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Advanced Electrical Circuits Analysis ( CCE 102 ) Prof. Dr. Fahmy El-khouly • • • • • • Weighting of assessments: -Quizzes 20 (Degrees) 20 % Activities 20 (degrees) 20 % -Mid-Term Exam 20 (Degrees) 20 % - Final-Term Exam 40 (Degrees) 40 % Total 100 (Degrees) 100 % • 6- List of References : • “ Electric Circuits “ James Nilsson and Susan Riedel, Ninth Edition COURSE OUTLINE Chapter1 : Response of First-Order RL and RC Circuit Chapter 2 : Sinusoidal Steady-State Analysis Chapter 3 : Active Filter Circuits Chapter 4: Two-Port Circuits Title date 15-2-2016 22-2-2016 29-2-2016 7-3-2016 14-3-2016 21-3-2016 28-3-2016 أسبوع امتحانات منتصف الفصل 4-4-2016 11-4-2016 18-4-2016 عطلة تحرير سيناء 25-4-2016 عطلة شم النسيم 2-5-2016 9-5-2016 أسبوع امتحانات نهاية الفصل 16-5-2016 Intended Learning Outcomes (ILOs): A : Knowledge and understanding: On completing this course, students will be able to: – – – – a1-Explain transients in reactive circuits (inductive and capacitive). a2- Describe the Response of RL and RC Circuits a3- Identify the meaning of active, reactive, and apparent power a4- Define the different theories that can be applied to AC electrical circuits B-Intellectual – – – – Skills: مهارات ذهنية b1- Analyze of AC circuits using Vectors. b2- Solve AC circuits and calculate active, reactive, and apparent power. b3- Solve AC circuits using circuits' theorems, Mesh and Nodal analysis b4- Calculate the response of first order RL-RC circuits. C-Professional and Practical Skills مهارات مهنية وعملية c1- Develop skills and practice in the design and fabrication of practical AC circuits used in electrical systems. c2- Design resonance circuits. D- General and Transferable Skills مهارات عامة d1- Use IT effectively. d2- Work in a team or individually d3- Communicate effectively using written, oral, graphical, and presentational skills Chapter 1 Response of First-Order RL and RC Circuit 1-The Natural Response of an RL Circuit the natural response is the currents and voltages that arise when stored energy in an inductor or capacitor is suddenly released to a resistive network. This happens when the inductor or capacitor is abruptly disconnected from its dc source. The switch is closed for long time Then Is reach its final value Is=constant Then VL=0 because di/dt=0 Then VRo=VR=0 Then IRo=IR=0 Then iL=Is When switch is opened If we use 0- to denote the time just prior to switching, and 0+ for the time immediately following switching, then Natural response of an RL circuit is: We derive the power dissipated in the resistor from any of the following expressions: The energy delivered to the resistor during any interval of time after the switch has been opened is The Significance of the Time Constant Using the time-constant concept, we write the expressions for current, voltage, power, and energy as a long time: implies that five or more time constants have elapsed. also means the time it takes the circuit to reach its steady-state value. the time constant is that it gives the time required for the current to reach its final value if the current continues to change at its initial rate −𝑡 𝑑𝑖 1 = − 𝐼𝑜 𝑒 τ 𝑑𝑡 τ When i=0 t= τ The above Equation indicates that i would reach its final value of zero in τ seconds Example 1 Determining the Natural Response of an RL Circuit The switch in the circuit shown in Fig. 7.7 has been closed for a long time before it is opened at t = 0. Find a) iL(t) for t > 0, b) io(t) for t > 0+, c) v0(t) for t > 0+, d) the percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 Ω resistor. Solution (a) the voltage across the inductor must be zero at t = 0-. Therefore the initial current in the inductor is 20 A at t = 0-. Hence, IL(0+) also is 20 A = 0.2 (b) (c) (d) The initial energy stored in the 2 H inductor is Therefore the percentage of energy dissipated in the 10 Ω resistor is Example 2 In the circuit shown in Fig. 7.8, the initial currents in inductors L1 and L2 have been established by sources not shown. The switch is opened at t = 0. a) Find i1 , i2, and i3 for t ≥ 0. b) Calculate the initial energy stored in the parallel inductors. c) Determine how much energy is stored in the inductors as t —> ∞. d) Show that the total energy delivered to the resistive network equals the difference between the results obtained in (b) and (c). 10∗15 + 10+15 10∗40 𝑅𝑒𝑞 = 50 𝑅𝑥 = 4 = 10 =8 Solution 𝐿𝑒𝑞 = 20∗5 20+5 =4 ( a ) the initial value of i(t) is 12 A and the time constant is 4/8, or 0.5 s. Therefore (b) (d) 2- The Natural Response of an RC Circuit Example 3 The switch in the circuit shown in Fig. 7.13 has been in position x for a long time. At t = 0, the switch moves instantaneously to position y. Find a) vc(t) for t > 0, b) Vo(t) for t > 0+, c) i0(t) for t > 0+, and d) the total energy dissipated in the 60 kΩ resistor. Solution a) capacitor will charge to 100 V and be positive at the upper terminal. We can replace the resistive network connected to the capacitor at t = 0+ with an equivalent resistance of 80 kΩ. Hence the time constant of the circuit is (0.5 X 10-6)(80 X 103) or 40 ms. Then, b) c) d) 3- The Step Response of RL and RC Circuits when either dc voltage or current sources are suddenly applied. The response of a circuit to the sudden application of a constant voltage or current source is referred to as the step response of the circuit. 3-1 The Step Response of an RL Circuit If the current were to continue to increase at its initial rate, it would reach its final value at t= τ that is, because If the current were to continue to increase at this rate at t= τ The voltage across an inductor is Ldi/dt, When the initial inductor current is zero Example The switch in the circuit shown in Fig. 7.19 has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a make-before-break type; that is, the connection at position b is established before the connection at position a is broken, so there is no interruption of current through the inductor. a) Find the expression for i(t) for t ≥ 0. b) What is the initial voltage across the inductor just after the switch has been moved to position b? c) How many milliseconds after the switch has been moved does the inductor voltage equal 24 V? d) Does this initial voltage make sense in terms of circuit behavior? e) Plot both i(t) and v(t) versus t. solution : a) Io is -8 A When the switch is in position b, the final value of i will be 24/2, or 12 A. The time constant of the circuit is 200/2, or 100 ms. . c) Yes; in the instant after the switch has been moved to position b, the inductor sustains a current of 8 A counterclockwise around the newly formed closed path. This current causes a 16 V drop across the 2 Ω resistor. This voltage drop adds to the drop across the source, producing a 40 V drop across the inductor. d) We find the time at which the inductor voltage equals 24 V by solving the expression Differential equation form Solution form Differential equation form Solution form The Step Response of an RC Circuit For current A General Solution for Step and Natural Responses where the value of the constant K can be zero Because the sources in the circuit are constant voltages and/or currents, the final value of x will be constant when x reaches its final value, the derivative dx/dt must be zero. Hence where xf represents the final value of the variable. When computing the step and natural responses of circuits, it may help to follow these steps: 1. Identify the variable of interest for the circuit. For RC circuits, it is most convenient to choose the capacitive voltage; for RL circuits, it is best to choose the inductive current. 2. Determine the initial value of the variable, which is its value at t o. Note that if you choose capacitive voltage or inductive current as your variable of interest, it is not necessary to distinguish between t = to- and t = to+. This is because they both are continuous variables. If you choose another variable, you need to remember that its initial value is defined at t = to+. 3. Calculate the final value of the variable, which is its value as t —> ∞. 4. Calculate the time constant for the circuit. Example The switch in the circuit shown in Fig. 7.22 has been in position 1 for a long time. At t = 0, the switch moves to position 2. Find a) v0(t) for t > 0 and b) io{t) for t > 0+. Solution a) The switch has been in position 1 for a long time, so the initial value of v0 is 40(60/80), or 30 V we find the Norton equivalent with respect to the terminals of the capacitor for t > 0. To do this, we begin by computing the open-circuit voltage, which is given by the -75 V source divided across the 40 kΩ and 160 k Ω resistors: The value of the Norton current source is the ratio of the opencircuit voltage to the Thevenin resistance, or -60/(40 X 103) = 1.5 mA.