I am struggling to understand how to take the KCL equation from time domain to Phasor domain. The time domain KCL equation is easy to understand and represented here:

RC circuit Time domain

The phasor domain equation looks like this:

Phasor domain

And the book i am using describes the phasor domain equation like this:

Phasor Domain

They tell me that "the time factor $e^{jwt}$ has disappeared because it was contained in all three terms". I am having a hard time convincing myself that this is true and was hoping someone could provide me with some clarity on this.

Thank you!


1 Answer 1


Unless, your point is entirely different, this should be fairly trivial.

I'll start from you first equation

$$ R\cdot Re\left\{ (\tilde{I}e^{j\omega t})\right\} +\frac{1}{C}Re\left\{\frac{\tilde{I}}{j\omega}e^{j\omega t}\right\} = Re\left\{\tilde{V}_se^{j\omega t}\right\}$$

because, $\tilde{I}, \tilde{V}_s$ are real quantities we can write:

$$ R\tilde{I}\cdot Re\left\{ e^{j\omega t)}\right\} +Re\left\{\frac{1}{j\omega C}\right\}\tilde{I}Re\left\{e^{j\omega t}\right\} = \tilde{V}_sRe\left\{e^{j\omega t}\right\}$$

because $Re\left\{ e^{j\omega t)}\right\}$ is a non zero, positive quantity for all t, we can divide both sides of the equation with it: so

$$\frac{1}{Re\left\{ e^{j\omega t)}\right\}} \left[R\tilde{I}\cdot Re\left\{ e^{j\omega t)}\right\} +Re\left\{\frac{1}{j\omega C}\right\}\tilde{I}Re\left\{e^{j\omega t}\right\} \right]= \frac{1}{Re\left\{ e^{j\omega t)}\right\}} \left[\tilde{V}_s Re\left\{e^{j\omega t}\right\}\right]$$

$$R\tilde{I} + Re\left\{\frac{1}{j\omega C}\right\} \tilde{I}= \tilde{V}_s $$


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