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I've been studying DC circuits recently, but I've run into some confusion concerning how batteries deliver energy to circuit elements.

From what I've read online, it seems that energy in circuits comes from waves that propagate through charges rather than the charges themselves. However, my book seems to convey that charges "pick up" potential energy at a battery and lose it in circuit elements.

These models of how energy is transferred seem to contradict each other. Which is right? And if the wave theory is right, then what does current or voltage have to do with it? Rather, how do current, voltage, and resistance explain the strength and properties of this wave?

If this matters, I'm using Halliday-Resnick-Krane 5th Edition Physics textbook.

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marked as duplicate by Aaron Stevens, Community Jul 21 at 3:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I've never heard anything about waves in DC circuits $\endgroup$ – Aaron Stevens Jul 20 at 21:36
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    $\begingroup$ In a DC circuit, the voltages and currents are constant which implies that the associated electric and magnetic fields are constant. As @AaronStevens points out, there aren't any propagating waves associated with a DC circuit. However, there is a non-zero poynting vector, i.e., there is an energy flow associated with the fields. Perhaps this is what you were thinking about rather than waves? $\endgroup$ – Alfred Centauri Jul 20 at 21:45
  • $\begingroup$ I see. Seems like I misunderstood something. So assuming charge carriers are having their energy reduced through circuit elements, what is that energy, exactly? I know the voltage drop means there's a dip in potential energy per charge between those points, but what is that a energy converted into? Also, if a circuit has no elements, does the charge not lose any charge? That seems intuitive, but I know that goes against the KVL. $\endgroup$ – wagboi Jul 21 at 0:26
  • $\begingroup$ That duplicate question you referenced cleared up a lot of my questions. Thanks boss. $\endgroup$ – wagboi Jul 21 at 2:18
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So assuming charge carriers are having their energy reduced through circuit elements, what is that energy, exactly? I know the voltage drop means there's a dip in potential energy per charge between those points, but what is that a energy converted into?

Like @Aaron Stevens I've never heard anything about waves in DC circuits. I confess to not knowing anything about non zero poynting vectors that @Alfred Centauri mentioned.

Some of the potential energy that a battery supplies to electrons is inevitably converted to heat in the resistance that all circuit elements have according to

$$E_{R}=I^{2}Rt$$

Some of potential energy of the electrons may be converted to energy stored in the magnetic field of any inductance in the circuit, according to

$$E_{L}=\frac{LI^2}{2}$$

And finally some of the potential energy of the electrons may be stored as potential energy in the electric fields of capacitors according to

$$E_{C}=\frac{CV^2}{2}$$

Also, if a circuit has no elements, does the charge not lose any charge? That seems intuitive, but I know that goes against the KVL.

At a minimum, there has to be some circuit resistance. If nothing else, all real batteries have internal resistance. If there was no circuit resistance, the charge would accelerate without opposition under the influence of the electric field, resulting in limitless current.

Hope this helps.

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  • $\begingroup$ Ok, that makes sense. Another question: from what I understand, a positive charge has a higher potential near the positive terminal and a lower potential near the negative terminal due to laws of electrostatic repulsion. So, if a positive charge has zero potential energy by the time it reaches the negative terminal, how does the current not decrease? It seems like having less energy would result in less charge per second, even though I know that can't be true. $\endgroup$ – wagboi Jul 21 at 2:31
  • $\begingroup$ @wagboi The battery is constantly supplying potential energy to positive charge returning to the negative terminal thereby maintaining constant current $\endgroup$ – Bob D Jul 21 at 5:03
  • $\begingroup$ I see. Thanks for the help. $\endgroup$ – wagboi Jul 21 at 12:23

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