# What does KVL imply about about phasors in an AC circuit?

In a circuit with a DC source, KVL states that $$\sum_{i} V_{i}=0$$, so I am wondering if i would translate this law into AC circuits, would be the statement that the vectorial sum of phasors equal zero?

• Yes, but the moment you have inductors, KVL is not applicable and you should be using Faraday's Law directly. May 22 at 10:21
• @naturallyInconsistent that is not correct. Inductors as a lumped element are perfectly consistent with KVL. It is the lumped element, not the induction, that is the issue
– Dale
May 22 at 11:43
• @Dale That KVL can be hacked to accept inductors does not mean that KVL is applicable. The electric field can no longer be described by an electrostatic potential because things stop being conservative, and the assumptions inside KVL ceases to apply. May 22 at 11:52
• @naturallyInconsistent The definition of "voltage" in KVL and in theory of AC circuits is not integral of total electric field, but integral of its conservative component, thus, difference of the Coulomb-gauge electric potential. Thus KVL is always valid, trivially. May 22 at 17:12
• @naturallyInconsistent that is not the point or usual use of KVL. KVL does not provide expression for voltage on inductor, or make any statement on induction. KVL just states sum of potential drops in closed path is zero. Ideal inductors have potential drop $LdI/dt$, and this is often used with success. But this is additional assumption valid for ideal inductors only. Complicated induction scenarios with real inductors or with distributed parameters still obey KVL, but they are more complicated in the sense the potential drops are not expressible as $LdI/dt$. May 23 at 11:12

Yes, and it's routine to do so when designing linear circuits. However, it's often disguised by using the substitution $$i\omega\rightarrow s$$ to give the resulting polynomials real coefficients (Fourier versus Laplace transforms). That puts you on well-trodden algebraic paths.