To determine Find the expression for the voltage v o for t ≥ 0 . Explanation Given data: Refer to Figure given in the textbook. The value of the inductor is L = 6.25 H . The value of the capacitor is C = 25 μ F . The value of the resistor is R = 200 Ω . The value of the voltage source is 7.5 V. Formula used: Write the general expression for resonant radian frequency. ω 0 = 1 L C (1) Write the general expression for neper frequency. α = 1 2 R C (2) Write the general expression for damping frequency. ω d = ω 0 2 − α 2 (3) Write the general expression for current through the capacitor. i C ( t ) = C d v o ( t ) d t (4) Calculation: In the given circuit, no energy is stored in the capacitor initially. Therefore, v o ( 0 + ) = 0 V As t → ∞ , the capacitor acts as an open circuit and inductor behaves like a short circuit. Therefore, V f = 0 V Substitute 200 for R and 25 μ for C in equation (2) as follows, α = 1 2 × ( 200 ) × ( 25 × 10 − 6 ) { ∵ 1 μ = 10 − 6 } α = 100 rad s Squaring the above expression as follows, α 2 = 10000 rad s Substitute 25 μ for C and 6.25 for L in equation (1) as follows, ω 0 = 1 ( 25 × 10 − 6 ) ( 6.25 ) { ∵ 1 μ = 10 − 6 } ω 0 = 80 rad s Squaring the above expression as follows, ω 0 2 = 6400 rad s As α 2 > ω 0 2 , the output response is overdamped. Hence, the output equation will be as follows, v o ( t ) = V f + A 1 ′ e s 1 t + A 2 ′ e s 2 t (5) The general expression for roots of the characteristic equation is, s 1 , 2 = − α ± α 2 − ω 0 2 Substitute 100 for α , 10000 for α 2 , and 6400 for ω 0 2 in the above equation as follows, s 1 , 2 = − 100 ± 10000 − 6400 = − 100 ± 60 Therefore, the roots are, s 1 = − 40 rad s s 2 = − 160 rad s Substitute 0 for V f , − 40 for s 1 , and − 160 for s 2 in equation (5) as follows, v o ( t ) = A 1 ′ e − 40 t + A 2 ′ e − 160 t (6) Under steady state condition, for t < 0 the switch is kept open and the inductor is short circuited. In Figure 1, the initial current flowing through the inductor is, i L ( 0 − ) = i L ( 0 + ) = 7.5 200 = 0.0375 A For t > 0 , the switch gets closed, and the modified circuit is shown in Figure 2. In Figure 2, apply Kirchhoff’s current law at node V as follows, 7.5 − v o ( 0 + ) 200 = i L ( 0 + ) + i C ( 0 + ) (7) Substitute 0 for v o ( 0 + ) and 0

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ISBN: 9780134746968

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Electric Circuits. (11th Edition)

Ch. 8.1 - The resistance and inductance of the circuit in...Ch. 8.2 - Use the integral relationship between iL and v to...Ch. 8.2 - Prob. 3AP Ch. 8.2 - Prob. 4AP Ch. 8.2 - Prob. 5AP Ch. 8.3 - Prob. 6AP Ch. 8.4 - Prob. 7AP Ch. 8.4 - Prob. 8AP Ch. 8.4 - Repeat Assessment Problems 8.7 and 8.8 if the 80 Ω...Ch. 8 - The resistance, inductance, and capacitance in a...

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