Explanation Given data: Refer to Figure given in the textbook. The value of the inductor is L = 20 mH . The value of the capacitor is C = 500 nF The value of the voltage source is 36 V. The value of the current source is 20 mA. Formula used: Write the general expression for resonant radian frequency as follows, ω 0 = 1 L C (1) Write the general expression for neper frequency as follows, α = 1 2 R C (2) Write the general expression for damoing frequency as follows, ω d = ω 0 2 − α 2 (3) Write the general expression for current through the inductor as follows, i L ( t ) = L d i L ( t ) d t (4) Calculation: In the given circuit, no energy is stored in the capacitor initially. Therefore, v o ( 0 + ) = 0 V . The switch in the given circuit has been open a long time before closing at t < 0 . Therefore, the capacitor acts as an open circuit and inductor behaves like a short circuit The initial current through the inductor i L ( 0 − ) is calculated as follows, i L ( 0 − ) = 36 300 = 0.12 A As no energy is stored in the capacitor initially, therefore v C ( 0 − ) = 0 V . The given circuit is modified as shown in Figure 2 for t > 0 . In the given circuit, the voltage source with series resistance is converted into current source with parallel resistance using source transformation method. The value of the current source is calculated as follows, I = 36 300 = 0.12 A The modified circuit is shown in Figure 3. In Figure 3, the 300 ohms and 150 ohms resistor are connected in parallel. Therefore, the equivalent resistance for the parallel connection is, 300 Ω | | 150 Ω = 300 Ω × 150 Ω 300 Ω + 150 Ω = 100 Ω And, The equivalent current source is, I 1 = 0.12 A − 20 × 10 − 3 A { ∵ 1 m = 10 − 3 } = 0.1 A The modified circuit is shown in Figure 4. Substitute 100 for R and 500 n for C in equation (2) as follows, α = 1 2 × ( 100 ) × ( 500 × 10 − 9 ) { ∵ 1 n = 10 − 9 } = 10000 rad s Squaring the above expression as follows, α 2 = 10000 2 rad s Substitute 500 n for C and 20 m for L in equation (1) as follows, ω 0 = 1 ( 500 × 10 − 9 ) ( 20 × 10 − 3 ) { ∵ 1 n = 10 − 9 , 1 m = 10 − 3 } ω 0 = 10000 rad s Squaring the above expression as follows, ω 0 2 = 10000 2 rad s As α 2 = ω 0 2 , the output response is critically damped. Hence, the output equation will be as follows, i L ( t ) = I f + D 1 t e − α t + D 2 e − α t (5) In Figure 4, the value of I f is, I f = 0.1 A Substitute 0.1 for I f and 10000 for α in equation (5) as follows, i L ( t ) = 0

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Chapter 8, Problem 37P

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Find the expression for the inductor current iL(t) for t≥0.

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Chapter 8, Problem 36P

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Chapter 8 Solutions

EBK ELECTRIC CIRCUITS

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 - The switch in the circuit shown has been in...Ch. 8.4 - Prob. 8AP Ch. 8 - Prob. 1P Ch. 8 - Prob. 2P

Ch. 8 - The resistance in Problem 8.1 is decreased to80 Ω...Ch. 8 - Prob. 4P Ch. 8 - Prob. 5P Ch. 8 - The natural voltage response of the circuit in...Ch. 8 - Prob. 7P Ch. 8 - Prob. 8P Ch. 8 - The natural response for the circuit shown in Fig....Ch. 8 - Prob. 10P Ch. 8 - The two switches in the circuit seen in Fig.P8.11...Ch. 8 - The resistor in the circuit of Fig. P8.11 is...Ch. 8 - The resistor in the circuit of Fig.P8.11 is...Ch. 8 - The switch in the circuit of Fig. P8.17 has been...Ch. 8 - The inductor in the circuit of Fig. P8.17 is...Ch. 8 - The inductor in the circuit of Fig. P8.17 is...Ch. 8 - Design a parallel RLC circuit (see Fig. 8.1) using...Ch. 8 - Prob. 18P Ch. 8 - Prob. 19P Ch. 8 - Find υ(t) for t ≥ 0 in the circuit in Problem 8.19...Ch. 8 - Prob. 21P Ch. 8 - Prob. 22P Ch. 8 - The initial value of the voltage υ in the circuit...Ch. 8 - Prob. 24P Ch. 8 - Prob. 25P Ch. 8 - Prob. 26P Ch. 8 - Prob. 27P Ch. 8 - Prob. 28P Ch. 8 - Prob. 29P Ch. 8 - Prob. 30P Ch. 8 - The switch in the circuit in Fig. P8.31 has been...Ch. 8 - Prob. 32P Ch. 8 - There is no energy stored in the circuit in Fig....Ch. 8 - For the circuit in Fig. P8.30, find υo for t ≥...Ch. 8 - The switch in the circuit in Fig. P8.36 has been...Ch. 8 - Prob. 36P Ch. 8 - Prob. 37P Ch. 8 - Prob. 38P Ch. 8 - Prob. 39P Ch. 8 - Find the voltage across the 80 nF capacitor for...Ch. 8 - The initial energy stored in the 31.25 nF...Ch. 8 - In the circuit in Fig. P8.42, the resistor is...Ch. 8 - Design a series RLC circuit (see Fig. 8.3) using...Ch. 8 - Change the resistance for the circuit you designed...Ch. 8 - Prob. 45P Ch. 8 - Prob. 46P Ch. 8 - The switch in the circuit shown in Fig. P8.48 has...Ch. 8 - The switch in the circuit in Fig. P8.48 has been...Ch. 8 - The initial energy stored in the circuit in Fig....Ch. 8 - The resistor in the circuit shown in Fig. P8.50 is...Ch. 8 - The resistor in the circuit shown in Fig. P8.50 is...Ch. 8 - Prob. 52P Ch. 8 - The two switches in the circuit seen in Fig. P8.53...Ch. 8 - Prob. 55P Ch. 8 - Prob. 57P Ch. 8 - Prob. 58P Ch. 8 - Prob. 59P Ch. 8 - Prob. 60P Ch. 8 - Prob. 61P Ch. 8 - Derive the differential equation that relates the...Ch. 8 - The voltage signal of Fig. P8.63(a) is applied to...Ch. 8 - The circuit in Fig. P8.63 (b) is modified by...Ch. 8 - Prob. 65P Ch. 8 - Prob. 66P Ch. 8 - Prob. 67P Ch. 8 - Prob. 68P

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