Transient Response of First Order Circuits: Source-Free RC and RL Circuits, Study notes of Electronic Circuits Analysis

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ELL 100 - Introduction to Electrical Engineering LECTURE 9: TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS (NATURAL RESPONSE) 2 SOURCE-FREE RC CIRCUITS EXAMPLE Fluid-flow analogy: water tank emptying through a small pipe Electrical circuit: Capacitor discharging through resistance 5 SOURCE-FREE RC CIRCUITS APPLICATIONS Oscillators 555 Timer circuits Timers Camera Flash SOURCE-FREE RC CIRCUITS APPLICATIONS Rn S&S & wr —“o— yh an + 70V = Nov = C == 04 pF Neon lamp A Delay Circuits Warning Blinkers 7 SOURCE FREE RC CIRCUITS APPLICATIONS Computer Circuits Digital and Time delay circuits 10 TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS • A first-order circuit is characterized by a first-order differential equation. • Example : • a circuit comprising a resistor and capacitor (RC circuit) • a circuit comprising a resistor and an inductor (RL circuit) Applying Kirchhoff’s laws to RC or RL circuit results in differential equations involving voltage or current, which are first-order. 11 TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS EXCITATION There are two ways to excite the circuits. • Initial conditions of the storage elements– Source-Free Circuits (Energy stored in the capacitor, Energy stored in the inductor) • Independent sources – Forced Excitation circuits (DC sources, Sinusoidal sources, Exponential Sources) 12 TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS NATURAL RESPONSE • The natural response of a circuit refers to the behavior (in terms of voltage or current) with no external sources of excitation. • The circuit has a response only because of the energy initially stored in the energy storage elements (i.e. capacitor or inductor). 15 SOURCE-FREE RC CIRCUIT DERIVATION Applying KCL at the top node of the circuit, yields By definition, iC = C dv∕dt and iR = v ∕ R. Thus, or This is a first-order differential equation. C Ri + i = 0 0 dv v C dt R   0 dv v dt RC   16 SOURCE-FREE RC CIRCUIT DERIVATION => Integrating both sides, we get => => But from the initial conditions, v(0) = A = V0. Hence, (Exponentially Decaying) 0 dv v dt RC   1dv dt v RC   ln ln A t v RC    ln v t A RC   /( ) t RCv t A e / 0( ) V t RCv t e 17 SOURCE-FREE RC CIRCUIT VOLTAGE RESPONSE • As t increases, the voltage decreases exponentially towards zero. The rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by τ. 20 SOURCE-FREE RC CIRCUIT TIME CONSTANT 21 SOURCE-FREE RC CIRCUIT POWER DISSIPATION The power dissipated in the resistor is The energy absorbed by the resistor up to time t is 2 2 /0V ( ) t Rp t vi e R   2 2 /0 0 0 2 2 / 2 2 /0 0 0 2 0 ( ) ( ) 1 | (1 ), RC 2 2 1 , ( ) 2 t t R t R V w t p d e d R V e CV e R t w CV                           – e-2t/τ) 22 SOURCE-FREE RL CIRCUIT • A circuit with series connection of a resistor and inductor • Current i(t) through the inductor is considered as response of this system. At t = 0, assume that the inductor has an initial current I0, or Initial energy stored in the inductor 0(0)i I 2 0 1 w(0) 2 LI 25 SOURCE FREE RL CIRCUIT POWER DISSIPATION Voltage across the resistor is The power dissipated in the resistor is The energy absorbed by the resistor is - / 0( ) t Rv t iR I Re   2 -2 / 0 t Rp v i I Re   2 0 1 , ( ) 2 Rt w LI   2 2 / 0 0 0 2 2 / 2 2 / 0 0 0 ( ) ( ) 1 | (1 ), / 2 t t R t w t p d I Re d I Re LI e L R                        26 SOLVING NUMERICALS Points to remember : Elements DC steady state Continuous quantity (from t=0- to t=0+) R R - L Short-circuit (v = 0) Current i C Open-circuit (i = 0) Voltage v 27 SOURCE-FREE RC CIRCUIT Q1. Consider the circuit below. Let vC (0)=15 V. Find vc , vx and ix for t > 0. 2.5( ) 15 tv t e V 30 SOURCE-FREE RC CIRCUIT we can use voltage division to get vx 2.5t 2.5t 2.5 12 0.6(15 ) 9 12 8 0.75 12 x tx x v v e e V v i e A          /0.4( ) 15 tv t e V C C / 0( ) V tv t e  31 SOURCE-FREE RC CIRCUIT Q2. The switch in the circuit below is closed for a long time, and then opened at t = 0. Find v(t) for t ≥ 0. Also calculate the energy stored in the capacitor before opening of the switch. 32 SOURCE-FREE RC CIRCUIT Solution: For t < 0, the switch is closed and the capacitor is an open circuit in steady state, as represented in Fig.(a). Using voltage division Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at t = 0− is the same at t = 0+, or 9 ( ) (20)=15 , 0 9 3 Cv t V t   0(0) V 15Cv V  35 SOURCE-FREE RL CIRCUIT Solution: There are two ways we can solve this problem Method -1: The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with vo = 1 V at the inductor terminals a-b, as shown below 36 SOURCE-FREE RL CIRCUIT Applying KVL to the two loops, 1 2 1 2 2 1 1 2 1 1 2( ) 1 0 (1) 2 5 6 2 3 0 (2) 6 i i i i i i i i i             Substituting Eq. (2) into Eq. (1) gives i1= -3A, i0= - i1 =3A The time constant is The current through the inductor is e 1 R 3 o q Th o v R i    => e 1 32 1R 2 3 q L s    / ( 2/3)( ) (0) 10 , 0t ti t i e e A t    37 SOURCE-FREE RL CIRCUIT Method-2: Applying KVL to the circuit For loop 1, For loop 2, Substituting above into Eq. (3) gives 1 1 2 1 2( ) 0 (3) 2 di i i dt    2 1 1 2 1 5 6 2 3 0 6 i i i i i     1 1 2 0 3 di i dt   (2/3) (2/3) 0 (t) 2 ln | ( ) (0) 10 , 0 (0) 3 t t ti t i t i e e A t i       => Ω 40 SOURCE-FREE RL CIRCUIT Solution: For t < 0, the switch is closed, and the inductor acts as a short circuit in steady state. The 16-Ω resistor is short-circuited; the resulting circuit is shown in Fig (a). We obtain i(t) from i1 using current division, e 4 12 R 2 5 4 12 q      1 40 8 5 i A  1 12 ( ) 6 , 0 12 4 i t i A t    Ω 41 SOURCE-FREE RL CIRCUIT Since the current through an inductor cannot change instantaneously, For t > 0, the switch is open and the voltage source is disconnected. We now have the source-free RL circuit in Fig.(b). Combining the resistors, we have The time constant is (0) (0 ) 6i i A  eR (12 4) ||16 8q     e 2 1 R 8 4q L s    / 4( ) (0) 6t ti t i e e A   42 SOURCE-FREE RL CIRCUIT Q5. In the circuit shown below, find io, vo, and i for all t > 0, assuming that the switch was open for a long time and closed at t = 0. 45 SOURCE-FREE RL CIRCUIT Because the inductor is in parallel with the 6-Ω and 3-Ω resistors, Thus for all time, ( ) 2( 2 ) 4 , t 0t t o L di v t v L e e V dt           t2 ( ) e A, t 0 6 3 L o v i t     0 0 ( ) 2 , 0 3 o t A t i t e t        6 0 ( ) 4 , 0 o t V t v t e t     2 0 ( ) 2 , 0t A t i t e t     46 SOURCE-FREE RL CIRCUIT Q6. The switch ‘S’ is kept in position ‘1’ for a long time and then suddenly changed to position ‘2’ at t = 0 as shown Compute the value of vL and iL i. At the instant just prior to the switch changing (t = 0-) ii. At the instant just after the switch changes (t = 0+) Also find the rate of change of current through the inductor at t = 0+ 47 SOURCE-FREE RL CIRCUIT Solution : At t = 0- the current through and the voltage across the inductor are (inductor acts as short-circuit) At t = 0+, (KVL) The rate of change of current through inductor at time t = 0+ is L 10 (0 ) 10 5 ; v (0 ) 0 V 10 10 Li A      L(0 ) 5 ; v (0 ) (10 10) 5 100 VLi A        0 0 ( ) ( ) 100 100 25 / 4 l l t t di t di t L V A s dt dt           50 SOURCE FREE RC CIRCUIT Q2. Switch ‘S’ shown in fig. is kept in position ‘1’ for a long time. When the switch is thrown in position ‘2’, find at steady state condition (i) the voltage across the each capacitor (ii) the charge across the each capacitor (iii) the energy stored by the each capacitor Answer: (i) V/2 (ii)CV/2 (iii)CV2/8 51 SOURCE FREE RC CIRCUIT Q3. In the circuit shown in Fig. v(t) = 56 e −200t V, t > 0 i(t) = 8 e −200t mA, t > 0 (a) Find the values of R and C. (b) Calculate the time constant τ. (c) Determine the time required for the voltage to decay half its initial value at t = 0. Answer: (a) 0.7143 μF, (b) 5 ms, (c) 3.466 ms 52 SOURCE FREE RC CIRCUIT Q4. In the circuit shown in Fig. Determine the charge lost by the capacitor from 25μs to 100 μs in coulombs. Consider, v(0) = 4 V, C=5μC, R=5Ω. Answer: 7 μC 55 SOURCE FREE RL CIRCUIT Q7. Find i and vx in the circuit of Fig. Let i(0) = 7 A. Answer: 7 e −2t A, −7 e −2t V, t > 0. 56 SOURCE FREE RL CIRCUIT Q8. For the circuit in Fig., find i(t) for t > 0.. Answer: 2 e −2t A, t > 0. 57 SOURCE FREE RL CIRCUIT Q9. For the circuit in Fig, find io for t > 0. Answer: 1.2 e −3t A, t > 0.