In an RLC series circuit let applied EMF be given $V=V_0\sin\omega t$, $$Z=Z_C+Z_R+Z_L=R+i\left(\frac{1}{\omega C}-\omega L\right)$$ $$|Z|=\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}$$

Then $$i(t)=\frac{V(t)}{Z}=\frac{V_0e^{i\omega t}}{R+i\left(\frac{1}{\omega C}-\omega L\right)}$$

Its given in my book that $$i(t)=\frac{V_0(\sin\omega t+\phi)}{\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}}$$

Why are they considering a phase difference of $\phi$?

Also, why are they taking modulus of $Z$ and only the imaginary part of applied voltage?

What is the difference between the first $i(t)$ and the second $i(t)$?

  • $\begingroup$ The first one is Complex Impedance : RLC SERIES CIRCUIT RESONANCE(Complex impedance) $\endgroup$
    – hxri
    May 4, 2016 at 6:33
  • 2
    $\begingroup$ The use of complex numbers in RLC circuits with a sinusoidal voltage source is for mathematical convenience, at the end of the day you only measure real currents and real voltages. $\endgroup$
    – jim
    May 4, 2016 at 11:06
  • $\begingroup$ You have sign errors in the capacitive and inductive impedances. $Z_L = i\omega L$ and $Z_C=-i/\omega C$. While it won't affect the magnitude, it will affect the sign of the phase and determine whether the current leads the voltage or vice-versa. $\endgroup$
    – Bill N
    May 4, 2016 at 15:18
  • $\begingroup$ For circuit analysis, one cannot use $i$ as the imaginary unit $\sqrt{-1}$, because it is used to represent time-varying current. Therefore the imaginary unit is denoted by $j$. (And electrical engineers tend to use $j$ at all times, whether current is a variable in the current work or not) $\endgroup$
    – Ben Voigt
    May 4, 2016 at 19:25

2 Answers 2


Why are they considering a phase difference of $\phi$?

Your calculations are not totally correct. The voltages across different impedants $V_C,V_R,V_L$ have a phase relationship between them and hence the different impedances $Z_C,Z_R,Z_L$ are not directly linearly related as you have done.

Consider the following phasor diagram:

enter image description here

I hope it is clear from the phasor diagram above as to why you have a phase difference of $\theta$, or $\phi$ as in your case.

Also, why are they taking modulus of $Z$?

The vector $V_S$ as in the above phasor diagram has a magnitude $\sqrt{(V_L-V_C)^2+V_R^2}$ and from there you can calculate the value of $Z$. So you see how the absolute value of $Z$ comes into play.

Also, why are they taking only the imaginary part of applied voltage?

You need to remember that $e^{i\omega t}$ is a complex number and contains an imaginary part. But current and voltage are all real quantities and hence they cannot contain the imaginary unit $i$. But again such calculations are much more simplified when we use complex numbers. So we usually use complex numbers while calculating and then while obtaining the final result, we take either the imaginary part of the answer or the real part, depending on the nature of the supply voltage. Here the voltage changes sinusoidally as given in the question, so the current also changes sinusoidally. In general, however, when it is not mentioned whether the voltage changes sinusoidally or varies with $\cos$ it makes no difference if you take the real part instead of the imaginary part.

  • $\begingroup$ So I cant add the impedences directly in terms of complex numbers ? $\endgroup$
    – Aditya Dev
    May 4, 2016 at 14:38
  • $\begingroup$ @HariPrasad 's pdf mentions the same expression I got. $\endgroup$
    – Aditya Dev
    May 4, 2016 at 14:58
  • $\begingroup$ Observe carefully. HariPrasad's pdf mentions the expression your book contains, not the one that you have deduced. Also, you can add the impedances directly in terms of complex numbers. That's why they are used, for mathematical simplicity. But you have forgotten to include the phase lag in your calculation, for which your solution gives an inaccurate result. $\endgroup$ May 4, 2016 at 18:35

Schrodinger's Cat explains well why only part of the impendance is taken.

Why are they considering a phase difference of $\phi$ ... Also, why are they taking modulus of Z

Here's an algebraic explanation:

For the RLC circuit, we can write the total impedance in a general form, $$Z_T=R + j Z_r,$$ where $Z_r$ is the total reactive impedance of the $L$'s and $C$'s, $R$ is the total effective external resistance, and $j^2=-1$, to avoid confusion with the variable we use for current.

Any complex number can be written in at least two forms: $$z=a+jb \text{ or } z=|z|e^{j\phi}$$ where $$\phi = \arctan\left(\frac{b}{a} \right)\text{ and }|Z|=\sqrt{a^2+b^2}.$$

For the RLC that means that $$Z_T=\left(\sqrt{R^2+Z_r^2}\right)e^{j\phi}$$ where $\phi = \arctan\left(\frac{Z_r}{R} \right)$

Using this form for $i(t)$ we get $$i(t)=\frac{V(t)}{Z_T}=\frac{Ve^{j\omega t}}{|Z_T|e^{j\phi}}=\frac{Ve^{j\left(\omega t-\phi\right)}}{\left(\sqrt{R^2+Z_r^2}\right)}$$

We can then extract the appropriate part of this complex voltage based on the source voltage, as Schrodinger's Cat describes.


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