140310 Circuit Theory Chapter 7 Response of First-Orer RL an R Circuits 140310 Circuit Theory Chapter 7 Response of First-Orer RL an RC Circuits Chapter Objectives Be able to etermine the natural response of both RL an RC circuits. Be able to etermine the step response of both RL an RC circuits. Prof. Anan Gutub Full Creit of theses slies are given to Dr Imran Tasauq whom generously share them for acaemic benefit 1 Be able to etermine Initial Energy Store in Inuctor an Capacitor Course Learning Outcome (CLO) No. Analysis of RL an RC circuits CLO An ability to analyze first orer an secon orer circuits by applying electrical circuit laws Chapters/Topics covering the CLO Chapters 6 an 7 Inuctors, capacitors,natural response of RL an RC circuits, step response of RL an RC circuits Response on v, i, P as DC source abruptly (suenly) isconnecte Natural Response Response of on v, i, P when DC source suenly connecte Step Response RL an RC circuits are known as First-Orer Circuits because of their first orer ifferential equations 3 4 Recall Inuctor & Capacitor v & I Switches 5 (a) Switch was close for a long time. It was opene at t = 0 (b) Switch was open for a long time. It was close at t = 0 (c) Switch was in position (a) for a long time. It move to position (b) at t = 0 () Switch was in position x for a long time. It move to position y at t = 0 6 Instructor: Prof. Anan Gutub 1
140310 Circuit Theory Chapter 7 Response of First-Orer RL an R Circuits Natural Response of an RL Circuit Natural Response of an RL Circuit In the figure shown, assume that the switch is close for a long time an it is opene at t = 0 After solving, following is the result: This is calle Natural Response of an RL Circuit In the above equation, L/R is calle the time constant enote by τ, i.e., τ= L/R I 0 is the initial current in the inuctor an is given by: i(0 ) = i(0 + ) = I 0 Using KVL: Above eqn. can be solve using first orer ifferential equations Therefore, the above equation can also be written as: i(t) = I 0 e t/τ 7 8 Example 7.1 Problem 7.4 a) 0 e -5t A Fin b) -4 e -5t A c) -160 e -5t V ) power issipate in the 10 Ohm resistor ) 560 e -10t W 9 11 Problem The switch shown in the above circuit was close for a long time. At t = 0, the switch was opene. Fin the following: (a) 0 I (b) i(t), t 0 a) 10 ma b) 10 e -(0000)t ma 1 13 Instructor: Prof. Anan Gutub
140310 Circuit Theory Chapter 7 Response of First-Orer RL an R Circuits Natural Response of an RC circuit τ= RC is the time constant 14 15 Example 7.3 Problem 7.1 Fin a) 100 e -5 t V b) 60 e -5 t V c) 1 e -5t ma ) 60 e -50t mw a) 8 kω b) 0.5 µf c) ms ) power issipate in the 60 kohm resistor 16 17 Assessment Problem 7.3/page 46 Energy Store in Inuctor an Capacitor a) V 0 = 00 V b) τ = 0 ms c) v(t) = 00 e t/0.0 = 00 e 50t V Inuctor Recall:i(t) = I 0 e t/τ = I 0 e t(r/l) Initial Energy (w 0 ) store in Inuctor L : w 0 = ½ L I 0 Capacitor Recall: v(t) = V 0 e t/τ = V 0 e t/rc Initial Energy (w 0 ) store in Capacitor C: w 0 = ½ C V 0 18 19 Instructor: Prof. Anan Gutub 3
140310 Circuit Theory Chapter 7 Response of First-Orer RL an R Circuits Initial Energy From Ex. 7.4 (p.3) The initial voltages on Capacitors C1& C have been establishe by sources not shown. The switch is close at t = 0. Calculate the initial energy store in Capacitors a) C1 b) C c) total energy initially store? ) v(t) numerical expression at 50kΩ? e) i(t) numerical expression relate to v(t) & 50kΩ? w(0)= ½ C V 0 a) w 1 (0)= ½(5 10-6 )(4) w 1 (0)= 40µ J b) w (0)= ½(0 10-6 )(4) w (0)= 5760µ J c) w total (0) = w 1 (0)+w (0) =40µ J+5760µ J = 5800µ J ) v(t) =0 e -t/(50k 4µ) =0 e -t e) i(t) = v(t)/50k= 80 e -t µa 0 From Assessment Problem 7.3/page 46 Fin the initial energy store in the capacitor w 0 = ½ C ( V 0 ) w 0 = ½(0.4 10-6 )(00) w 0 = 8 mj 1 From Problem 7.6 page 50 For t >0. Fin: a) Initial energy store? b) v(t) numerical expression at 1kΩ c) i(t) base on v(t)? a) w 0 = ½ C ( V 0 ) w 0 for every capacitor is neee. w 0 = ½ C ( V 0 ) C 300nF :: w 0 = ½(300n)(96) 1.4mJ C 600nF :: w 0 = ½(600n)(7) 1.5mJ Total initial energy: w 0.9 mj b) v(t) = 4 e -5000t V c) i(t) = v(t) / 1k Ω = 4 e -5000t ma Assessment Problem 7.1 page 0 ::Fin a) initial value of i? b) initial energy store in inuctor? c) time constant of the circuit for t > 0? ) numerical expression for i(t) as t 0? a) i 0 = -1.5 A b) w 0 = 65 mj c) τ = 4 ms ) i(t) = -1.5 e -50t A 3 Step Response of RL an RC Circuits When a DC voltage or current source is suenly applie to inuctor or capacitor, the current an voltages that arise are calle Step Response Step response of an RL circuit From graph, it is clear that V s /Ris stable current value i.e., i = V s / R Response of on v, i, P when DC source suenly connect Step Response RL Circuits RC Circuits i( t) = i + ( I0 i ) e ( R/ L) t τ= L/R 4 5 Instructor: Prof. Anan Gutub 4
140310 Circuit Theory Chapter 7 Response of First-Orer RL an R Circuits Remember Derivatives : Inuctor ( f( x) = e ( e x + x) x + x fin: ) = ( e x + x f( x) ) ( x + x) = ( e Capacitor f (x) x + x )(x+ ) 6 Example 7.5 a) Fin expression of i(t) for t 0? b) What is initial voltage across inuctor just after switch move to position b? c) How many millisecons after switch move oes the inuctor voltage equal 4 V? a) i 0 = - 8A i = V s / R =4/=1A i(t) = 1 + (-8-1)e t/0. = 1-0 e 10t A b) L = 0. recall: ( e x + x ) = ( e x + x V(t) = 0. (00e -10t ) = 40e -10t V 0 = 40 V )(x+ ) c) 4 = 40 e -10t t = ⅟₁₀ ln(⁴⁰ ₂₄) = 51.08 ms 7 Assessment Problem 7.5 a) i 0 = 1 A i = -8 A (a) i(0) = 4/ = 1 A b) V 0 = -00 V c) τ =L/R = 00/10 = 0 ms (c) For t> 0, only the 10 Ohm resistor is connecte to the inuctor. Therefore, τ = L/R = 00 x 10 3 /10 = 0 ms ) i(t) = -8+[1-(-8)]e t/τ = -8+0 e -50t () e) From: V(t) = - 00e -50t 9 Since, i(t) = 8 + [1 ( 8)]e t/0.0 = 8 + 0e 50t A, t 0 (e) 30 Another way to solve Part () When the switch moves to position a, the circuit is as shown in the following figure: Step Response of an RC Circuit To use the equation for i(t), we nee to convert the circuit as known earlier. We can use source transformation to o so. Now the circuit looks like as shown below: Since: Therefore, from part (a), I 0 = 1 A From the circuit shown, V s = 80 V, R = 10 Ohm Therefore: i(t) = 8 + [1 ( 8)]e t/0.0 = 8 + 0e 50t A, t 0 31 v( t) = v + ( V0 v ) e t/ RC As I s Ris the final value of voltage, i.e., v C ( ) = I s R 3 Instructor: Prof. Anan Gutub 5
140310 Circuit Theory Chapter 7 Response of First-Orer RL an R Circuits The switch has been at position 1 for long time. At t = 0, the switch moves to position. Fin: v c (t) an i(t) for t 0 recall: x ( e + x x ) = ( e + x )(x+ ) First, calculate V 0 for t<0 v C (0) = V 0 = 30 V v C ( ) = I s R = -1.5 ma* 40 kω = -60 V τ = RC = 40 10 3 0.5 10-6 = 10 10-3 v C = I s R + (V 0 -I s R)e -t/τ v C = -60+[30-(-60)]e t/0.01 = -60 + 90 e -100t i(t) = C(v C /t) = 0.5 10-6 (-60 + 90 e -100t )/t = 0.5 10-6 (-9000 e -100t ) = -.5 e -100t ma 33 Example 7.6/page 5 The switch has been at position 1 for long time. At t = 0, the switch moves to position. Fin: v c (t) for t 0 i(t) for t 0 v C (0) = V 0 = 30 V v C ( ) = I s R = -1.5 ma* 40 kω = -60 V τ = RC = 40 10 3 0.5 10-6 =10 msec v C = I s R + (V 0 -I s R)e -t/τ v C = -60+[30-(-60)]e t/0.01 = -60 + 90 e -100t i(t) = C(v C /t) = 0.5 10-6 (-60 + 90 e -100t )/t = 0.5 10-6 (-9000 e -100t ) = -.5 e -100t ma 34 Problem 7.55 After long time, at t=0 switch is move to position b. Fin: v C (0), v C ( ), τ for t >0 i(0), v C (t)for t >0, i(t) for t >0 vc i( t) = C t v C (0) = V 0 = 50 V v C ( ) = I s R = -4 V τ = R eq C = 4 5 10-9 = 0.1 µsec i(0) = -18.5 A v C = I s R + (V 0 -I s R)e -t/τ v C = -4+[50-(-4)]e t/0.1µ = -4 + 74 e -10000000t V i(t) = C(v C /t) =5 10-9 (-4+74e -10000000 t )/t =5 10-9 (-74 10 7 e -10000000t ) = -18.5 e -100t A 36 Example Practice Problems Examples 7.1, 7.3, 7.5(a, b, c), 7.6, 7.7(a, b, c,, e, f) Assessment Problems 7.1(a,b, c, ), 7.(a), 7.3(a, b, c, ), 7.4, 7.5 En of Chapter Problem 7.1, 7.55 40 Instructor: Prof. Anan Gutub 6