I imagine before a circuit flows the electrons are still. Some work needs to be done to establish a current. Otherwise we just arrive at a situation where a battery gives 8 Joules of energy per Coulomb of electrons and a resistor takes that exact same amount. Am I right? I ask because this seems to violate Kirchhoff's law and I can't find anywhere on the internet about this.
6
-
3$\begingroup$ "Where is the energy to start the electrons moving in the first place" The battery...? $\endgroup$– Vincent ThackerCommented Sep 22 at 19:13
-
1$\begingroup$ You're completely right, and I have no idea why this question was closed. The extra kinetic energy you're talking about is real, and it leads to an effect called kinetic inductance. However, for ordinary sized circuits, the amount of energy required is very small since electrons are very light, so we don't bother talking about it in introductory physics. In microscopic superconducting circuits, kinetic inductance effects can be very important. $\endgroup$– knzhouCommented Sep 30 at 20:51
-
1$\begingroup$ To answer your question more directly: yes, since the naive Kirchoff's loop rule is about the energy of charges, it is violated when you have to account for the kinetic energy of the charges. But when those effects matter, we usually just move that term to the other side of the equation and say it's the "voltage drop due to kinetic inductance" instead, so that we don't have to change the form of Kirchoff's loop rule. $\endgroup$– knzhouCommented Sep 30 at 20:54
-
$\begingroup$ @knzhou KVL is about sum of potential differences in a closed path being zero, not "about energy of charges". KVL is not violated when "kinetic energy of the charges" is taken into account; KVL does not claim any definite value for potential difference on an inductor. Kinetic inductance is just another contribution to net self-inductance, this has nothing to do with validity of KVL. $\endgroup$– Ján LalinskýCommented Oct 1 at 0:46
-
$\begingroup$ @JánLalinský Aren't you just repeating what I said in my second comment? A change in kinetic energy is not actually a "potential difference", unless you're defining the word "potential" in a very unusual way. But we can keep KVL in the same form if we pretend it is one, so that's what we often do. $\endgroup$– knzhouCommented Oct 1 at 1:00
|
Show 1 more comment
2 Answers
$\begingroup$
$\endgroup$
Current is made of moving massive charged particles, so it carries some kinetic energy. So indeed, in addition to energy that is being lost to resistance, transferred to load, or stored in magnetic field (inductors), some energy has to be put into kinetic energy of the current as well, to get it flowing. This kinetic energy is usually many times smaller than other energies in the circuit, so usually it is neglected.
This affects self-inductance $L$ of circuit elements. In case of an inductor, it adds a contribution to $L$ that is proportional to mass of the charged particles, but almost insensitive to geometry of its coils. So instead of net self-inductance being equal to contribution $L_{ind}$ due to induced electric field, which is often calculated as magnetic flux per unit current, there is additional contribution $L_k$, due to non-zero inertial mass of the charge carriers, so we have $L= L_{ind}+L_k$. Then potential difference on the inductor is $V =(L_{ind}+L_k)\frac{dI}{dt}$, and net energy stored in the inductor is $\frac{1}{2}(L_{ind} + L_k)I^2$.
This does not affect the KVL rule in any way; it is valid, independently of whether the charge carriers have zero or non-zero mass, because it just states that sum of potential differences in closed path equals zero and this holds in any case. Just use the correct net self-inductance $L = L_{ind}+L_k$ to express potential difference on inductor $V = L\frac{dI}{dt}$.
$\begingroup$
$\endgroup$
Otherwise we just arrive at a situation where a battery gives 8 Joules of energy per Coulomb of electrons and a resistor takes that exact same amount.
There is nothing wrong with that. The electrons also have some KE, but that KE is constant over time at any location in a steady circuit.
I ask because this seems to violate Kirchhoff's law
Kirchoff’s laws are an approximation to Maxwell’s equations. In the condition where they hold, they are valid and self consistent. However, one of those assumptions is that we are not interested in very short time scales where the finite speed of light is importantly.
In particular, as a switch is closed we ignore the initial transient. During this time the KE of the charge carriers increases throughout the circuit and not the full amount of current is going through the resistor so not the full voltage is dropped over the resistor
