Explanation Formula used: The expression for the exponential damping coefficient or the neper frequency for series R L C circuit is as follows: α = R 2 L (1) Here, α is the exponential damping coefficient or the neper frequency of series R L C circuit, R is the resistance of a series R L C circuit, C is the capacitance of a series R L C circuit. The expression for the resonant frequency for series R L C circuit is as follows: ω 0 = 1 L C (2) Here, ω 0 is the resonant frequency and L is the inductance of a series R L C circuit. The expression for the two solutions of the characteristic equation of a series 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 series R L C circuit. The expression for the natural response for series R L C circuit is as follows: v ( t ) = A 1 e S 1 t + A 2 e S 2 t V (5) Here, v ( t ) is the natural response of the series R L C circuit, A 1 and A 2 are arbitrary constants and t is the time. Calculation: The capacitor and the inductor are connected in the circuit for long time. So, the capacitor behaves as open circuit and the inductor behaves as short circuit. The redrawn circuit diagram is given in Figure 1 for t = 0 − . Refer to the redrawn Figure 1: The expression for the current flowing through the 30 Ω resistor is as follows: i ( 0 − ) = v 1 R 1 (6) Here, i ( 0 − ) is the current flowing through the 30 Ω resistor, v 1 is the voltage across 9 V independent source and R 1 is the resistance of 30 Ω resistor. Substitute 9 V for v 1 and 30 Ω for R 1 in equation (6). i ( 0 − ) = 9 V 30 Ω = 0.3 A The expression for the voltage across the capacitor at t = 0 − is as follows: v C ( 0 − ) = v 1 − v 2 (7) Here, v 2 is the voltage across the 2 i dependent voltage source. Substitute 2 i for v 2 in equation (7). v C ( 0 − ) = v 1 − 2 i (8) Substitute 9 V for v 1 and 0.3 A for i in equation (8). v C ( 0 − ) = 9 V − 2 ( 0.3 ) V = 9 V − 0.6 V = 8 .4 V The voltage across the capacitor at t = 0 − is 8 .4 V . The right hand mesh is open circuited so, the current flowing through the inductor at t = 0 − is 0 A . The capacitor does not allow sudden change in the voltage and the inductor does not allow sudden change in the current. So, v C ( 0 + ) = v C ( 0 − ) And, i L ( 0 + ) = i L ( 0 − ) Therefore, the voltage across the capacitor at t = 0 is 8 .4 V and the current flowing through inductor at t = 0 is 0 A . The redrawn circuit diagram is given in Figure 2 at t = 0 + . Refer to the redrawn Figure 2: The current flowing in the 30 Ω resistor at t = 0 + is 0 A . To find equivalent resistance connect 1 A test source across the 40 μ F capacitor which is open circuited and the inductor is short circuited. The redrawn circuit diagram is given in Figure 3: Refer to the redrawn Figure 3: Apply KVL in the circuit. v 0 = i R 1 + i R 2 − v 2 (9) Here, v 0 is the equivalent voltage across the 40 μ F capacitor, i is the is the current flowing in the circuit, R 2 is the resistance of 100 Ω resistor. Substitute 2 i for v 2 in equation (9). v 0 = i R 1 + i R 2 − 2 i (10) Substitute 1 A for i , 30 Ω for R 1 and 100 Ω R 2 in equation (10). v 0 = ( 1 A ) ( 30 Ω ) + ( 1 A ) ( 100 Ω ) − 2 ( 1 A ) = 30 V+100 V − 2 V =128 V The expression for the equivalent resistance is as follows: R e q = v 0 i (11) Here, R e q is the equivalent resistance across the 40 μ F capacitor. Substitute 128 V for v 0 and 1 A for i in equation (11). R e q = 128 V 1 A = 128 Ω Substitute 128 Ω for R and 90 mH for L in equation (1). α = 128 Ω ( 2 ) ( 90 mH ) = 128 Ω ( 2 ) ( 90 × 10 − 3 H ) { ∵ 1 mH = 10 − 6 H } = 711 .1 s − 1 Substitute 90 mH for L and 40 μ F for C in equation (2). ω 0 = 1 ( 90 mH ) × ( 40 μ F ) = 1 ( 90 × 10 − 3 H ) × ( 40 × 10 − 6 F ) { ∵ 1 mH = 10 − 3 H ∵ 1 μ F = 10 − 6 F } = 1 3.6 × 10 − 6 s − 1 = 527.1 rad/s Here, the exponential damping coefficient is greater thanthe exponential damping coefficient, So, the response of the parallel R L C circuit is over-damped. Substitute 711 .1 s − 1 for α and 527.1 rad/s for ω 0 in equation (3). s 1 = − ( 711 .1 s − 1 ) + ( 711 .1 s − 1 ) 2 − ( 527.1 rad/s ) 2 = ( − 711 .1 s − 1 ) + ( 477

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf

9th Edition

ISBN: 9781259989452

Author: Hayt

Publisher: Mcgraw Hill Publishers

See similar textbooks

Concept explainers

Graph theory

It is that part of network analysis where the configuration is called graph or structure of the network system when all of the elements (e.g., R, L, and C) in a network system are substituted with lines having circles/rings or dots across both endpoints.…

Question

Chapter 9, Problem 47E

To determine

Obtain an equation for vC as labeled in the circuit of Fig. 9.50 valid for all t>0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9, Problem 46E

Chapter 9, Problem 48E

Students have asked these similar questions

Assignment #2 A 110-V, three-phase, Y-connected, 8 pole, 48-slot, 6000-rpm, double-layer wound, synchronous generator has 12 turns per coil. If one side of the coil is in slot 1, the other side is in slot 6. There are 4 parallel paths. When the generator delivers the rated load at a line voltage of 110 V, the voltage regulation is 5%. What is the flux per pole? Draw two consecutive phasegroups of one of the phase windings and connect them (a) in series and (b) in parallel showing the Start (S) and Finish (F) of both connections. (A separate drawing for each connection)

3-4 Transmissiva Live of 120km has R= 0.2 ~2/15 X= 0.8 -2/km Y = 15H/6 5/km The line is supplies a load of 45 kV, SOMW, 0.8 lead p.f find sending voltage, Sending Current p.f. Sanding Voltage Regulation ⑨Voltage 5 Ⓒ charching coming! изу usy π cct ले

A (medium) single phase transmission line 100 km long has the following constants : Resistance/km = 0.25 Q; Susceptance/km = 14 × 10° siemen ; Reactance/km = 0.8 Receiving end line voltage = 66,000 V Assuming that the total capacitance of the line is localised at the receiving end alone, determine (i) the sending end current (ii) the sending end voltage (iii) regulation and (iv) supply power factor. The line is delivering 15,000 kW at 0.8 power factor Lead Draw the phasor diagram to illustrate your calculations.

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.

Obtain an equation for v C as labeled in the circuit of Fig. 9.50 valid for all t > 0 .

Solution Summary: The author explains the equation for v_C as labeled in the circuit of Fig. 9.50.

Concept explainers

Obtain an equation for vC as labeled in the circuit of Fig. 9.50 valid for all t>0.

Want to see the full answer?

Chapter 9 Solutions

Obtain an equation for v C as labeled in the circuit of Fig. 9.50 valid for all t &gt; 0 .

Solution Summary: The author explains the equation for v_C as labeled in the circuit of Fig. 9.50.

Concept explainers

Obtain an equation for vC as labeled in the circuit of Fig. 9.50 valid for all t>0.

Want to see the full answer?

Chapter 9 Solutions

Obtain an equation for v C as labeled in the circuit of Fig. 9.50 valid for all t > 0 .