Frequency response 2

The resistor with the capacitor are a voltage divider. Let's call $Z_c$ the capacitor's impedance and $R$ the resistor's impedance, we can calculate the voltage across the resistor as

$V_o = V_i \cdot \left(\frac{R}{R + Z_c}\right)$

We can calculate $Z_c$ in function of the frequency as $Z_c = \frac{-j}{2\pi f C}$ with $j = \sqrt{-1}$ . It's easy to see that the capacitor's impedance will get infinitesimally small as the frequency goes to infinity, which means that

$\left(\frac{R}{R + Z_c}\right) = \frac{R}{R} = 1$ so $V_o = V_i$

Since the voltages will be equal, the voltage effective voltage across the resistor will be equal to the effective voltage of the power source, which is $\frac{V_p}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10V$