Final answer:
The output voltages of the circuit in the figure can be calculated for different input voltages using Ohm's Law and the formula for the reactance of a capacitor.
Step-by-step explanation:
The circuit shown in the figure is a low-pass filter that consists of a resistor and a capacitor. In this type of filter, low-frequency signals are passed through while attenuating high-frequency signals. To calculate the output voltage, we need to apply Ohm's Law and the formula for the reactance of a capacitor.
(a) When Vin = 0V, the capacitor acts as an open circuit to DC signals. Consequently, the output voltage is zero V.
(b) When Vin = -1V, we need to calculate the reactance of the capacitor using the formula Xc = 1 / (2πfC), where f is the frequency and C is the capacitance. Since the frequency is not given, we assume it to be 1 Hz. Xc = 1 / (2π * 1 * 1e-6) = 159.16 Ω. Applying Ohm's Law, Vout = Vin * (Xc / (Xc + R)) = -1 * (159.16 / (159.16 + 10k)) = -1.57 mV.
(c) When Vin = -2V, the calculation is similar to (b) with the exception that Vin is doubled. Vout = Vin * (Xc / (Xc + R)) = -2 * (159.16 / (159.16 + 10k)) = -3.14 mV.