# How to understand Kirchhoff's Voltage Law?

I'm having a hard time trying to understand why Kirchhoff's Voltage Law is true. I looked for an answer on the forum but I couldn't find a convincing one.

So my question is : How to "physically" understand Kirchhoff's Voltage Law, with a precise microscopical interpretation and if possible without analogies (with gravity for example).

I understand that a battery creates a potential difference in the circuit, but I don't have the least idea as to why, say for a simple circuit composed of an ideal voltage generator and a resistor, the voltage of the resistor has to be equal to the voltage of the generator.

What is happening when electrons flowing from the negative terminal to the positive terminal of the generator passe through the resistor? I know that they are somehow moving from the a lower to a higher electrical potential, but what are the precise "energy trades" made there?

Sorry if the question is confusing or unclear, and thanks a lot to the people that will take the time to answer <3

• Not sure if it's worth our time, since you have already have found a satisfactory answer: but I disagree in a fundamental way with all of these answers. You cannot physically understand the voltage law without considering Faraday-Lenz' law. – zhutchens1 Dec 18 '18 at 4:00
• May I please ask you to help me understand that? – s89ne Dec 18 '18 at 7:40
• See my posted answer. – zhutchens1 Dec 18 '18 at 15:33

Fundamentally, it's because of the conservation of energy. Voltage is the same as electrical potential; an electron moving through a potential difference of 1 V gains or loses precisely 1 eV = $$1.6\times 10^{-19} J$$ of energy. We know that if an electron travels all the way around a circuit and returns to its starting point, it must have the same energy it started with. Therefore, it must have traveled through a total voltage difference of zero.

Note that Kirchhoff's voltage law breaks down when a changing magnetic field is present inside the loop. This is a consequence of the Maxwell-Faraday equation.

• Presented this way, it is a very convincing answer. Thanks a lot. But for some reason, I still have some trouble seeing the full picture. Would you mind continuing the discussion tomorrow, after I think this over? <3 – s89ne Dec 17 '18 at 20:52
• Sure, feel free to ask more questions later. I'll be happy to answer if I have time. – Thorondor Dec 17 '18 at 20:57
• Good morning, what forms of energy does an electron carry. I know it carries some Electric potential energy but in what other form of energy is it "converted"? Is it, as for the case of gravitational energy, kinetic energy? – s89ne Dec 18 '18 at 8:02

It is because of how voltage is defined: as difference of electric potential. Electric potential is a number for each point of space.

Choose two points in space, say, on terminals of a resistor you mentioned. At each of these two points, electric potential has some value. Voltage between them is defined uniquely as difference of those two potentials.

At the same time, most usually any short wire can be assumed to have the same potential throughout, and if connected to any other wire, their potential is, in stable state, the same.

So voltage across resistor must be the same as voltage across the voltage source, because their points of contact can be used as those two points used to define that voltage.

I understand that a battery creates a potential difference in the circuit, but I don't have the least idea as to why, say for a simple circuit composed of an ideal voltage generator and a resistor, the voltage of the resistor has to be equal to the voltage of the generator.

It's a feedback loop. If the resistor has a lower voltage drop than the voltage source, then charges would gain energy each time through the loop. This energy accelerates the charges and allows the current in the circuit to increase. This increased current causes a greater voltage drop in the resistor. When the voltage drop is equal, there's no additional energy to accelerate the charges and the current stops changing.

• Could you please specify what forms of energy an electron carries. I know it carries some Electric potential energy but in what other form of energy is it "converted"? Is it, as for the case of gravitational energy, kinetic energy? – s89ne Dec 18 '18 at 7:47
• It certainly has some non-zero kinetic energy, but in a circuit, most of the energy of the current is in an induced magnetic field. The wires in the circuit have a positive inductance and as an inductor, stores energy as current flows ($E = 0.5LI^2$) – BowlOfRed Dec 18 '18 at 8:09

@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.