Question B.18: Finding the effective time constant The circuit in Fig. B.48...

If the source is shorted, the effective resistance seen by capacitor C is the parallel combination of R_{1}, R_{2}, \text { and } R_{3} . The effective resistance R is given by

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} \text { or } R=\frac{R_{1}}{3}=\frac{6 k }{3}=2 k \Omega

(a) The effective time constant is

\tau=C R=2 k \Omega \times 0.1 \mu F =0.2 ms

(b) The cutoff frequency is

\omega_{ o }=\frac{1}{\tau}=\frac{1}{0.2 ms }=5000 rad / s \text {, or } 795.8 Hz

(c) At ω = 0, capacitor C is open-circuited, and the output voltage has a finite value. At a high frequency, tending to infinity (ω = ∞), capacitor C is short-circuited, and the output voltage becomes zero. This is a lowpass circuit with f1 = 0 and f2 = fo = 795.8 Hz. Thus, the bandwidth is

BW =f_{2}-f_{1}=795.8 Hz