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

Electrical Circuits and Modified MasteringEngineering - With Access

10th Edition

ISBN: 9780133992793

Author: NILSSON

Publisher: PEARSON

See similar textbooks

Question

Chapter 8, Problem 37P

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8, Problem 36P

Chapter 8, Problem 38P

Students have asked these similar questions

Find the Q-points for the diodes in the circuit. Assume ideal diodes, and startwith the assumption that D is OFF, and D2 is ON for both circuits.

please see the following image to answer questions thanks

5- Find the Eigenvalues and the corresponding Eigenvectors. [4 5 31 A=0 4 6 Lo 0 1

Chapter 8 Solutions

Electrical Circuits and Modified MasteringEngineering - With Access

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

Knowledge Booster

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:9780078028229
Author:Charles K Alexander, Matthew Sadiku
Publisher:McGraw-Hill Education
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON
Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,