Frequency response 3

When $\omega \to 0$ , the voltage source becomes a DC (direct-current) source. Then the capacitor will take on the voltage of the voltage source when fully charged that is $\boxed{10} \text{ V}$ .

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 capacitor as

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

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 go to infinity as the frequency goes to zero, which means that

$\lim_{Z_c \to 0}\left(\frac{Z_c}{R + Z_c}\right) = \lim_{Z_c \to 0}\left(\frac{1}{1 + \frac{R}{Z_c}}\right) = 1$ so $V_o = V_i$

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