How to calculate capacitator voltage over time for a RC circuit with AC power source?

Question

I know with DC power source, the voltage for capacitator is expressed as:

https://www.electronics-tutorials.ws/rc/rc_1.html

Can the same equation be applied to AC power source?

Answer 1

You can analyze a RC-series circuit with an arbitrary input voltage \$\text{V}_\text{i}\left(t\right)\$, by using Laplace transform and applying the convolution property:

$$ \begin{alignat*}{1} \text{V}_\text{C}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{v}_\text{i}\left(\text{s}\right)}{\displaystyle\text{R}+\frac{\displaystyle1}{\text{sC}}}\cdot\frac{1}{\displaystyle\text{sC}}\right]_{\left(t\right)}\\ \\ &=\int\limits_0^t\mathscr{L}_\text{s}^{-1}\left[\text{v}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle1}{\displaystyle\text{R}+\frac{\displaystyle1}{\text{sC}}}\cdot\frac{1}{\displaystyle\text{sC}}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int\limits_0^t\text{V}_\text{i}\left(\tau\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle1}{\displaystyle\text{sCR}+\frac{\displaystyle\text{sC}}{\text{sC}}}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int\limits_0^t\text{V}_\text{i}\left(\tau\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle1}{\displaystyle\text{sCR}+1}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int\limits_0^t\text{V}_\text{i}\left(\tau\right)\cdot\frac{\displaystyle\exp\left(-\frac{\displaystyle t-\tau}{\displaystyle\text{CR}}\right)}{\displaystyle\text{CR}}\space\text{d}\tau\\ \\ &=\frac{1}{\text{CR}}\int\limits_0^t\text{V}_\text{i}\left(\tau\right)\exp\left(\frac{\displaystyle\tau-t}{\displaystyle\text{CR}}\right)\space\text{d}\tau \end{alignat*} \tag1 $$

So, let's do a few examples:

• When \$\text{V}_\text{i}\left(t\right)\$ is a stable DC-voltage for \$t\ge0\$ equal to \$\hat{\text{u}}_\text{i}\$, we get: $$\text{V}_\text{C}\left(t\right)=\frac{1}{\text{CR}}\int\limits_0^t\hat{\text{u}}_\text{i}\exp\left(\frac{\displaystyle\tau-t}{\displaystyle\text{CR}}\right)\space\text{d}\tau=\hat{\text{u}}_\text{i}\left(1-\exp\left(-\frac{t}{\text{CR}}\right)\right)\tag2$$
• When \$\text{V}_\text{i}\left(t\right)\$ is given by \$\sin\left(t\right)\$ for \$t\ge0\$, we get: $$ \begin{alignat*}{1} \text{V}_\text{C}\left(t\right)&=\frac{1}{\text{CR}}\int\limits_0^t\sin\left(\tau\right)\exp\left(\frac{\displaystyle\tau-t}{\displaystyle\text{CR}}\right)\space\text{d}\tau\\ \\ &=\frac{\displaystyle\text{CR}\left(\exp\left(-\frac{t}{\text{CR}}\right)-\cos\left(t\right)\right)+\sin\left(t\right)}{\displaystyle1+\left(\text{CR}\right)^2} \end{alignat*} \tag3 $$
• When \$\text{V}_\text{i}\left(t\right)\$ is given by \$\exp\left(-t\right)\$ for \$t\ge0\$, we get: $$\text{V}_\text{C}\left(t\right)=\frac{1}{\text{CR}}\int\limits_0^t\exp\left(-\tau\right)\exp\left(\frac{\displaystyle\tau-t}{\displaystyle\text{CR}}\right)\space\text{d}\tau=\frac{\displaystyle\exp\left(-\frac{t}{\text{CR}}\right)-\exp\left(-t\right)}{\displaystyle\text{CR}-1}\tag4$$