find current, voltage and power at each branch using fourier series analysis
First, I will present a general method of finding your transfer function. I will use Mathematica to do it.
Well, we are trying to analyze the following circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_1}{\text{R}_3}\\
\\
\text{I}_2=\frac{\text{V}_2}{\text{R}_4}
\end{cases}\tag2
$$
Now, it is not hard to solve for \$\text{V}_1\$:
$$\text{V}_1=\frac{\text{V}_\text{i}\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)+\text{R}_3\left(\text{R}_2+\text{R}_4\right)}\tag3$$
Where I used Mathematica-code to solve for that:
In[1]:=FullSimplify[
 Solve[{I1 == I2 + I3, I1 == (Vi - V1)/R1, I2 == (V1 - V2)/R2, 
   I3 == V1/R3, I2 == V2/R4}, {I1, I2, I3, V1, V2}]]

Out[1]={{I1 -&gt; ((R2 + R3 + R4) Vi)/(R3 (R2 + R4) + R1 (R2 + R3 + R4)), 
  I2 -&gt; (R3 Vi)/(R3 (R2 + R4) + R1 (R2 + R3 + R4)), 
  I3 -&gt; ((R2 + R4) Vi)/(R3 (R2 + R4) + R1 (R2 + R3 + R4)), 
  V1 -&gt; (R3 (R2 + R4) Vi)/(R3 (R2 + R4) + R1 (R2 + R3 + R4)), 
  V2 -&gt; (R3 R4 Vi)/(R3 (R2 + R4) + R1 (R2 + R3 + R4))}}


When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

$$\text{R}_1=\frac{1}{\text{sC}_1}\tag4$$
$$\text{R}_2=\text{sL}\tag5$$
$$\text{R}_4=\frac{1}{\text{sC}_2}\tag6$$

Using the convolution theorem of the Laplace transform we can find that:
$$\text{V}_1\left(t\right)=\int_0^t\text{V}_\text{i}\left(\tau\right)\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{\frac{1}{\text{sC}_2}\left(\text{sL}+\text{R}_4\right)}{\frac{1}{\text{sC}_1}\left(\text{sL}+\frac{1}{\text{sC}_2}+\text{R}_4\right)+\frac{1}{\text{sC}_2}\left(\text{sL}+\text{R}_4\right)}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag7$$
Using Mathematica, we can see that:
$$\text{V}_1\left(t\right)=\int_0^t\left\{\sum_{\text{n}\ge1}\frac{4\cos\left(\text{n}\tau\right)}{\pi\text{n}^3}\right\}\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{\frac{1}{\text{sC}_2}\left(\text{sL}+\text{R}_4\right)}{\frac{1}{\text{sC}_1}\left(\text{sL}+\frac{1}{\text{sC}_2}+\text{R}_4\right)+\frac{1}{\text{sC}_2}\left(\text{sL}+\text{R}_4\right)}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag8$$