ELL 100 - Introduction to Electrical Engineering
LECTURE 9: TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS (NATURAL RESPONSE) SOURCE-FREE RC CIRCUITS EXAMPLE
Fluid-flow analogy: Electrical circuit: water tank emptying Capacitor discharging through a small pipe through resistance
2 SOURCE-FREE RC CIRCUITS
EXAMPLE
Mechanical system Electrical equivalent (velocity decays (charge dissipated through damping) through resistance)
3 SOURCE-FREE RC CIRCUITS
APPLICATIONS
High pass filter Low pass filter
4 SOURCE-FREE RC CIRCUITS
APPLICATIONS
Timers
Camera Flash 555 Timer Oscillators circuits
5 SOURCE-FREE RC CIRCUITS
APPLICATIONS
Delay Circuits Warning Blinkers
6 SOURCE FREE RC CIRCUITS
APPLICATIONS
Computer Circuits Digital and Time delay circuits
7 SOURCE FREE RC CIRCUITS APPLICATIONS
Pacemakers Timing device in automobile intermittent wiper system
8 SOURCE-FREE RL CIRCUIT
APPLICATIONS
Pulse Generators
Electronic filter
Tubelight choke
9 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.
10 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)
11 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).
12 SOURCE-FREE RC CIRCUIT
• A source-free RC circuit occurs when its DC source is suddenly disconnected. • The energy already stored in the capacitor(s) is released to the resistor(s) & dissipated. • RC source-free circuit is analyzed from its
initial voltage v(0) = V0 and time constant τ
13 SOURCE-FREE RC CIRCUIT
DERIVATION • Assume the voltage v(t) across the capacitor. • Since the capacitor is initially charged, Assume that at time t = 0, the initial voltage is,
vV(0) 0
with the corresponding value of the energy stored as 1 w(0) CV 2 2 0
14 SOURCE-FREE RC CIRCUIT DERIVATION Applying KCL at the top node of the circuit, yields
i CR+ i = 0
By definition, iC = C dv∕dt and iR = v ∕ R. Thus, dvv dvv C 0 or 0 dtR dtRC
This is a first-order differential equation.
15 SOURCE-FREE RC CIRCUIT DERIVATION dvv dv 1 0 => dt dtRC vRC t Integrating both sides, we get lnlnv A RC vt => ln ARC => v( tA ) e t/ RC
But from the initial conditions, v(0) = A = V0. t/ RC Hence, v( te ) V 0 (Exponentially Decaying)
16 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. 17 SOURCE-FREE RC CIRCUIT TIME CONSTANT The time constant of a circuit is the time required for the response to decay to a factor of 1/e or 36.8 percent of its initial value.
tRC/ vte() V 0
- /RC-1 V000 e = V e0.368 V RC
t/ v( t ) V0 e
18 SOURCE-FREE RC CIRCUIT TIME CONSTANT
t v(t)/V0 τ 0.36788 2τ 0.13534 3τ 0.04979 4τ 0.01832 5τ 0.00674
Graphical determination of the time constant τ from the response curve. 19 SOURCE-FREE RC CIRCUIT TIME CONSTANT
20 SOURCE-FREE RC CIRCUIT POWER DISSIPATION The power dissipated in the resistor is V 2 p( t ) vi 0 e2/t R R The energy absorbed by the resistor up to time t is ttV 2 w()() t p d 0 e2/ d R 00R V 2 1 0 e2 / |t CV 2 (1(1 – ee- 22 t/ /τ ) ), RC 22R 00 1 t ,() w CV 2 R 2 0 21 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 iI(0) 0 1 Initial energy stored in the inductor w(0) LI 2 2 0
22 SOURCE-FREE RL CIRCUIT
RESPONSE OF THE CIRCUIT
Applying KVL around the loop, vvLR0 vL = L di/dt and vR = iR. Thus, didiR LRii00 dtdtL it()diR t => dt iL I0 0
i() t Rt Rt/ L => ln i ( t ) I0 e IL0
23 SOURCE FREE RL CIRCUIT RESPONSE OF THE CIRCUIT
• Current through inductor decays Rt /L exponentially itIe() 0
• Time constant for the RL circuit is L R
t/ i() t I0 e
24 SOURCE FREE RL CIRCUIT POWER DISSIPATION -/t Voltage across the resistor is vtiRIReR ()0 2-2 / t The power dissipated in the resistor is pviIReR 0
The energy absorbed by the resistor is tt w()() t p d I2 Re 2 / d R 0 00 1 I2 Re 2 / |t LI 2 (1 e 2 / ), L / R 0 02 0 1 t ,() w LI 2 R 2 0
25 SOLVING NUMERICALS Points to remember :
Elements DC steady state Continuous quantity (from t=0- to t=0+) R R -
L Short-circuit Current i (v = 0) C Open-circuit Voltage v (i = 0)
26 SOURCE-FREE RC CIRCUIT
Q1. Consider the circuit below. Let vC (0)=15 V. Find vc , vx and ix for t > 0.
27 SOURCE-FREE RC CIRCUIT
Solution : • We first convert the given circuit into a simple R-C circuit. • Find the equivalent resistance or the Thevenin resistance at the capacitor terminals.
28 SOURCE-FREE RC CIRCUIT
205 R4 eq 205
The time constant is Req Cs 4(0.1) 0.4
29 SOURCE-FREE RC CIRCUIT
vte() V t/ t/0.4 2.5t C 0 vteV() 15 v( t ) 15 e V we can use voltage division to get vx 12 v v 0.6(15 e2.5t ) 9 e 2.5t V x 12 8 C v ix 0.75 e2.5t A x 12 30 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.
31 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 9 vtVt( ) (20)=15 ,0 C 93
Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at t = 0− is the same at t = 0+, or
vVC (0) V0 15
32 SOURCE-FREE RC CIRCUIT Solution: For t > 0, the switch is open, and we have the RC circuit shown in Fig. (b),
R eq 1 + 910
The time constant is 3 = RCeq 1020100.2s
Thus, the voltage across the capacitor for t ≥ 0 is t/ t/0.2 5t v( t ) vC (0)e 15e V 15 e V
The initial energy stored 11 w (0)Cv2 (0) 20 10 3 15 2 2.25 J in the capacitor is: C 22C 33 SOURCE-FREE RL CIRCUIT
Q3. Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the circuit below
34 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
35 SOURCE-FREE RL CIRCUIT Applying KVL to the two loops, 1 2()10(1)iiii 1212 2 5 6230(2)iiiii 21121 6
Substituting Eq. (2) into Eq. (1) gives vo 1 i = -3A, i = - i =3A => R eqThR 1 0 1 i 3 1 o L 3 2 s The time constant is 1 R2eq 3 The current through the inductor is i( t ) i (0) ett/ 10 e ( 2/3) A , t 0 36 SOURCE-FREE RL CIRCUIT Method-2: Applying KVL to the circuit For loop 1, 1 di 1 2()0(3)ii 2 dt 12
For loop 2, 5 6i 2 i 3 i 0 i i 2 1 1 26 1 di 2 Substituting above into Eq. (3) gives 1 i 0 dt 3 1 i(t) 2 => ln t |t i ()(0) t i e(2/3) t 10 e (2/3) t A , t 0 i(0) 3 0
37 SOURCE-FREE RL CIRCUIT The voltage across the inductor is
di 210 (2/3)(2/3)tt vLeeV 0.5(10) dt 33
Since the inductor and the 2-Ω resistor are in parallel, v i( teAt )1.6667,0 (2/3)t x 2
38 SOURCE-FREE RL CIRCUIT Q4. The switch in the circuit below is closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.
39 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). 412 R25 Ω eq 412 Ω 40 iA8 1 5
We obtain i(t) from i1 using current division, 12 i( t ) i 6 A , t 0 12 4 1 40 SOURCE-FREE RL CIRCUIT Since the current through an inductor cannot change instantaneously, iiA(0)(0)6 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
Req (12 4) ||16 8
The time constant is L 21 s Req 8 4
i( t ) i (0) ett/4 6 e A 41 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.
42 SOURCE-FREE RL CIRCUIT Solution : It is better to first find the inductor current i and then obtain other quantities from it. For t < 0, the switch is open. Since the inductor acts like a short circuit to DC, the 6-Ω resistor is short-circuited, so that we have the circuit shown
Hence, io = 0 and 10 i( t ) 2 A t 0 23 vto ( ) 3i(t) 6V t 0 43 SOURCE-FREE RL CIRCUIT Thus, i(0) = 2 A For t > 0, the switch is closed, so that the voltage source is short-circuited We now have a source-free RL circuit as shown. At the inductor terminals, RTh 3|| 6 2
Thus the time constant is L 1s RTh Hence, i( t ) i (0)20 eettt//
44 SOURCE-FREE RL CIRCUIT Because the inductor is in parallel with the 6-Ω and 3-Ω resistors, di v( t ) v L 2( 2 ett ) 4 e V , t 0 oLdt v 2 it( )eA,t0L t o 63
Thus for all time, 60Vt vt() o t 00At 4et , 0 it() o 2 t 20At et,0it() 3 t 2et , 0 45 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+ 46 SOURCE-FREE RL CIRCUIT Solution : At t = 0- the current through and the voltage across the inductor are 10 iA(0 )105 ;v (0 )0V (inductor acts as short-circuit) L 1010 L
+ At t = 0 , iAL (0 ) 5 ;v (0 )(10L 10) 5100V (KVL)
The rate of change of current through inductor at time t = 0+ is di()() t di t 100 Lll 100 V 25 A / s dttt00 dt 4
47 PRACTICE PROBLEMS
48 SOURCE FREE RC CIRCUIT
Q1. Calculate time constants of the following circuits.
(a) (b) ( c )
Answer: (a) 6μs (b) 1ms ( c) 0.25 sec
49 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 50 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.
(a) 0.7143 μF, (b) 5 ms, (c) 3.466 ms Answer: 51 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.
7 μC Answer: 52 SOURCE FREE RC CIRCUIT
Q5. Refer to the circuit in Fig. Let vC (0) = 60 V. Determine vC, vx, and io for t ≥ 0.
Answer: 60e−0.25t V, 20e−0.25t V, −5e−0.25t A. 53 SOURCE FREE RC CIRCUIT
Q6. If the switch in Fig. opens at t = 0, find v(t) for t ≥ 0 and wC (0).
Answer: 8 e −2t V, 5.333 J.
54 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.
55 SOURCE FREE RL CIRCUIT
Q8. For the circuit in Fig., find i(t) for t > 0..
Answer: 2 e −2t A, t > 0.
56 SOURCE FREE RL CIRCUIT
Q9. For the circuit in Fig, find io for t > 0.
Answer: 1.2 e −3t A, t > 0.
57