Transients in RC and RL Circuits: Solving First-Order and Second-Order Circuits, Exams of Electrical and Electronics Engineering

Download Transients in RC and RL Circuits: Solving First-Order and Second-Order Circuits and more Exams Electrical and Electronics Engineering in PDF only on Docsity! 1 Chapter 4 Transients First-Order RC Circuits DC Steady State RL Circuits RC and RL Circuits with General Sources Second-Order Circuits Chapter 4 Transients 1. Solve first-order RC or RL circuits. 2. Understand the concepts of transient response and steady-state response. 3. Relate the transient response of first-order circuits to the time constant. 4. Solve RLC circuits in dc steady-state conditions. 5. Solve second-order circuits. 6. Relate the step response of a second-order system to its natural frequency and damping ratio. Transients The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integrodifferential equations. 2 Discharge of a Capacitance through a Resistance ( ) ( ) 0=+ R tv dt tdv C CC ( ) ( ) 0=+ tv dt tdv RC C C ( ) stC Ketv = 0=+ stst KeRCKse RC s 1−= ( ) RCtC Ketv −= ( ) RCt iC eVtv −= ( ) iC Vv =+0 5 4. Substitute the solution into the differential equation to determine the values of K1 and s . (Alternatively, we can determine K1 by solving the circuit in steady state as discussed in Section 4.2.) 5. Use the initial conditions to determine the value of K2. 6. Write the final solution. RL Transient Analysis ( ) LtReKti −+= 22 Time constant is R L=τ Figure 4.9 The circuit analyzed in Example 4.3, Figure 4.10 The current and voltage for the circuit of Figure 4.9, Figure 4.11 The circuit for Exercise 4.5. 02 anf Figure 4.12. The circuit for Exercise 4.6. lov 7 RC AND RL CIRCUITS WITH GENERAL SOURCES The general solution consists of two parts. The particular solution (also called the forced response) is any expression that satisfies the equation. In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution. The homogeneous equation is obtained by setting the forcing function to zero. The complementary solution (also called the natural response) is obtained by solving the homogeneous equation. 10 SECOND-ORDER CIRCUITS ( ) ( ) ( ) ( ) ( )tvvdtti C tRi dt tdi L sC t =+++ ∫ 0 1 0 L R 2 =α LC 1 0 =ω ( ) ( ) ( ) ( )tfti dt tdi dt tid =++ 202 2 2 ωα 0ω αζ = 2 0 2 1 ωαα −+−=s 2 0 2 2 ωαα −−−=s 11 1. Overdamped case (ζ > 1). If ζ > 1 (or equivalently, if α > ω0), the roots of the characteristic equation are real and distinct. Then the complementary solution is ( ) tstsc eKeKtx 21 21 += In this case, we say that the circuit is overdamped. 2. Critically damped case (ζ = 1). If ζ = 1 (or equivalently, if α = ω0 ), the roots are real and equal. Then the complementary solution is ( ) tstsc teKeKtx 11 21 += In this case, we say that the circuit is critically damped. 3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently, if α < ω0), the roots are complex. (By the term complex, we mean that the roots involve the square root of –1.) In other words, the roots are of the form nn jsjs ωαωα −−=+−= 21 and in which j is the square root of -1 and the natural frequency is given by 22 0 αωω −=n In electrical engineering, we use j rather than i to stand for square root of -1, because we use i for current. For complex roots, the complementary solution is of the form ( ) ( ) ( )teKteKtx ntntc ωω αα sincos 21 −− += In this case, we say that the circuit is underdamped. = 10mit \=1V CR ein Lae (0)=0 Figure 4.21 The circuit for Example 45, Figure 4.22 The equivalent circuit for Figure 4.21 under steady-state conilions The nduetor has been regacea by a shor cireult ana the capacitor by an open circu. gl = IOV 002K Figure 4.23 Solution for R — 300 Voltage. 15, ¢ git) = WV 0 Kye 0 Ke 10 0 o 02 Os 06 Os 10 Figure 4.24 Solution for R = 240 2, 12