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Assist .Prof. Sibel ÇİMEN Electronics and Communication Engineering University of Kocaeli Course Book : • Fundamentals of Electric Circuits, by Charles K. Alexander and Matthew N. O. Sadiku, McGraw Hill; 3rd edition (2007) Reference Books: 1) Electric Circuits, by James W. Nilsson and Susan Riedel, Prentice Hall; 8th edition (2007) 2) Schaum's Outline of Electric Circuits, by Mahmood Nahvi and Joseph Edminister, McGraw-Hill; 4th edition (2002) 3) Introduction to Electric Circuits, by Richard C. Dorf and James A. Svoboda, Wiley, 7th edition (2006) 4) Schaum's Outline of Basic Circuit Analysis, by John O'Malley and John O'Malley, McGraw-Hill; 2nd edition (1992) Course Outline 1) Second Order DC Circuits (Fund. of Electric Circuits, CH 8) 2) Sinusoids and Phasors (Fund. of Electric Circuits, CH 9) 3) Sinusoidal Steady-State Analysis (Fund. of Electric Circuits, CH 10) 4) AC Power Analysis (Fund. of Electric Circuits, CH 11) 5) Frequency Response (Fund. of Electric Circuits, CH 14) 6) Laplace Transform (Fund. of Electric Circuits, CH 15 1.1. Introduction 1.1. Introduction Keep in mind; Capacitor voltage always continuous… Inductor current always continuous… EXAMPLE 1.1. • The switch in Figure has been closed for a long time. İt is open at t=0. • Find; a) (0+ ), (0+ ) b) ((0+ )/dt, ((0+ )/dt, c) (∞) , (∞) EXAMPLE 1.1. /Solution: • (a) If the switch is closed a long time before t = 0, it means that the circuit has reached dc steady state at t = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor acts like an open circuit, so we have the circuit in Fig. 8.(a) at t = 0−. Thus, As the inductor current and the capacitor voltage cannot change abruptly, EXAMPLE 1.1. /Solution: • (b) At t = 0+, the switch is open; the equivalent circuit is as shown in Fig. 8 (b). The same current flows through both the inductor and capacitor. We now obtain by applying KVL to the loop EXAMPLE 1.1. /Solution: (c) Fort > 0, the circuit undergoes transience. But as t →∞, the circuit reaches steady state again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the circuit becomes that shown in Fig. 8(c), from which we have EXAMPLE 1.2. • In figure calculate; a) (0+ ), (0+ ), (0+ ) b) ((0+ )/dt, ((0+ )/dt c) ∞ , ∞ , (∞) EXAMPLE 1.2. /Solution: • (a) For t < 0, 3u(t) = 0. At t = 0−, since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. (a). From this figure we obtain EXAMPLE 1.2. /Solution: • For t > 0, 3u(t) = 3, so that the circuit is now equivalent to that in Fig. (b). Since the inductor current and capacitor voltage cannot change abruptly, Applying KCL at node a in Fig. (b) gives Applying KVL to the middle mesh in Fig.(b) yields EXAMPLE 1.2. /Solution: But applying KVL to the right mesh in Fig. (b) gives EXAMPLE 1.2. /Solution: • (c) As t →∞, the circuit reaches steady state. We have the equivalent circuit in Fig.(a) except that the 3-A current source is now operative. • By current division principle, 1.3. The Source-Free Series RLC Circuits (1.a) (1.b) 1.3. The Source-Free Series RLC Circuits Applying KVL around the loop; (2) To eliminate the integral, we differentiate with respect to t and rearrange terms:…. (3) 1.3. The Source-Free Series RLC Circuits with initial values equation (2)… (4) or Bobinin uçlarından akan akımın exp. Karakteristiği olduğunu biliyoruz… equation (3) becomes… or i 0 1.3. The Source-Free Series RLC Circuits (5) Known as characteristic equation İt’s roots; Where; 1.3. The Source-Free Series RLC Circuits 0 : resonant frequency or undamped natural frequency (rad/s) ∝: neper frequency or damping factor (Np/s) İn terms of 0 ∝ equation (5) gets… (6) The two values of s in Eq. (5) indicate that there are two possible solutions in Eq. (6); that is, Natural response of series RLC; 1.3. The Source-Free Series RLC Circuits There are three types of solutions; Overdamped Case (∝> ) İn this situation; 1 2 negative and real 1.3. The Source-Free Series RLC Circuits Critically Damped Case (∝= ) 1.3. The Source-Free Series RLC Circuits Under Damped Case (∝< ) EXAMPLE 1.3. R=40 Ω, L=4H and C=1/4 F. Calculate the characteristic roots of the circuit. Is the natural response overdamped, underdamped, or critically damped? Solution: Since α > ω0, we conclude that the response is overdamped. This is also evident from the fact that the roots are real and negative. EXAMPLE 1.4. Find i(t) in the circuit. Assume that the circuit has reached steady state at t=0− . Solution: For t < 0, the switch is closed. The capacitor acts like an open circuit while the inductor acts like a shunted circuit. The equivalent circuit is shown in Fig. (a). Thus, at t = 0, EXAMPLE 1.4./ Solution • For t > 0, the switch is opened and the voltage source is disconnected. The equivalent circuit is shown in Fig. 8.11(b), which is a source-free series RLC circuit. Hence, the response is underdamped (α < ω); that is, EXAMPLE 1.4./ Solution • We now obtain 1 and 2 using the initial conditions. At t = 0,