A typical circuit for a low-pass filter is shown in the figure. An AC input V_i = 10 mV is applied at the left end, while the output V₀ is received at the right end. Determine the output voltages for frequencies = 10 kHz , 100 kHz , 1.0 MHz , and 10.0 MHz . Observe that as the frequency increases, the output decreases, which accounts for the name low-pass filter.

Step-by-step solution

The output voltage V₀ across the capacitor in an RC series circuit is given by: V₀ = V_i X_C Z = V_i ( 1 C ) R^2 + ( 1 C )^2 = V_i 1 + ( RC)^2 Given values are: V_i = 10 mV R = 1.0 k = 10^3 C = 10 nF = 10 10⁻⁹ F = 10⁻⁸ F The time constant is RC = 10^3 10⁻⁸ = 10⁻⁵ s . Also, = 2 . For = 10 kHz = 10^4 Hz : RC = 2 10^4 10⁻⁵ = 0.2 0.628 V₀ = 10 1 + (0.628)^2 = 10 1 + 0.394 8.47 mV 8.5 mV For = 100 kHz = 10^5 Hz : RC = 2 10^5 10⁻⁵ = 2 6.28 V₀ = 10 1 + (6.28)^2 = 10 1 + 39.4 1.57 mV 1.6 mV For = 1.0 MHz = 10^6 Hz : RC = 2