Let’s consider an RLC circuit connected with a sinusoidal source, my goal is to find the current \$I(t)\$, this is done by solving the following differential equation: \$ \frac{{d^2I}}{{dt^2}} + \frac{R}{L}\frac{{dI}}{{dt}} + \frac{1}{LC}I = -\omega V_0\sin(\omega t)\$ (1).
The general solution of (1) is of the form :

\$I(t) = I_h(t)+I_p(t)\$ (2)

Where \$I_h\$ and \$I_p\$ are the homogeneous and particular solutions respectively,In the steady-state situation (\$t\$ is very large) the homogeneous solution will vanish and we only left with the particular solution therefore (2) becomes \$I(t) = I_p(t)\$

So we need to seek a particular solution of (1), fortunately \$I_p(t)=A\cos(\omega t+\phi)\$

is a good candidate, so in order to keep the algebra simple we can turn \$I_p(t)\$ into a complex form and using Euler formula we get : \$I_p(t) = Ae^{j\omega t +\phi}\$

Now we can define impedance to be \$Z = V/I\$ I understand that impedance is kind of generalisation of Ohm’s law so that inductors and capacitors can be treated as simple resistors and we can apply preceding formula about resistors like voltage divider or the formulas of equivalent impedances instead of resistors, this is valid only where the current across each component is sinusoidal. The problem is when we have a circuit(contains only resistors, capacitors and inductors) connected with a sinusoidal source that contains many nodes and branches, why is it taken for granted that the current in each component of the circuit is sinusoidal? Should I justify that in each circuit by deriving the differential equation associated with it? Or is there a general demonstration that in steady-state the current of each component will behave this way?

  • \$\begingroup\$ Before I say anything else, did you look at the dimensions of your first equation -- left side vs right side? I suspect you forgot something on the left. \$\endgroup\$ Oct 21, 2023 at 10:59
  • \$\begingroup\$ You failed to make even the more basic dimensional cross-check of your first equation. In the time domain, it is \$\frac{\text{d}}{\text{d}t}v_{t}=L\left[\frac{\text{d}^2}{\text{d}t^2}+\frac{R}{L}\frac{\text{d}}{\text{d}t}+\frac1{L\,C}\right]i_{t}\$. Not as you expressed it at the outset. But in general, if you can find an annihilator for whatever \$\frac{\text{d}}{\text{d}t}v_{t}\$ happens to be (always possible if you can find its Laplace), then the resulting equation is readily solved. Part of that solution will include the non-homogeneous part with undetermined but findable coefficients. \$\endgroup\$ Oct 21, 2023 at 11:50
  • \$\begingroup\$ I forgot to divide the right-hand side by L \$\endgroup\$ Oct 21, 2023 at 16:10
  • \$\begingroup\$ Always check your work with dimensional analysis as your first go-to. It's the first thing I do every single time I do anything or read anything from someone else. And the answer is that you do not need to justify each case. The reasoning in large part has to do with superposition, complex analysis, and the ability to find winding frequencies (Fourier) which are always sums of sines and cosines and phases. \$\endgroup\$ Oct 21, 2023 at 19:34

1 Answer 1


Welcome to EE.SE

I understand that impedance is kind of generalisation of Ohm’s law ... why is it taken for granted that the current in each component of the circuit is sinusoidal?

Because impedance as a concept is limited to steady state (your particular solution), single frequency, sinusoidal operation.

By the late 19th century it was realised that RLC circuits could be described by differential equations. Unfortunately it requires considerable mathematical competence to solve them directly so there was a search for simplified solutions.

Impedance is one such simplified solution. Named by Oliver Heaviside, though contributed to by various people, it allows for solutions of RLC circuits by extending Ohm's law from DC and resistors to include (single frequency) AC and LC components.

However a restriction of impedance, which is usually unstated, is that it only provides a solution for steady state, single frequency, sinusoidal operation. If you want a description of the transient (homogeneous) or non-sinusoidal operation you'll need to use a more advanced method to provide a complete solution of the differential equation of the circuit.

The phasor approach you mention is one such method; another popular method is Laplace transforms.

  • \$\begingroup\$ My question was if I have a complicated circuit with many nodes and branches, let’s say there is a current passing through some branch we denote it by \$I_n(t)\$, why is it always taken for granted that \$I_n(t)=Ae^{j\omega t+\phi}\$? \$\endgroup\$ Oct 21, 2023 at 15:21
  • \$\begingroup\$ @user9276043 We can only answer the questions you actually write; not ones that are running round in your head. \$I_n(t)=Ae^{j\omega t+\phi}\$ is just the phasor representation of some current \$A \angle \phi\$ A (where \$A\$ is the peak current and the frequency is \$\omega\$ rad/s). Why is everything sinusoidal? Well, your exitation is sinusoidal - \$V_0\sin(\omega t)\$ from 1. \$\endgroup\$
    – Graham Nye
    Oct 21, 2023 at 22:28

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.