An AC source is connected with a resistance (R) and an uncharged capacitance (C), in series. The potential difference across the resistor is in phase with the initial potential difference across the capacitor for the first time at the instant (assume that at $t=0$, emf is zero)

Let $V=V_0\sin \omega t$  (as $V=0$ at t = 0 ) Then $V_R=V_0\sin \omega t$ and $V_C=V_0\sin (\omega t-\pi /2)$

V and $V_R$ are in same phase. While $V_C$ lags V (or $V_R$ ) by ${90}^{\circ }.$ Now $V_R$ is in same phase with initial potential difference across the capacitor for the first time when,

Phase of potential difference across the resistor $=\omega t$

Phase of potential difference across the capacitor$=\omega \left(0\right)-\frac{\pi }{2}$

They will be equal when,

$\omega t=-\frac{\pi }{2}+2\pi =\frac{3\pi }{2}, \therefore t=\frac{3\pi }{2\omega }$