Explanation Given data: The branch voltage in an R L C circuit is given as, d 2 v d t 2 + 4 d v d t + 8 v = 24 (1) The initial value of voltage ( v ( 0 ) ) is 0 . The value of d v ( 0 ) d t is 0 . Formula used: Write an expression to find the voltage response with the step input, if the roots of characteristic equation are real and imaginary. v ( t ) = V s + ( A 1 cos ω d t + A 2 sin ω d t ) e − α t . (2) Here, V s is the step response voltage, α is the neper frequency, ω d is the damped natural frequency, and A 1 and A 2 are constants. Write a general expression for the roots of characteristic equation when the roots are real and imaginary. s 1 = − α + j ω d (3) s 2 = − α − j ω d (4) Write an expression to solve quadratic equation. s 1 , 2 = − b ± b 2 − 4 a c 2 (5) Here, a is the coefficient of second order term, b is the coefficient of first order term, and c is the coefficient of constant term. Calculation: From equation (1), the characteristic equation is written as follows. s 2 + 4 s + 8 = 0 (6) From the equation (6), a = 1 b = 4 c = 8 Substitute 1 for a , 4 for b , and 8 for c in equation (5) to find s 1 , 2 . s 1 , 2 = − 4 ± ( 4 ) 2 − 4 ( 1 ) ( 8 ) 2 = − 4 ± 16 − 32 2 = − 4 ± − 16 2 = − 4 ± j 4 2 Simplify the equation. s 1 , 2 = − 4 + j 4 2 , − 4 − j 4 2 = − 2 + j 2 , − 2 − j 2 Therefore, the roots of characteristic equations are real and imaginary. s 1 = − 2 + j 2 s 2 = − 2 − j 2 Comparing the values of roots with the equation (3) and equation (4). α = 2 ω d = 2 Substitute 2 for α , and 2 for ω d in equation (2) to find v ( t ) . v ( t ) = V s + ( A 1 cos 2 t + A 2 sin 2 t ) e − 2 t (7) At t = ∞ , the R L C circuit reaches the steady state condition and rate of change of voltage with time equals to zero. d v ( ∞ ) d t = 0 Therefore, d v 2 ( ∞ ) d t 2 = 0 Substitute v ( ∞ ) for i in equation (1). d 2 v ( ∞ ) d t 2 + 4 d v ( ∞ ) d t + 8 v ( ∞ ) = 24 (8) For the step response R L C circuit, v ( ∞ ) = V s Substitute 0 for d v ( ∞ ) d t , 0 for d v 2 ( ∞ ) d t 2 and V s for v ( ∞ ) in equation (8) to find V s . 0 + 4 ( 0 ) + 8 V s = 24 V s = 24 8 = 3 V Substitute 3 V for V s in equation (7) to find v ( t ) . v ( t ) = 3 V + ( A 1 cos 2 t + A 2 sin 2 t ) e − 2 t (9) Substitute 0 for t in equation (9) to find v ( 0 ) . v ( 0 ) = 3 V + ( A 1 cos 2 ( 0 ) + A 2 sin 2 ( 0 ) ) e − ( 0 ) = 2 A+ ( A 1 ( 1 ) + A 2 ( 0 ) ) ( 1 ) { ∵ cos 0 ° = 1 , and sin 0 ° = 0 } v ( 0 ) = 3 V + A 1 (10) Substitute 0 V for v ( 0 ) in equation (10) to find A 1

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ISBN: 8220102801448

Author: Alexander

Publisher: YUZU

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Chapter 8, Problem 27P

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Chapter 8 Solutions

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS

Ch. 8.2 - The switch in Fig. 8.4 was open for a long time...Ch. 8.2 - For the circuit in Fig. 8.7, find: (a) iL(0+),...Ch. 8.3 - If R = 10 , L = 5 H, and C = 2 mF in Fig. 8.8,...Ch. 8.3 - The circuit in Fig. 8.12 has reached steady state...Ch. 8.4 - In Fig. 8.13, let R = 2 , L = 0.4 H, C = 25 mF,...Ch. 8.4 - Refer to the circuit in Fig. 8.17. Find v(t) for t...Ch. 8.5 - Having been in position a for a long time, the...Ch. 8.6 - Find i(t) and v(t) for t 0 in the circuit of Fig....Ch. 8.7 - Determine v and i for t 0 in the circuit of Fig....Ch. 8.7 - For t 0, obtain v0(t) in the circuit of Fig....

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Find the expression for v ( t ) .

Solution Summary: The author explains how to find the expression for v(t), if the roots of characteristic equation are real and imaginary.

Find the expression for v(t).

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Chapter 8 Solutions