Explanation Given Data: Value of initial current of inductor i L ( 0 − ) is 6 A and Value of initial voltage of capacitor v C ( 0 + ) is 0 V . Formula used: The expression for the exponential damping coefficient is as follows: α = 1 2 R C (1) Here, α is the exponential damping coefficient, R is the resistance of a parallel R L C circuit and C is the capacitance of a parallel R L C circuit. The expression for the resonant frequency is as follows: ω 0 = 1 L C (2) Here, ω 0 is the resonant frequency and L is the inductance of a parallel R L C circuit. The expression for the two solutions of the characteristic equation of a parallel R L C circuit is as follows: s 1 = − α + α 2 − ω 0 2 (3) s 2 = − α − α 2 − ω 0 2 (4) Here, s 1 and s 2 is the solutions of the characteristic equation of a parallel R L C circuit. Here, i L ( t ) is the natural response of the parallel R L C circuit, A 1 and A 2 are arbitrary constants and t is the time. Calculation: Substitute 1.5 k Ω for R and 4 pF C in equation (1). α = 1 2 × 3 Ω × 4 pF = 1 2 × 3 Ω × 4 × 10 − 12 F { ∵ 1 pF = 10 − 12 F } = 1 2.4 × 10 13 s − 1 α = 4.167 × 10 10 s − 1 Substitute 4 pF for C and 625 pH for L in equation (2). ω 0 = 1 625 pH × 4 pF { ∵ 1 pF = 1 × 10 − 12 F 1 pH = 1 × 10 − 12 H } = 1 625 × 10 − 12 H × 4 × 10 − 12 F = 1 25 × 10 − 22 ω 0 = 1 5 × 10 − 11 = 2 × 10 10 rad/s The exponential decaying coefficient α is greater than the resonant frequency ω 0 . Therefore, the circuit is over-damped. Calculation: The redrawn circuit at t = 0 − is shown in Figure 1 as follows: Refer to the Figure 1: At t = 0 − s when the switch is opened the capacitor is short circuited and inductor gets short circuited as per the given data. At t = 0 + the capacitor does not allow sudden change in the voltage. Therefore, v C ( 0 + ) = v C ( 0 − ) So, v C ( 0 − ) = 0 V Refer to the FIGURE 9.8 in the textbook. The expression for the natural response of voltage for capacitor in the parallel R L C circuit is as follows: v C ( t ) = A 1 e s 1 t + A 2 e s 2 t (5) Substitute 0 s for t and 0 V for v C ( 0 − ) in equation (5). 0 V = A 1 e S 1 × 0 − s + A 2 e S 2 × 0 − s 0 V = A 1 + A 2 Rearrange for A 1 . A 1 = − A 2 (6) Differentiate equation (5) both the side with respect to time t is: d v C ( t ) d t = A 1 s 1 e s 1 t + A 2 s 2 e s 2 t (7) The expression for the voltage across capacitor at time t = 0 + is: i C ( 0 + ) = C d v C ( t ) d t | t = 0 + (8) Substitute 0 s for t in equation (7). d v C ( t ) d t | t = 0 + = A 1 s 1 e s 1 × 0 + + A 2 s 2 e s 2 × 0 + V/s = ( A 1 s 1 + A 2 s 2 ) V/s Substitute 4.167 × 10 10 s − 1 for α and 2 × 10 10 rad/s for ω 0 in equation (3). s 1 = − 4.167 × 10 10 s − 1 + ( 4.167 × 10 10 s − 1 ) 2 − ( 2 × 10 10 rad/s ) 2 = − 4.167 × 10 10 s − 1 + 17

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf

9th Edition

ISBN: 9781259989452

Author: Hayt

Publisher: Mcgraw Hill Publishers

See similar textbooks

Videos

Question

Chapter 9.2, Problem 3P

To determine

Find the current iR.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.2, Problem 2P

Chapter 9.2, Problem 4P

Students have asked these similar questions

Example 9.2-1 (See Example 9.2-1 in the textbook for the solution to a similar problem.) 0.25H 9Ω Ω 0.20H This circuit can be represented by the differential equation d i2 + aoi2. dvs b1 dt + + a1 dt dt2 Determine the values of the coefficients b,, a, and a- |and a, = a1 = b1 Click if you would like to Show Work for this question: Open Show Work

9.20. For Fig. 9.54, biased at 2 mA, and if Vcc=6V, B = 125 for Q₁ and Q2, find voltage gain. Load is considered as ideal cur- rent source. (|VA|=50V) I Vcc 1₁ - Vout V₂1 Q₂ Vin Q₁ TIT Figure 9.54

If /1 goes to the right through R1, 12 goes to the right through R2, and /3 goes to the right through R3, what is the resulting equation resulting from applying Kirchhoff's junction rule at point b? Please show solution not only ansSwer. 8.0V R 9.0 V a

Chapter 9 Solutions

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf

Ch. 9.1 - A parallel RLC circuit contains a 100 2 resistor...Ch. 9.2 - After being open for a long time, the switch in...Ch. 9.2 - Prob. 3P Ch. 9.2 - Prob. 4P Ch. 9.3 - (a) Choose R1 in the circuit of Fig. 9.14 so that...Ch. 9.4 - Prob. 6P Ch. 9.5 - Prob. 7P Ch. 9.5 - Prob. 8P Ch. 9.6 - Let is = 10u(t) 20u(t) A in Fig. 9.31. Find (a)...Ch. 9.6 - Let vs = 10 + 20u(t) V in the circuit of Fig....

Ch. 9.7 - Alter the capacitor value and voltage source in...Ch. 9 - For a certain source-free parallel RLC circuit, R...Ch. 9 - Element values of 10 mF and 2 nH are employed in...Ch. 9 - If a parallel RLC circuit is constructed from...Ch. 9 - Prob. 4E Ch. 9 - You go to construct the circuit in Exercise 1,...Ch. 9 - A parallel RLC circuit has inductance 2 mH and...Ch. 9 - Prob. 7E Ch. 9 - A parallel RLC circuit has R = 1 k, L = 50 mH. and...Ch. 9 - Prob. 9E Ch. 9 - Prob. 10E Ch. 9 - The current flowing through a 5 resistor in a...Ch. 9 - For the circuit of Fig.9.40, obtain an expression...Ch. 9 - Consider the circuit depicted in Fig. 9.40. (a)...Ch. 9 - With regard to the circuit represented in Fig....Ch. 9 - (a) Assuming the passive sign convention, obtain...Ch. 9 - With regard to the circuit presented in Fig. 9.42,...Ch. 9 - Obtain expressions for the current i(t) and...Ch. 9 - FIGURE 9.43 Replace the 14 resistor in the...Ch. 9 - Design a complete source-free parallel RLC circuit...Ch. 9 - For the circuit represented by Fig. 9.44, the two...Ch. 9 - Prob. 21E Ch. 9 - Prob. 22E Ch. 9 - A critically damped parallel RLC circuit is...Ch. 9 - A source-free parallel RLC circuit has an initial...Ch. 9 - A critically damped parallel RLC circuit is...Ch. 9 - For the circuit of Fig. 9.45, is(t) = 30u(t) mA....Ch. 9 - Prob. 27E Ch. 9 - The circuit of Fig. 9.44 is rebuilt such that the...Ch. 9 - Prob. 29E Ch. 9 - Prob. 30E Ch. 9 - The source-free circuit depicted in Fig. 9.1 is...Ch. 9 - (a) Graph the current i for the circuit described...Ch. 9 - Analyze the circuit described in Exercise 31 to...Ch. 9 - A source-free parallel RLC circuit has capacitance...Ch. 9 - Prob. 35E Ch. 9 - Obtain an expression for vL(t), t 0, for the...Ch. 9 - For the circuit of Fig. 9.47, determine (a) the...Ch. 9 - (a) Design a parallel RLC circuit that provides a...Ch. 9 - The circuit depicted in Fig. 9.48 is just barely...Ch. 9 - When constructing the circuit of Fig. 9.48, you...Ch. 9 - The circuit of Fig. 9.22a is constructed with a...Ch. 9 - Prob. 42E Ch. 9 - Prob. 43E Ch. 9 - The simple three-element series RLC circuit of...Ch. 9 - Prob. 45E Ch. 9 - Prob. 46E Ch. 9 - Prob. 47E Ch. 9 - With reference to the series RLC circuit of Fig....Ch. 9 - Obtain an expression for i1 as labeled in Fig....Ch. 9 - The circuit in Fig. 9.52 has the switch in...Ch. 9 - For the circuit in Fig. 9.52, determine the value...Ch. 9 - In the series circuit of Fig. 9.53, set R = 1 ....Ch. 9 - Evaluate the derivative of each current and...Ch. 9 - Consider the circuit depicted in Fig. 9.55. If...Ch. 9 - Prob. 55E Ch. 9 - In the circuit shown in Fig. 9.56, (a) obtain an...Ch. 9 - Prob. 57E Ch. 9 - For the circuit represented in Fig. 9.57, (a)...Ch. 9 - FIGURE 9.57 Replace the 1 resistor in Fig. 9.57...Ch. 9 - A circuit has an inductive load of 2 H, a...Ch. 9 - (a) Adjust the value of the 3 resistor in the...Ch. 9 - Determine expressions for vC(t) and iL(t) in Fig....Ch. 9 - The capacitor in the LC circuit in Fig. 9.60 has...Ch. 9 - Suppose that the switch in the circuit in Fig....Ch. 9 - The capacitor in the circuit of Fig. 9.63 is set...Ch. 9 - The physical behavior of automotive suspension...Ch. 9 - A lossless LC circuit can be used to provide...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Find the current i R .

Solution Summary: The author calculates the exponential damping coefficient, resistance, and capacitance of a parallel RLC circuit.

Videos

Find the current iR.

Want to see the full answer?

Chapter 9 Solutions