# Phase difference between voltage across a capacitor and the source in a low pass filter

Consider I have a low pass filter like that in the figure.

simulate this circuit – Schematic created using CircuitLab

I have considered the voltage source to be $$\V\$$ and the voltage across the capacitor to be $$\V_c\$$. Since, this source is an AC source, we know that the reactance of the capacitor is $$\X_C\$$.
What I want to do is, I want to establish an equation expressing the phase difference between the voltage drop across the capacitor and the voltage source.
Here, I shall start by considering:

$$V_C=i_CX_C=\frac {i_C}{\omega C}$$ In order to calculate $$\i_C\$$, we can use the equation: $$i_C=\frac {dq}{dt}=C\frac{dV_C}{dt}$$ I have tried putting $$\V_C=V-V_R\$$ but have no idea of what to do next. I couldn't calculate any derivative since the function seems to be bizzare to me.

• Are you familiar with Laplace-domain circuit analysis? This is one of those things that it makes trivial. Nov 20 at 16:57

Well, the transfer function of the circuit is given by:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right):=\frac{\displaystyle\underline{\text{V}}_{\space\text{o}}}{\displaystyle\underline{\text{V}}_{\space\text{i}}}=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}}{\displaystyle\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}+\text{R}}=\frac{\displaystyle\frac{\displaystyle\text{j}\omega\text{C}}{\displaystyle\text{j}\omega\text{C}}}{\displaystyle\frac{\displaystyle\text{j}\omega\text{C}}{\displaystyle\text{j}\omega\text{C}}+\text{j}\omega\text{CR}}=\frac{\displaystyle1}{\displaystyle1+\text{j}\omega\text{CR}}\tag1$$

So, for the argument we get:

\begin{alignat*}{1} \arg\left(\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(\frac{\displaystyle1}{\displaystyle1+\text{j}\omega\text{CR}}\right)\\ \\ &=\arg\left(1\right)-\arg\left(1+\text{j}\omega\text{CR}\right)\\ \\ &=0-\arctan\left(\frac{\displaystyle\omega\text{CR}}{\displaystyle1}\right)\\ \\ &=-\arctan\left(\omega\text{CR}\right) \end{alignat*}\tag2

• Which is also equals to $\mathrm{-arctg(1/f_0)}$. Nov 21 at 13:44

Mixing frequency domain with time domain won’t work. Pick one or the other.

The impedance of capacitance $$\Z_C=-jX_C\$$ is a frequency domain quantity: $$Z_C(j\omega)=\frac{-j}{\omega C}$$

The first equation should be written: $$V_C(j\omega)=I_C(j\omega)Z_C(j\omega)= -j\frac{I_C(j\omega)}{\omega C}$$

This is called the defining relation for capacitance in the frequency domain.

The second equation is correct, but let me write it so that its time domain nature is clear. $$i_C(t)=\frac{dq_C(t)}{dt}=C\frac{dv_C(t)}{dt}$$

This is called the defining relation for capacitance in the time domain.

Clearly these equations cannot be made to work together.

Notice that uppercase variables are used for the frequency domain and constants. Lowercase is used for time domain. Using this formalism allows writing the equations without the arguments because they are implied. $$V_C=I_CZ_C= -j\frac{I_C}{\omega C}$$

And

$$i_C=\frac{dq_C}{dt}=C\frac{dv_C}{dt}$$

So choose either frequency domain or time domain

Frequency domain

Use phasors notation with the voltage divider rule. Or other network techniques.

Time domain

Use KVL, then manipulate into a differential equation in $$\v_C(t)\$$. Solve using standard techniques. Start by assuming a solution $$\v_C(t)=V_A\text{sin}(\omega t+\theta)\$$.

Just use Voltage Divider Law on the two components....

Vc/V = ZC/(ZC+ZR) = (1/jwC)/(R+(1/jwC)) = 1/(1+jwCR)

Then use standard phase (and magnitude if you wanted to) equation, arctan(imaj/real).

Angle of Vc = -arctan(wCR)

There is a negative sign in the above expression because the factor 1+jwCR is the denominator which results in a lagging phase angle of Vc compared to V which is what we would expect for a low pass filter.

This yields the same transfer function as the first method above so you can now use the standard equation as before to find an expression for the phase.