Input capacitor stage of ADC
Such a circuit is called a switched capacitor resistor. To see why it behaves as a resistor, consider the circuit in steady state. It is assumed that the switching frequency is low enough so that the capacitor has time to charge to full potential which is applied to it.



simulate this circuit &ndash; Schematic created using CircuitLab
SW1 is closed and SW2 is open:
The capacitor was connected to voltage V2 so had initial charge \$q_i = CV_2\$. Now the capacitor has been connected to voltage V1 so has final charge \$q_f = CV_1\$. The amount of charge transferred from node V1 is:
$$q_{tr} = C(V_1-V_2)$$
SW2 is closed and SW1 open:
The capacitor was connected to voltage V1 so had initial charge \$q_i = CV_1\$. Now the capacitor has been connected to voltage V2 so has final charge \$q_f = CV_2\$. The amount of charge transferred to node V2 is:
$$q_{tr} = C(V_1-V_2)$$
On average, a charge of \$q_{tr}\$ is transferred from node V1 to node V2 in one switching period. If the switching frequency is \$f\$, then average charge transferred per unit time or average current is equal to:
$$i = fC(V_1-V_2)$$
$$\frac{V_1-V_2}{i} = \frac{1}{fC}$$
$$Z_{eff} = \frac{1}{fC}$$
Thus, the switched capacitor acts as an average resistor over the switching period with resistance \$\frac{1}{fC}\$.