A  capacitor and resistor are connected with an AC source as shown in figure. Reactance of capacitor is  $X_C$ = 3 $\Omega$ and resistance of resistor is 4 $\Omega$. Phase difference between current $I$ and $I_1$ is    $\left[{\tan }^{-1}\left(\frac{3}{4}\right)=37°\right]$

Currents in two branches are,

$I_2=\frac{V}{X_C}=\frac{V}{3}$  (here $V$= rms value of source voltage)

$I_1=\frac{V}{R}=\frac{V}{4}$

$I_{2 }$ is $90°$ ahead of applied voltage function and $I_1$ is in phase with it.

$\therefore \tan \varphi =\frac{{\displaystyle \raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=\frac{4}{3}$

$\Rightarrow \varphi ={\tan }^{-1}\left(\frac{4}{3}\right)=90°-{\tan }^{-1}\left(\frac{3}{4}\right)=90°-37°=53°$.