Explanation Given data: For series LR circuit: The value of the inductor ( L ) is 2 mH . The value of the resistor ( R ) is 10 k Ω . Formula used: Write a general expression to calculate the impedance of a resistor in frequency domain. Z R = R (1) Here, R is the value of the resistor. Write a general expression to calculate the impedance of an inductor in frequency domain. Z L = j ω L (2) Here, L is the value of the inductor. Calculation: The series LR circuit with the output across the resistor is drawn as Figure 1. The frequency response circuit of Figure 1 is drawn as Figure 2 using the equations (1) and (2). Apply voltage division rule for the circuit in Figure 2. V o ( s ) = R R + s L V i ( s ) (3) Here, V o ( s ) is the output voltage response of the system, and V i ( s ) is the input voltage response of the system. Rearrange the equation (3) to find V o ( s ) V i ( s ) . V o ( s ) V i ( s ) = R R + s L (4) Substitute j ω for s in equation (4) to find V o V i . V o V i = R R + j ω L = R R ( 1 + j ω L R ) V o V i = 1 ( 1 + j ω L R ) (5) Write a general expression to calculate the transfer function of a system. H ( ω ) = V o V i (6) Here, V o is the output response of the system, and V i is the input response of the system. Therefore, from the equation (5), the equation (6) becomes, H ( ω ) = V o V i = 1 ( 1 + j ω L R ) (7) To show that the series LR circuit is a low pass filter, consider the conditions used in low-pass filter. For low-pass filter, H ( 0 ) = 1 , H ( ∞ ) = 0 . Substitute 0 for ω , 2 mH for L and 10 k Ω for R in equation (7) to find H ( 0 ) . H ( 0 ) = 1 ( 1 + j ( 0 ) ( 2 mH ) ( 10 k Ω ) ) = 1 ( 1 + 0 ) = 1 Substitute ∞ for ω , 2 mH for L and 10 k Ω for R in equation (7) to find H ( ∞ ) . H ( ∞ ) = 1 ( 1 + j ( ∞ ) ( 2 mH ) ( 10 k Ω ) ) = 1 ( 1 + ∞ ) = 1 ∞ = 0 From the above calculations, it is clear and shown that the series LR circuit with the output across the resistor is a low pass filter

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 14, Problem 47P

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Show that a series LR circuit is a low pass filter if the output is taken across the resistor and find the value of the corner frequency.

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Chapter 14, Problem 46P

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Chapter 14 Solutions

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Show that a series LR circuit is a low pass filter if the output is taken across the resistor and find the value of the corner frequency.

Solution Summary: The author illustrates a series LR circuit with the output across the resistor and find the value of the corner frequency.

Show that a series LR circuit is a low pass filter if the output is taken across the resistor and find the value of the corner frequency.

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Chapter 14 Solutions