While analysing AC circuits, we write voltage, current etc all with complex numbers namely "phasors". While studying the same, I wondered if Kirchhoff's laws held good with current and voltage in their phasor form. And the internet said they did! They argued somewhat as follows:

$I=Re[\vec{I}]=Re[I_{max}e^{j(\omega t+\phi)}]$

Now, $\Sigma I=0$ [By normal Kirchhoff's law]

Or, $\Sigma Re[I_{max}e^{j(\omega t+\phi)}]=0$

Or, $Re[\Sigma I_{max}e^{j\omega t} e^{j\phi}]=0$

Now, $e^{j\omega t}\neq 0$

Therefore, $\Sigma I_{max}e^{j\phi}=0$

i.e. $\Sigma\vec{I}=0$

Here $I$ stands for scalar current and $\vec{I}$ for phasor current. Similar argument went on for the voltage law.

I didn't get what they did in the sixth step. The fact that real part of a complex number is zero doesn't always imply that the number itself is zero. Can anyone please explain (if this is correct at all!)? And I would be glad if anyone kindly provides any argument, appropriate and more lucid, for the same. Thanks.

P.S. Here is the link to what I found on the internet.


2 Answers 2


Note that the sum of two phasors is another phasor:

$Ae^{j(wt+\phi_1)} + Be^{j(wt+\phi_2)} = Ce^{j(wt+\phi_3)}$

Where A,B,C are real.

The only way for the real part of the right side to be equal to zero at all times is if $C=0$. In which case the whole thing is $0$, both the real and complex parts are 0.

So a sum of phasors (of the full sinusoidal form with time) with a zero real part at all times, must have a zero complex part.

EDIT: Note that here by phasor, I mean the full sinusoidal form with the time included: $Ae^{j(wt+phi_1)}$, not necessarily the shortened $Ae^{j\phi_1}$.

To be clear, we can have:

$Ae^{j\phi_1} + Be^{j\phi_2} = Ce^{j 90^\circ}$

where C is nonzero and in this situation we have a $0$ real part, and a nonzero complex part on the right hand side. This doesn't apply to our situation, because we're talking about time-signals that are 0 at all times. If I take $Ce^{j 90^\circ}$ and convert it to the time domain, I get $Ccos(wt+90^\circ)$ where C is nonzero. This isn't 0 at all times as we require. Only way to get $0$ at all times is if $C=0$.

The thing with the kirchoff laws is that they give $0$ for all time.

To put it simpler:

If some time signal $v(t)$ is $0$ for all t, then its phasor form must be $0$ (complex and real parts are both 0).

Another way to look at it:

Phasor transforms are linear and one-to-one. So a $v_2(t) = k v_1(t)$, where k is some real number, then in the phasor domain $V_2 = kV_1$. so if $k=0$ we immediately get a 0 phasor in the phasor domain for $v_2$.

  • $\begingroup$ What is a phasor transform? What do you mean by 'time signal'? I'm a mere curious high school student. So please don't use technical terms like that...I mean, I'll be really grateful if you provide links to where these beasts can be studied. However, I understood the answer. Many many thanks. $\endgroup$ Apr 20, 2016 at 12:11
  • $\begingroup$ By time-signal, I just mean a normal voltage or current as a function of time. When we turn this time-signal into a phasor, I'm calling that process a phasor transform. 'Time-signals' in the time-domain, with corresponding phasor signals in the phasor-domain. I highly recommend getting a text on Linear Circuits. If you know some calculus you should be able to handle it fine. There's a lot of college course notes online, but I'm not sure which are good. $\endgroup$ Apr 20, 2016 at 12:36

Great observation! The solution lies in the fact that the complex number $\sum I e^{j\phi}$ is rotating with a speed $\omega$ counterclockwise in the complex plane. Consider what the expression is telling you -

The real part of a complex number that is rotating counterclockwise in the complex plane is always zero or if we take the projection of the rotating complex number in the real axis we see nothing.

We can therefore conclude that the complex number is zero, i.e., $\sum I e^{j\phi} = 0 + j \cdot 0$.

  • $\begingroup$ Note that we do have a LaTeX editor (specifically MathJax), details of which can be found in help center (search for 'notation') $\endgroup$
    – Kyle Kanos
    Mar 10, 2019 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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