Finding phase of voltage divider

Question

I have this AC voltage divider here wish I want to find the exit voltage of. What I've done is I've applied the voltage divider formula:

$$ V_{out} = \frac{Z_2}{Z_1 + Z_2} V_{in}$$

Now that I have got the magnitude of \$ V_{out} \$, I am trying to figure out the sinusoidal function associated with it. But I'm not exactly sure how to do it

Diagram of this scenario:

$$ Z_{R-L} = \sqrt{ R^2 + w^2 L^2}$$

$$ Z_{C} = \frac{1}{wC}$$

hence,

$$ V_{out} = \frac{ V_{in}}{ 1+ wC \sqrt{R^2 + w^2 L^2} } $$

assume \$V_{in}\$ to be some generic sinusoidal function like \$ V_{in} = A sin( \omega t)\$

Answer 1

The question's values for impedances \$Z_{R-L}\$ and \$Z_C\$ are actually the magnitude of those impedances. I suspect that the approach was mixing time-domain and frequency-domain approaches; I will address both to make the distinction.

Since the inductor voltage depends on changing current and the capacitor current depends on changing voltage, a time domain analysis produces a differential equation.

$$ v_{OUT}(t) = v_C(t) = v_{IN}(t) - L\frac{d}{dt}i_{IN}(t) - Ri_{IN}(t) $$

and since the input current is split between the capacitor and output: $$ i_{IN}(t) = C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t) $$ putting them together: $$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) \\ &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\ \end{align}$$

This is still somewhat manageable, except that \$i_{OUT}\$ likely also depends on \$v_{OUT}\$. Flipping to the frequency domain, the diff-eq turns to algebra and we get a much better handle on the output load.

Let's show the output load explicitly, with an impedance of \$Z_{LOAD}\$. Also note that L and C have impedances \$j\omega L\$ and \$\frac{1}{j\omega C}\$, respectively. See this concise reference for more detail there. $$ $$

Preserving our output nets, we can lump together L with R (in series) and C with the load (parallel). This gives:

$$ \begin{align} Z_{LR} &= j\omega L + R \\ Z_{COUT} &= \bigg({\frac{1}{j\omega C}}^{-1}+{Z_{LOAD}}^{-1}\bigg)^{-1} \\ &= \frac{Z_{LOAD}}{j\omega C Z_{LOAD}+1} \end{align}$$

Now you can use the resistor divider rule to calculate a transfer function: $$ H = \frac{V_{OUT}}{V_{IN}} = \frac{Z_{COUT}}{Z_{LR}+Z_{COUT}} $$

And finally, the phase shift can be determined by comparing the real and imaginary parts of the transfer function: $$ \Theta = tan^{-1}\biggl(\frac{\mathfrak{Im}(H)}{\mathfrak{Re}(H)}\biggr)$$

Answer 2

If you need to know real values and want to examine different scenarios then use an on-line calculator to give the frequency response of this: -

Here's an example: -

You also get the formulas and derivation: -

Answer 3

You cant just add voltages in AC because the voltage drop of the resistor is in a different phase with the voltage drop of the capacitor and is in a different phase with the voltage drop of the inductor.

In AC generally:

Current through a resistor has the same phase with the voltage drop of the resistor.

In capacitors current leads voltage by pi/2 rad.

In inductors voltage leads current by pi/2 rad.

To find the total voltage drop you must draw this diagram:

The total voltage drop across all circuit elements will be sqrt(Vr^2+Vl^2+Vc^2).

To get the output voltage hide all the lines which represent the voltage drop of the resistor and the inductor:

Since we only have circuit elements connected in series the voltage will steal lead the current so the graph will look like that:

where I = Current.

I am a stubborn woman!