So Kirchhoff's second law states the sum of the emfs around a circuit is equal to the sum of potential drops. All explanations point to the conservation of energy. Fine. But why do electrons have to lose all their gained energy from batteries to the components. Conservation of energy would still hold if they lost a bit of the energy to components yet retained the remainder as kinetic energy as it passed back through the battery?

  • $\begingroup$ Intuitively, note that a battery has a fixed voltage. Thus, it theoretically always maintains the same difference in potential energy between its terminals. By definition, circuits just don't function in the way you described. The electrons will always lose the same amount of energy by virtue of the fact that they are traveling from one terminal to the other, but the way this is manifest varies based on the components they travel through. $\endgroup$ – user122423 Jun 6 '17 at 23:20

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More than conservation of energy, this law is related with the fact that electric field is a conservative one, indeed, if $V$ is the electric potential, then in a static situation

$$ \vect{E} = -\nabla V $$

therefore if $a$ and $b$ are two points and $\gamma$ is a path connecting these points, the work to move a charge $q$ from $a$ to $b$ is

$$ \int_\gamma{\rm d}\vect{l}\cdot (q\vect{E}) = -q\int_\gamma{\rm d}\vect{l}\cdot \nabla V = -q\int_a^b{\rm d}V = -q(V_a - V_b) $$

If $\gamma$ is a closed curve, then this last term vanishes and

$$ \oint_\gamma{\rm d}V = 0 $$

And this is actually the reason behind Kirchhoff's second Law. Imagine that you split the path $\gamma$ into a bunch of segments $p_1\to p_2 \to \cdots p_N \to p_1$. This last equations ensures that after you measure the potential between consecutive poitns $p_i p_{i+1}$ and then add the values up, the result should be exactly zero


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