circuits (consider first connecting one circuit and then the other to the same battery). It just means that the electrochemical processes need to progress at a quicker rate in the first case than in the second. That is as the charge are removed quicker from the poles more chemical processes runs to replenish the charge at the poles.

So I was right in my OP? These batteries in the different circuits producing energy at different rates imply one will be dead before the other?
 
Yes, if the batteries are not recharged the battery that is supplying charge at a higher rate will run out of chemical reactions before the other.
 

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