# Phasor representation of RC circuit

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:

The phasor domain equation looks like this:

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

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!

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

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

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