# 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!

## 1 Answer

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

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

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