# Voltage in the phasor domain

Given the below circuit (at regime) I have to calculate the voltage $v_u(t)$, by knowing that.

$I_0=3$ mA, $I_s(t)=I_m sin(2\pi f t)$, $I_m=2$ mA, $f=1$ kHz, $R=1$k$\Omega$, $C=100$ nF. simulate this circuit – Schematic created using CircuitLab

I solved it with the uperposition principle: $v_u(t)=v_u'+v_u''=v_{u_{\Big|I_0}}+v_{u_{\Big|I_s}}$

• $v_{u_{\Big|I_0}}=I_0\frac{R}{1/(j\omega C)+R}\cdot \frac{1}{j\omega C}=I_0\frac{1/(j\omega C)}{1/(j\omega C)+R}\cdot R=Z\cdot I_0$, where $Z=\frac{R \cdot 1/(j\omega C)}{R+1/(j\omega C)}=\frac{R}{j\omega RC+1}$

• $v_{u_{\Big|I_s}}=I_s\frac{R}{1/(j\omega C)+R}\cdot \frac{1}{j\omega C}=I_s\frac{1/(j\omega C)}{1/(j\omega C)+R}\cdot R=Z\cdot I_s$, where $I_S=I_m$ (by assuming the phase of $I_s$ as reference phase).

Thus the phase taken as reference is $\angle I_s=\angle sin(2\pi f t)=\angle cos(2\pi f t - \frac{\pi}{2})=-\frac{\pi}{2}$

$Z=|Z|e^{j\angle Z}$, $|Z|=|R|/|1+j\omega RC|=...=846.73 \Omega$, $\angle Z=\angle R - \angle (1+j\omega RC)=-arctan(\omega RC)=-arctan(2\pi fRC)=0.561$ rad

By substituing the values, I obtain: $v_u(t)=(3+2)Z=5\cdot 1.69 sin (\omega t - 0.561)$ V,(s,rad)

Afterwards, if I calculate the module and phase, I can write $v_u(t)$ as somethinng like $v_u(t)=A\cdot cos(\omega t + \phi)$

Is the whole process right?

Just in one thing I'm not so sure: my book calculates $v_{u_{\Big|I_0}}=R I_0$. Why?

Superimposition is OK, but you made a mistake: $I_0$ is a DC component, so it cannot pass through a capacitor (as you say, we are analyzing steady-state). Therefore there is no point in computing the impedance, because it is apparent that the capacitor won't contribute to it (at DC it has infinite impedance, so in the parallel with R is completely negligible, if you want to use this POV).

Therefore $I_0$ will pass through R alone, that's why your book computes $v_u'$ simply as $R I_0$.

At the end you will have a DC component in $v_u$ due to $I_0$ and a sinusoidal component with frequency $f$ due to $I_s$, so your output voltage will have this form:

$v_u(t) = V_0 + A \cos(2 \pi f t + \phi)$

EDIT

(in response to a comment)

The impedance of a capacitor in the phasor domain is

$Z = \dfrac{1}{j\omega C} = j \, \dfrac{-1}{\omega C} = j X_C$

where $X_C=\dfrac{-1}{\omega C}$ is called the reactance of the capacitor.

If you want to make a comparison with what happens in DC circuits, i.e. with resistance, you should take the modulus of the impedance $|Z|$, which expresses intuitively how much the current flow is impeded when trying to flow in the capacitor.

A you can see: $|Z| = \dfrac {1}{\omega C} \rightarrow \infty \quad\textrm{as}\quad \omega \rightarrow 0$

And since $I=\dfrac V Z \;\Rightarrow\; |I|= \dfrac{|V|}{|Z|} = 0$ when $\omega=0$

• So anytime there is a constant current (non-dependent from time) the capicitor will have infinite resistance? – sl34x May 27 '15 at 8:23