Someone can explain me the phenomena in rlc circuit. If we have an input voltage \$V_e = V_{max}\sin(wt)\$. Why the current resulting when it is written in books is not a complex number? And why the current that passes in every component is the same? I know that you are going to tell me that the circuit is in series so the current is the same but I see it as being a complex number

  • 1
    \$\begingroup\$ There is no actual complex number in the sense that I cannot obtain 100j electrons or pass them through an ammeter. Are you confusing a phasor representation with an actual current? \$\endgroup\$
    – nanofarad
    Commented Mar 12, 2023 at 19:59
  • \$\begingroup\$ Yes you're right, I hadn't thought about it @nanofarad \$\endgroup\$
    – Mouh Kramo
    Commented Mar 12, 2023 at 20:08

3 Answers 3


Why the current resulting when it is written in books is not a complex number?

There is only one frequency when the current in an RLC series circuit is purely resistive. This is when the reactive impedance of the inductor and, the reactive impedance of the capacitor are exactly equal in magnitude. However, their reactive impedances are always opposite in phase polarity and, this means that their reactances will cancel to zero.

Another name for this is circuit resonance; the reactive impedances cancel to zero leaving only the resistor to define the current that flows.

  • \$\begingroup\$ When the current is purely resistive? \$\endgroup\$
    – Hearth
    Commented Mar 12, 2023 at 20:18
  • \$\begingroup\$ @Hearth not sure what clarification you might be seeking but, the current from the applied source (the sinewave alluded to by the op) has a phase angle relative to the source voltage that is purely resistive. \$\endgroup\$
    – Andy aka
    Commented Mar 12, 2023 at 21:02
  • \$\begingroup\$ I've never heard current or voltage phase angles referred to as resistive or reactive before. I thought you had meant to say impedance. \$\endgroup\$
    – Hearth
    Commented Mar 12, 2023 at 23:16

The series RLC circuit current driven by the sinusoid is easily solved in the time domain by using the defining relations for the individual components and Kirchhoff's Voltage Law and by using the fact that elements in series must have the same current. This results in a differential equation in current:$$\frac{d^2i}{dt^2}+\frac{R}{L} \frac{di}{dt}+\frac{1}{LC}i=\frac{1}{L}\frac{d}{dt}\left(V_{\text{max}}\sin(\omega t)\right)$$ During sinusoidal steady state the current will be a scaled and phase shifted version if the input voltage. Substituting: $$ i(t)=I(\omega)\text{sin}(\omega t+\theta(\omega))$$ \$I(\omega)\$ and \$\theta(\omega)\$ can be found using algebra and some trigonometry, without the use of complex variables. This is what an oscilloscope will display using a current probe.

Complex variables are introduced with impedance, allowing the use of standard network tools such as the voltage divider rule.$$Z_R(j\omega)=R+j0=R\angle{0^0}$$ $$Z_L(j\omega)=0+jL\omega=L\omega\angle{90^0}$$ $$Z_C(j\omega)=0+\frac{1}{jC\omega}=\frac{1}{C\omega}\angle{-90^0}$$

Since \$Z(j\omega)\$ is valid for the sinusoidal domain only, a different representation is require for voltage and current. The frequency is the same throughout so omitting it for now is not a problem. So:$$V_{max}\angle{0^0} \text{ and } I(\omega)\angle{\theta}$$ are used. These values are constants in the time domain, but are variable in the frequency domain.

The current can now be calculated as the ratio of a voltage across one of the elements to the element impedance. The voltage can be calculated with the voltage divider rule:$$I\angle{\theta}=\frac{v_a}{Z_a}, a=R,L,\text{ or }C$$ which can the be expanded to the form $$ i(t)=I(\omega)\text{sin}(\omega t+\theta(\omega))$$


The reason we don't express that current as a complex value, is because it's real! It can be measured, it has a relationship with time, and there's nothing imaginary about it.

You expressed the time-varying voltage as

$$ V_e = V_{max}sin(\omega t) $$

and I'm telling you that the resulting current in the loop will have an equally real time varying value. I mean that in the real sense of real, in that it's a tangible, measurable, thing:

$$ I = I_{max}sin(\omega t + \phi) $$

Its amplitude is \$I_{max}\$, its frequency is \$\omega\$, and \$\phi\$ is some delay (I'm reluctant to call it "phase shift", because actually it's an offset in time, and has the unit of seconds) relative to the voltage. There are no imaginary terms.

While it is useful to express impedances and signals in terms of complex \$s\$, in the frequency domain, you don't have too. You can stick to the time domain, if you wish, with all those horrific integrals and derivatives, and solve them there. It's possible, but very difficult. Importantly, in the time domain everything is expressed in terms of time \$t\$. There are no imaginary values anywhere in any expression or equation written in terms of time, at least not for real-life measurable phenomena like voltage and current.

The real world, the one we actually perceive and are familiar with, is the time domain, and in it everything is function of time, and everything is real, by definition. That includes voltage, and that includes current.

You may think that a sinusoidal function of time has frequency, phase and amplitude, but those are merely interpretations of a graph, information extracted artificially. The function \$Asin(\omega t+\phi)\$ is only describing some state at all instants in time, and anything else you say about it, including its amplitude, frequency and phase are just you interpreting the signal in terms of attributes belonging to the frequency domain.

Complex numbers are useful to encapsulate amplitude and phase in a single value, and as such they are key when working in the frequency domain (where everything is terms of \$s\$), but they do not feature in time-domain expressions.


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