Explanation Calculation: The required diagram is shown in Figure 1. The inductive reactance X L is given by, X L = j ω L (1) Here, ω is the angular frequency. L is the inductance. The conversion from mH to H is given by, 1 mH = 10 − 3 H The conversion from 200 mH to H is given by, 200 mH = 200 × 10 − 3 H = 0 .2 H Hence, the inductance L is, L = 0.2 H The conversion from k Ω to Ω is given by, 1 k Ω = 1000 Ω The conversion from 100 k Ω to Ω is given by, 100 k Ω = 100 × 1000 Ω Hence, the resistor R is, R = 10 5 Ω The conversion from μ F to F is given by, 1 μ F = 10 − 6 F The conversion from 10 μ F to F is given by, 10 μ F = 10 × 10 − 6 F Hence, the capacitance C is, C = 10 × 10 − 6 F Substitute 0.2 H for L in equation (1). X L = j ω ( 0.2 H ) = j 0.2 ω Ω The capacitive reactance X C is given by, X C = 1 j ω C Substitute 10 × 10 − 6 F for C in the above equation. X C = 1 j ω ( 10 × 10 − 6 F ) = 1 j ω ( 10 × 10 − 6 ) Ω The steady state impedance Z between the terminals is given by, Z = [ ( 10 5 ∥ 1 j ω ( 10 × 10 − 6 ) ) + 2 ] ∥ [ 1 + j 0.2 ω ] = [ 10 5 ( 1 j ω ( 10 × 10 − 6 ) ) 10 5 + 1 j ω ( 10 × 10 − 6 ) + 2 ] ∥ [ 1 + j 0.2 ω ] = [ 10 5 + 2 ( 10 5 ) ( j ω ( 10 × 10 − 6 ) ) + 2 ( 10 5 ) ( j ω ( 10 × 10 − 6 ) ) + 1 ] ∥ [ 1 + j 0.2 ω ] Since, 10 5 ≫ 2 hence neglect 2 . Solve for the impedance Z as, Z = ( 10 5 + 2 j ω j ω + 1 ) ∥ [ 1 + j 0.2 ω ] = ( 10 5 + 2 j ω j ω + 1 ) ( 1 + j 0.2 ω ) 10 5 + 2 j ω j ω + 1 + 1 + j 0.2 ω = 10 5 + j 2 × 10 4 ω + 2 j ω − 0.4 ω 2 10 5 + 2 j ω + 1 + j ω + 0

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Chapter 15, Problem 24E

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The frequency at which the input impedance has the maximum magnitude.

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Chapter 15, Problem 23E

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Chapter 15 Solutions

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<

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The frequency at which the input impedance has the maximum magnitude.

Solution Summary: The inductive reactance X_L=jomega L is the angular frequency and the inductance is L=0.2.

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The frequency at which the input impedance has the maximum magnitude.

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