Why Does Kirchhoff's Voltage Law Apply to LR Circuits?

[Lost1ne]

1) Take a non-steady circuit such as an LR circuit. Why does Kirchoff's voltage law work when analyzing such a circuit? Is it because we're assuming that dI/dt and thus dB/dt are approximately zero thus meaning that curl E is approximately zero?
2) ε, the electromotive force, is the line integral of the force per unit charge integrated around a circuit. Although I feel many texts don't make this distinction clear (maybe because it's trivial), this is not necessarily equal to -dΦ/dt, Φ being the magnetic flux through our designated surface following Faraday's Law, correct? If so, we would already have a clear contradiction using an example like a steady circuit consisting of a battery and a resistor. -dΦ/dt is only the induced EMF, and this adds algebraically with a pre-existing EMF, right? In the end, the net EMF must follow the line integral definition.

 

[Dale]

Take a non-steady circuit such as an LR circuit. Why does Kirchoff's voltage law work when analyzing such a circuit?

Circuit theory is based on three assumptions. The first is that there is no net charge on any component. The second is that there is no magnetic flux outside any component. The third is that the circuit is small enough that the speed of light can be considered to be instantaneous.

With those three assumptions Maxwell’s equations reduce to Kirchoff’s laws. So the reason that Kirchoff’s laws work for a R.L. Circuit is simply that the circuit obeys those assumptions to a very good approximation.