Explanation Given data: Refer to Figure 16.62 in the textbook. Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (1) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (2) Here, L is the value of inductance. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (3) Here, C is the value of capacitance. Calculation: The given circuit is redrawn as shown in Figure 1. For a DC circuit, at steady state condition at time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the voltage across the capacitor is same as the source voltage ( v s ) Since the capacitor is open circuited, there is no current flows through the inductor. v C ( 0 − ) = 20 V i L ( 0 − ) = 0 A The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 0 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 20 V For time t > 0 , the circuit remains same as shown in Figure 1. Substitute 1 Ω for R 1 in equation (1) to find Z R 1 . Z R 1 = 1 Substitute 3 Ω for R 2 in equation (1) to find Z R 2 . Z R 2 = 3 Substitute 1 H for L in equation (2) to find Z L . Z L = s ( 1 H ) = s Substitute 1 F for C in equation (3) to find Z C . Z C = 1 s ( 1 F ) = 1 s Apply Laplace transform for i s ( t ) to find I s ( s ) . I s ( s ) = 40 s { ∵ L ( u ( t ) ) = 1 s } The voltage source is written as follows: v s = 20 ( 1 ) V = 20 u ( t ) V { ∵ u ( t ) = 1 for t > 0 } Apply Laplace transform for above equation to find V s ( s ) . V s ( s ) = 20 s { ∵ L ( u ( t ) ) = 1 s } Using element transformation methods with initial conditions convert the Figure 1into s -domain. Apply Kirchhoff’s voltage law for the circuit shown in Figure 3. ( 1 ) I ( s ) − 40 s + 3 I ( s ) + s I ( s ) + 20 s − v ( 0 ) s + 25 s I ( s ) = 0 4 I ( s ) + s I ( s ) + 25 s I ( s ) = − 20 s + v ( 0 ) s + 40 s I ( s ) [ 4 + s + 25 s ] = − 20 s + v ( 0 ) s + 40 s I ( s ) [ 4 s + s 2 + 25 s ] = − 20 s + v ( 0 ) s + 40 s Substitute 20 for v ( 0 ) in above equation. I ( s ) [ 4 s + s 2 + 25 s ] = − 20 s + 20 s + 40 s I ( s ) [ s 2 + 4 s + 25 s ] = 40 s Simplify the above equation to find I ( s ) . I ( s ) = 40 s ( s s 2 + 4 s + 25 ) I ( s ) = 40 s 2 + 4 s + 25 (4) From the equation (4), the characteristic equation is s 2 + 4 s + 25 = 0 (5) Write a general expression to calculate the roots of quadratic equation ( a s 2 + b s + c = 0 ) . s 1 , 2 = − b ± b 2 − 4 a c 2 a (6) Comparing the equation (5) with the equation ( a s 2 + b s + c = 0 ) . a = 1 b = 4 c = 25 Substitute 1 for a , 4 for b , and 25 for c in equation (6) to find s 1 , 2 . s 1 , 2 = − 4 ± ( 4 ) 2 − 4 ( 1 ) ( 25 ) 2 ( 1 ) = − 4 ± 16 − 100 2 ( 1 ) = − 4 ± − 84 2 = − 4 ± j 9.1652 2 Simplify the above equation to find s 1 , 2 . s 1 , 2 = − 4 + j 9.1652 2 , − 4 − j 9.1652 2 = − 2 + j 4.5826 , − 2 − j 4.5826 Substitute the roots of characteristic equation in equation (4) to find V ( s ) . I ( s ) = 40 ( s − ( − 2 + j 4.5826 ) ) ( ( s − ( − 2 − j 4.5826 ) ) ) = 40 ( s − ( − 2 + j 4.5826 ) ) ( s − ( − 2 − j 4.5826 ) ) Take partial fraction for above equation

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 39P

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Find the expression of current i(t) for t>0 in the circuit of Figure 16.62.

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Chapter 16, Problem 38P

Chapter 16, Problem 40P

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b) Draw the magnitude and phase bode plot c) Given Cdb=0.02pF, how will the frequency response change, draw the resulting magnitude and phase bode plotplz help me to solve part b and c.

Medium 1 is a lossless dielectric (ε₁, μ₁ = μo, σ₁ = 0) Medium 2 is a perfect electric conductor (PEC) ( 2 = 0, μ2 = μo, σ₂ = ∞) [ Moσ = 0] [ε0 μ₁ σ₂ = ∞ ] (J=σE is finite, E = 0) E(z) Exe² +Пe₁²] 1. For the case εr] = λι = = E2(z)-0 - 1 (vacuum), E₁x 1 V/m and a frequency f = 500 MHz determine: n₁ = 12= 2. Determine: r = T= 3. Using this I show that the total electric field E₁0(z) in region 1 can be written as: E(z) = -2jE, sin(2лz/λ)✰ 4. The magnitude E10(z) will show an interference pattern. The SWR (standing wave ratio) is the Emax/Emin ratio of the magnitude of the total electric field in region 1. What is the SWR? E (z) = 2|E|sin(2лz/2₁)| E" (z) SWR A Imax E(z) Imin 1+r 1-|| tot 5. Roughly SKETCH the magnitude of E10(z) and E20(z) on the graph below. E₁tot(z) tot E20(z) -0.40 -0.30 -0.ło z=0 +0.1b +0.20

Chapter 16 Solutions

Fundamentals of Electric Circuits

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Find the expression of current i ( t ) for t > 0 in the circuit of Figure 16.62.

Solution Summary: The author explains the formula used to calculate the impedance of a resistor in s -domain.

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Find the expression of current i(t) for t>0 in the circuit of Figure 16.62.

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Chapter 16 Solutions

Find the expression of current i ( t ) for t &gt; 0 in the circuit of Figure 16.62.

Solution Summary: The author explains the formula used to calculate the impedance of a resistor in s -domain.

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Find the expression of current i(t) for t>0 in the circuit of Figure 16.62.

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Chapter 16 Solutions

Find the expression of current i ( t ) for t > 0 in the circuit of Figure 16.62.