@Thorondor has provided an excellent answer but I would like to expand on it with regard to the Faraday-Lenz' law, for Lenz' law is the fundamental origin of Kirchhoff's law.
The Faraday-Lenz law states that
$$\oint \vec{E}\cdot d\vec{\ell} = \Delta V_{\rm loop} = -\frac{d\Phi_B}{dt},$$
or, in differential form
$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}.$ Qualitatively, it means that a changing magnetic field creates a nonconservative electric field. Note that it is a loop integral, which means that we have to integrate around a closed loop.
So let's look at simple example, taken from the Electrical Engineering SE.
We have a battery, resistor, and capacitor. Let's calculate the voltage drop over each component. For the entire circuit,
$$\oint \vec{E}\cdot d\vec{\ell} = -\frac{d\Phi_B}{dt} = 0,$$
since there is no changing magnetic field here. With this specification we now have derived Kirchhoff's voltage law
$$ \Delta V_{\rm loop} = 0 = -\oint \vec{E}\cdot d\vec{\ell}.$$
So we now know that if we integrate around the entire loop, we're gonna have zero as our final answer. So let's do just that.
Start at the bottom left corner of the loop and let's integrate clockwise around it. First we encounter battery. Within the battery, the electric field points from the positive to negative terminal. We know we are going to have a voltage rise over the battery, so let's call that $+V_{\rm batt}$.
Now to integrate along the resistor:
$$ V_R = - \oint \vec{E}\cdot d\vec{\ell} = -IR.$$
(Note here that $\vec{E}$ and $d\vec{\ell}$ are parallel for the resistor.) The electric field is constant everywhere in the circuit in the steady-state of the system (i.e., after the battery has been connected for a short time). The electrons (or positive holes) speed up as they pass through the resistor, and they give off more thermal energy.
Lastly for the capacitor:
$$ V_C = - \oint \vec{E} \cdot d\vec{\ell} = -\frac{Q}{C}.$$
In total we have:
$$ V_{\rm batt} + V_R + V_C = 0$$
$$ V_{\rm batt} - IR - \frac{Q}{C}=0.$$
This is the same result you would get from using Kirchhoff's loop rule, but we have done it using the more general Faraday law. Faraday-Lenz' law is a fundamental equation in the study of electromagnetism. Its application permeates throughout our everyday existence through its responsibility for electric power generation. It is one of four Maxwell equations which, in combination with the Lorentz force law, to the best of our knowledge give a complete theory of electromagnetism.
Now, since $\vec{E}$ is constant throughout the circuit in the steady-state, it means that the electric potential varies linearly between the positive and negative terminal. That is, if $E = \rm const.$, then given $\vec{E} = -\nabla V$, we have something like
$$ V = -Ex + V_0.$$
The Drude model treats electrons as having random, frequent collisions with the metal atoms of the conductor. These electrons start moving, stop when they collide with the conductor, and then immediately re-accelerate. This gives them a net "drift speed". This process of colliding generates thermal energy. In the Drude model we have the current as
$$ I = nAu\bar{v},$$
where: $I$ is the current, $n$ is the number density, $A$ is the cross-sectional area of the conductor/wire, $u$ is a constant that characterizes the mobility of the charges, and $\bar{v}$ is the electron drift speed. This allows $I$ to remain constant as $A$ and $\bar{v}$ vary throughout the circuit. For example, if the cross-section of the wire decreases, the electrons will have to speed up to maintain the constant $I$, which is what happens in resistors.
