Is it safe to apply Kirchhoff's voltage law to a closed loop containing an inductance with unsteady current? If I have a circuit that is just a battery in series with a resistor and an inductor, can I apply Kirchhoff's voltage law to that loop while the current has not reached its steady state value yet?

  • 1
    $\begingroup$ Stop using Kirchoff's law and use $\int E.dl = -\frac{d\phi}{dt}$. This is a maxwell's equation which is never wrong. $\endgroup$
    – Yashas
    Commented Feb 13, 2017 at 13:17
  • 1
    $\begingroup$ Walter Lewin explains this by himself youtube.com/watch?v=cZN0AyNR4Kw $\endgroup$ Commented Mar 27, 2017 at 12:26

5 Answers 5


Yes, the Kirchhoff Voltage Law (KVL):

Sum of voltages (here, meaning potential drops) across all elements connected via perfect conducting wire in series in to a closed circuit is zero.

is valid for lumped element RLC circuits, so also for inductors (for currents that do not change too fast, so voltage can be measured in practice). In practical circuits designed not to radiate, voltage can be measured across any element and KVL can be validated experimentally. It is valid for common frequencies, up to hundreds of MHz and even higher to GHz range if parasitic elements are added to the model.

The whole theory of RLC circuits with harmonic voltage sources is derived from KVL being valid all the time, while currents and voltages change.

Some people say KVL is not valid for a circuit with an inductor, since $\oint \mathbf E \cdot d \mathbf s \neq 0$ if ideal inductor is in the circuit. However, that is actually not a problem for KVL, because KVL is formulated using potential drops, not integrals of total electric field. Potential drop across inductor may be non-zero, even if total electric field in the wire is zero, because the drop is defined not by integral of total electric field, but by integral of electrostatic component of that field.

  • $\begingroup$ Hit's the core of the question (and the core of the common misconception people have with Kirchhoff's Law). $\endgroup$ Commented Jun 15, 2018 at 19:56
  • $\begingroup$ Clear cut explaination. $\endgroup$
    – Kashmiri
    Commented Oct 19, 2020 at 16:57

$\renewcommand{\vec}{\boldsymbol}$Yes, Kirchoff's law can be applied to inductive circuits.

Faraday's law of induction states that $$\oint \boldsymbol{E} \cdot \mathrm{d}\vec{l} = - \frac{\mathrm{d}\it\Phi_B}{\mathrm{d}t}\tag{1}$$ where $\vec{E}$ is the electric field and $\it\Phi_B$ is the magnetic flux passing through the loop at any given instant.

Now, the electric field generated in space is of two types, conservative and non-conservative. Therefore, $$\oint \vec{E} \cdot \mathrm{d}\vec{l} = \oint \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} + \oint \vec{E}_\text{non-conservative} \cdot \mathrm{d}\vec{l}.\tag{2}$$

The conservative field is produced by the accumulated charges in the circuit. Kirchhoff's second law states that the algebraic sum of all differences in potential around a complete circuit loop must be zero*, i.e. $$\oint \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} = 0 \tag{3}$$ which is to be expected by the definition of conservative fields. It states about the 'differences in potential'. We define a potential for conservative fields. The law does not make any statements about non-conservative fields. So, Kirchhoff's second law is correct and you can apply it to inductive circuits.

Applying the law to inductive circuits

Consider a closed circuit that has resistive wires of resistance $R$, a generator and an inductor of inductance $L$. Let us assume that the generator is creating a potential difference of $\mathcal{E}$ across its ends at some instant. At the same time, the potential differences across the resistor and the inductor are $iR$ and $ L\frac{\mathrm{d}i}{\mathrm{d}t}$.

How are these potential differences created? There are magnetic fields in the inductor and the generator. Consider the inductor; when the current passing through it is decreasing, an electric field is induced in it such that $\int \vec{E}_\text{non-conservative} \cdot \mathrm{d}\vec{l} = L\frac{\mathrm{d}i}{\mathrm{d}t}$ when integrated across the length of the inductor. (You can prove this using Faraday's law) Now, assuming that the charges in the circuit distribute themselves quickly, the charges distribute themselves across the ends of the inductor such that the non-conservative electric field is balanced by the electric field due to the accumulated charges. And the potential difference due to these accumulated charges is what you calculate when you apply Kirchhoff's second law to inductive circuits. Thus, $$\int \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} = L\frac{\mathrm{d}i}{\mathrm{d}t}$$ across the inductor. Similarly, charges get accumulated across the ends of the generator. And thus you can apply Kirchhoff's voltage law to a circuit containing inductors with unsteady currents.

*This statement is taken from Physics by Halliday, Resnick and Krane, 5th ed., Vol. 2. I cannot read German so I do not know what were the actual words used by Kirchhoff.

I did not include batteries in my discussion because the non-conservative forces in the battery which maintain the potential difference across the ends require quantum mechanics to explain them. Maxwell's equations alone cannot account for it. But still, $\oint \vec{E}_\text{conservative} \cdot \mathrm{d}\vec{l} = 0$ is valid by the definition of conservative fields. In case of a battery, you can assume that some non-conservative (chemical in nature) forces maintain a constant potential difference across the ends by accumulating charges. See this.


Golden Rule: Don't use Kirchhoff's law. Use maxwell's equations. They are always correct.

Kirchhoff's law is just a special case of the Maxwell–Faraday equation. $$\nabla \times E = -\frac{\partial B}{\partial t}$$ or in a simpler way (strictly speaking, this is also a special case)

$$\oint E.dl = -\frac{d\phi_B}{dt}$$

The above formula can be interpreted as, if you go around a circuit and sum up the potentials ($E.dx = dV$), it must be equal to $-\frac{d\phi_B}{dt}$. In most circuits, the right hand side of the integral evaluates to zero. This is Kirchhoff's law.

Whenever you are in doubt, refer to the Maxwell's equations. They will always work.

  • $\begingroup$ I do not think that Maxwell's equations are always correct. For example, they cannot explain the non-conservative forces in a chemical battery which give rise to emf. $\endgroup$ Commented Jun 12, 2018 at 9:40
  • 1
    $\begingroup$ @ApoorvPotnis, Maxwell's equations by themselves do not explain non-conservative forces in chemical battery, but those do not contradict Maxwell equations either. It is believed Maxwell's equations are correct even in battery, they just do not explain the forces that push the charges against electric force. $\endgroup$ Commented Jun 15, 2018 at 0:15
  • $\begingroup$ @JánLalinský Does this mean that if I take a loop which passes via the terminals and the inside of the battery, then the magnetic flux linking with that loop must be changing? $\endgroup$ Commented Jun 15, 2018 at 5:06
  • $\begingroup$ @JánLalinský Wikipedia states that "Some observed electromagnetic phenomena are incompatible with Maxwell's equations." $\endgroup$ Commented Jun 15, 2018 at 5:12
  • $\begingroup$ You say Maxwell equations do not work but you have used it in your answer. Kirchhoff's laws do not work because they are incomplete. They were developed to work with DC circuits. In your answer, you have used the same Maxwell's equation which I stated but then wrongly claim that it is equivalent to Kirchhoff's law. $\endgroup$
    – Yashas
    Commented Jun 15, 2018 at 7:58


Voltage between two points is just a difference of potentials of these points. So, sum of voltages in in closed loop must always be zero, it's just mathematics:

$$(p_2 - p_1) + (p_3 - p_2) + ... + (p_1 - p_n) = 0$$


If we have some electric circuit consisting of batteries, inductors, diodes, etc., the electric field is "potential". That means that it is possible to assign some number ("potential") to each point and the voltage between any two points would be the difference of above-mentioned potentials.

If the electric field is potential, the Kirchhoff's voltage law IS just a simple mathematical sequence of the fact that field is potential.

Electric field is NOT always potential. If there is a changing magnetic field, or some parts of an electric circuit are moving through magnetic field the electric field is not potential and you can not use KVL out of the box. Usually the effect of magnetic field can be substituted with some additional batteries and after that you are welcome.

Can you use KVL if there is an inductance in a circuit? Inductance coil is about changing magnetic field! But you can use KVL even in this case. The effect of magnetic field inside the coil is the same as an additional battery producing $L * \delta{I}/\delta{t}$, the effect of magnetic field on voltages outside the coil is much much smaller (approximately number-of-turns-of-the-coil times smaller).

How exactly you can use KVL in Alternating Current circuit? At any moment you can calculate potentials and the voltages on any element of the circuit. And the voltages at this (and any other) moment would obey KVL. But let's say peak voltages or root-mean-square voltages would not.

  • 2
    $\begingroup$ No, it can't be just mathematics since KVL is based on the fact that an electron can't gain or lose potential energy in moving a complete round in a closed loop. If what you are saying were true, we would be able to apply KVL in alternating current circuits, which is not the case. $\endgroup$ Commented Feb 13, 2017 at 9:43
  • $\begingroup$ If we have a loop with nontrivial inductance in a changing magnetic field, as in the OP, then no, the electric field isn't "potential". You can box that up into a single inductance if you do things right, but that's not what you're doing in this answer. $\endgroup$ Commented Feb 13, 2017 at 20:10

Yes, it is okay to apply Kirchhoff's law in an inductor+resistor circuit even if the current has not reached steady value yet. This is because Kirchhoff's law is a direct consequence of the conservation of energy, which always holds. At any instant, the back emf due to inductor and the potential drop across resistor must add up to voltage. If Kirchhoff's law does not hold, then it means that there has been a loss or gain in energy, which is not possible as voltage due to battery remains constant and an electron travelling through the circuit must always lose that much energy.

  • $\begingroup$ Why then can't it be applied to alternating current circuits? $\endgroup$ Commented Feb 13, 2017 at 9:47
  • 1
    $\begingroup$ It can be applied to alternating circuits. The applied ac voltage at any instant must be equal to the sum of instantaneous voltages across the inductor,resistor or capacitor. It has to hold, it must hold at any particular instant. $\endgroup$
    – TheFool
    Commented Feb 13, 2017 at 9:52
  • $\begingroup$ If you can post any link from literature, this would be greatly appreciated. $\endgroup$ Commented Feb 13, 2017 at 9:55
  • 1
    $\begingroup$ This is wrong,Electric field is nonconservative when magnetic field is changing and you cannot apply Kirchhoff's law then. $\endgroup$
    – Paul
    Commented Feb 13, 2017 at 10:03
  • 3
    $\begingroup$ This answer is completely incorrect. Kirchhoff's voltage law is not a "direct consequence of the conservation of energy", it is a restatement of Faraday's law, $\oint_{\partial S} \mathbf E\cdot\mathrm d\mathbf l = -\frac{\mathrm d}{\mathrm dt}\iint_S\mathbf B\cdot\mathrm d\mathbf S$ in magnetostatic situations in which the time derivative of the magnetic flux through the loop is zero and you get $\oint_{\partial S} \mathbf E\cdot\mathrm d\mathbf l = 0$ (a.k.a. "the sum of voltages in a loop is zero"). The OP's situation is precisely where this is no longer the case. $\endgroup$ Commented Feb 13, 2017 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.