What is the cutoff frequency for the low-pass circuit shown if R1 = 1 k , R2 = 100 k , and C = 200 pF? R2/sCR2 +1/sCV.(s) = V;(s)-R1R2/SCRi +R2 +1/sCR2V.(s)A, (s) =V (s)R1 + R2R21+

Given R1=1 kΩR2=100 kΩC=200 pF The angular upper cut-off frequency for low pass filter is given by…

Answered: R2/sC R2 +1/sC V.(s) = V;(s)- R1 R2/SC…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Question

What is the cutoff frequency for the low-pass circuit shown if R₁ = 1 k , R₂ = 100 k , and C = 200 pF?

R2/sC
R2 +1/sC
V.(s) = V;(s)-
R1
R2/SC
Ri +
R2 +1/sC
R2
V.(s)
A, (s) =
V (s)
R1 + R2
R2
1+

Expert Solution

Step 1

Given

$\begin{array}{rcl} R_{1} & = & 1 kΩ \\ R_{2} & = & 100 kΩ \\ C & = & 200 pF \end{array}$

The angular upper cut-off frequency for low pass filter is given by

$ω_{H} = \frac{1}{(R_{1} ∥ R_{2}) C} . . . . . (1)$

The total resistance is given by

$\begin{array}{rcl} (R_{1} ∥ R_{2}) & = & \frac{R_{1} R_{2}}{R_{1} + R_{2}} \\ = & \frac{(1 \times 100) 10^{6}}{(1 + 100) 10^{3}} \\ = & 0.99 kΩ \end{array}$

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