0 ECE 222 Electric Circuit Analysis II Chapter 6 First-Order RL Circuits Herbert G. Mayer, PSU Status 5/4/2016 For use at CCUT Spring 2016

1 Syllabus Definition Natural Response RL Time Constant τ Example 1 Example 2 Bibliography

2 Definition Circuits with resistors and inductors only are denoted as RL circuits Circuits with resistors and capacitors only are denoted as RC circuits Both RL and RC circuits are known as first-order circuits since they can be described by a first-order differential equation DE Here we focus on analysis of first-order circuits with inductors

3 Definition If a first-order circuit contains multiple interconnected capacitors, they can all be replaced by (reduced to) an equivalent single capacitor C eq Similarly, if a first-order circuit contains multiple interconnected inductors, they can all be replaced by (reduced to) an equivalent single inductor L eq The reduction follows the rules of parallel and series resistors or conductors Also uses the Thevénin Equivalent for finding a corresponding voltage source Or the Norton Equivalent for a current source

4 Definition Below left we see a first-order RL circuit, in which all resistances are substituted by an equivalent resistor R eq and all inductors by an equivalent L eq At right we have a first-order RC circuit with one equivalent C eq

5 Natural Response with Inductor

6 Natural Response i(t) Analyze the natural response of first-order RL circuits: Assume that a constant current source has been connected in the circuit below for a long time Only 2 resistors and one inductivity are parallel to the current source; R0 protects the current source Hence the generated voltage is 0 V All components are ideal, so the inductivity has resistance 0 Ω No current flows through the 2 resistors, hence their voltage drop is 0 V The only current generated by the source of size i runs through the ideal L Generating a voltage drop across L of 0 V

7 Natural RL Response i(t) Current of i [A] generated after a long time And i is running through L for t ≤ 0 [s] No current flows through R0 or R, until time t = 0

8 Natural RL Response i(t) After a long time before time 0, current i is constant That means, L di/dt = 0 So the magnetic field has been built up in L. It does not change any longer; it is stable and constant Now comes the new step: we open the switch S, the source is disconnected, and the circuit is a pure and simple RL circuit at time t > 0

9 Natural RL Response i(t) When switch S is opened at time t=0, the RL circuit is separated from the current source The magnetic field in L breaks down L produces a current i(t) in the pure RL circuit Current i(t) generates a voltage drop v(t) along R But we know that the initial current i(t) at time 0, named i(0), is the current generated by the constant current source: i. It was flowing through L before time t=0 Now we can use Kirchhoff’s Voltage Law KVL: L di/dt + R i(t) = 0 [V], with i(0) = i from the CCS

10 Natural RL Response i(t) L di/dt = -R i(t) This is a first order differential equation, as the variable i occurs only in the first power; other units are constant, e.g. R and L L di/dt=-R i(t) di / i(t)=-R/L dt di / i(t)=-R/L dt -- for time 0 to t ln( i(t) / i 0 )=-R/L * t -- account for i at time 0, i 0 i(t)=i 0 e -t R/L

11 Natural RL Response i(t) This is a key, first order equation for the current function over time i(t) = i 0 e -t R/L With the current i at time t = 0 being the one from the original constant current source i0i0 i(t) t

12 Natural RL Response v(t) To compute the voltage v(t) over time, we use Ohm’s Law at the resistor R i(t)=i 0 e -t R/L v(t)=R * i 0 e -t R/L v(0)=R * i, from original i = i(0) of CCS

13 Natural RL Response p(t) The power p(t) is just current * voltage p(t)=v * i p(t)=R * i * i p(t)=v * v / R-- must generate identical result p(t)=R * i 0 e -t R/L * i 0 e -t R/L p(t)=R * i 0 2 e -2t R/L for t >= 0

14 Natural RL Response w(t) The energy w(t) is the integral of power over time: w(t)=p(t) dt-- from time 0 to t w(t)=R * i 0 2 e -2t R/L dt w(t)= R * i 0 2 e -2t R/L dt-- from time 0 to t w(t)= -R * i 0 2 * (L/2R) e -2t R/L + R * i 0 2 * (L/2R) w(t)=½ L i 0 2 ( 1 - e -2t R/L ) For t >= 0 For t -> ∞ w(t) approaches 0, i.e. no more energy is being dissipated At which time the energy dissipated by the magnetic field equals the energy stored while the CCS was connected!

15 Significance of RL Time Constant τ

16 RL Time Constant τ We frequently see ratio R/L, or the term -t (R/L) Coefficient R/L is the rate, at which its factor approaches 0, for time towards ∞ Factor can be current, or voltage, or power etc. i(t) = ~ e -t (R/L) –For current v(t) = ~ e -t (R/L) –For voltage p(t) = ~ e -2t (R/L) –For power Reciprocal of ratio R / L is the time constant τ τ has SI unit of seconds [s] τ = 1 / (R/L) = time constant = L / R

17 RL Time Constant τ We re-write the formulae for i(t), v(t), etc. with τ as: i(t) = I 0 e -t/τ – current v(t) = I 0 R e -t/τ – voltage p(t) = I 0 2 R e -2 t/τ – power w(t) = ½ L I 0 2 ( 1 - e -2 /τ ) – energy Formulae indicate: electric units decrease rapidly, when negative exponent -t/τ is large I.e. when τ is very small, or a long time t has elapsed Or for small L or large R values The period of the electric unit still being large ( >> 1% of original) we call this state a: transient response Afterwards, when the unit is << 1%, we refer to this as the: steady-state!

18 RL Time Constant τ 1 time constant in a first-order circuit, after L has released stored energy to its R, i(t) is reduced to e -1 With e = …, e -1 = 1/e = …: After 5 times τ, remaining current i is << 1 %, i.e. only has some transient value left That is within most measuring error variations, so often can be ignored in engineering

19 RL Time Constant τ We see, τ characterizes a first-order RL circuit It defines, how quickly the current (or other unit) reaches its final value of 0 [A] The current in an RL circuit changes at the rate: di / dt = - I 0 R / L = - I 0 / τ If it were to change linearly (which it does not!), we can use the value as a way to characterize the time constant τ in an RL circuit i(t) = I 0 - t I 0 / τ Which would cause 0 [A] for i(t) at τ seconds Sample next page: L = 0.1 [H], R = 8 [Ω], I 0 = 10 [A]

20 RL Time Constant τ Example: I 0 = 10 A, R = 8 Ω, and L = 0.1 H, τ = R/L = 80 [s] i 1 (t) = I 0 - t * ( I 0 / τ ),i 2 (t) = I 0 e -t/τ i 1 (t) = 10 - t * ( 10 / 80 ),i 2 (t) = 10 e -t/80 i = I 0 e -t/τ τ i = I 0 -t I 0 /τ

21 RL Time Constant τ Key learning and procedure for time constant τ in a first-order RL circuit: 1. Identify the initial current I 0 that creates the magnetic field in L 2. Compute time constant τ, with τ = R / L 3. To compute the current function over time, use equation: i(t) = I 0 e -t/τ for a first-order circuit!

22 Example 1 for Inductor

23 Example 1 – The RL Circuit

24 Example 1 Switch S in Example 1 has been closed for a long time, and current I 0 has flown steadily through L L poses no resistance, is assumed ideal Resistors R 1 and R 8 have no current, hence no voltage drop All 10 A flow through L of 0.1 [H], coming from the current source At time t = 0, suddenly S is opened, and the energy stored in L is released, flowing through L and resistor R 8 Initial current when S is opened is identical to current flowing before through L, i.e. I 0 = 10 A Current source and R 1 are no longer part of circuit

25 Example 1 – Compute: a.) show τ b.) compute current i L (t) c.) current i 8 (t) d.) voltage v 8 (t) e.) power p(t) in R 8

26 a.) Example 1 – Show τ τ is the inverse of R / L = 1 / ( R / L ) τ=L / R τ=0.1 / 8 -- unit of [s] τ= [s] = 12.5 [ms] Note: in current and voltage equations we use the inverse of τ, with τ = L / R Quick SI unit check of τ, must be [s]: [τ] = [V s A -1 ] / [V A -1 ] = [s]

27 b.) Example 1 – Current i L (t), c.) i 8 (t) i L =I 0 e -R/L t i L =10 e -R/L t i L =10 e -80 t -- for t from 0.. ∞ i 8 =-i L i 8 =-10 e -R/L t i 8 =-10 e -80 t -- for t from 0.. ∞

28 d.) Example 1 – voltage v 8 (t) Using Ohms Law: V = R * I, and using the current i 8 through R 8 of 8 Ω, we compute the voltage v(t): i 8 =-10 e -80 t v(t)=8 * -10 e -80 t v(t)=-80 e -80 t

29 e.) Example 1 – power p(t) in R 8 p(t)=v(t) * i(t) -- in general p(t)=v(t) 2 / R -- and in this example with v 8 (t) and R 8 = 8 Ω p(t)=v 8 (t) 2 / R 8 p(t)=(-80) 2 e -2*80 t / 8 p(t)=800 e -160 t

30 Example 2 for Inductor

31 Example 2 – The RL Circuit

32 Example 2 Example 2 taken from [1], p. 218 Switch S has been closed for a long time, and current I 0 has flown steadily through L L poses no Ohmian resistance Resistors 2 Ω, 10 Ω, and 40 Ω have no current, hence no voltage drop All 20 A flow through L of 2 Henry [H], coming from the constant current source At time t = 0, suddenly S is opened, and the energy stored in L is released, flowing through L and three resistors Initial current when S is open is equal to current flowing before through L, and that is I 0 = 20 A Current source and 1Ω resistor no longer in circuit

33 Example 2 – Compute: a.) resistance R eq b.) current i L (t) c.) current i R40 thru 40 Ω d.) voltage v(t) across 40 Ω e.) power p(t) in 10 Ω f.) energy w 10 in 10 Ω g.) energy w L in L h.) percent w in 10 Ω

34 a.) Example 2 – Resistance R eq 10 Ω and 40 Ω parallel: 10 * 40 / ( ) = 8 Ω 8 Ohm series to 2 Ohm = 10 Ω R eq = 10 Ω

35 b.) Example 2 – Current i L (t) Inductivity is 2 H R eq is 10 Ω Initial current I 0 is the same as the one delivered through the current source, before S was opened, and that is = 20 A i L (t)=I 0 e -(R/L) t i L (t)=20 e -5 t

36 c.) Example 2 – Current i 0 thru 40 Ω We use current division: i L (t) through L is the same as through R eq, just of opposite sign i R40 = - i L i R40 flows through 2 Ω, then divides through 10 Ω and 40 Ω i R40 =- i L 10 / ( ) i R40 =- 1/5 i L i R40 =- 4 e -5 t

37 d.) Example 2 – Voltage v(t) thru 40 Ω We know current through R of 40 [Ω] = -4 e -5 t Ohms Law: V = R * I v(t)=40 * -4 e -5 t --across 40 [Ω] resistor v(t)=-160 e -5 t --for t >= 0 [s]

38 e.) Example 2 – Power p(t) in 10 Ω p=v * i P=v * v / R P=v 2 / R -- applied here, using v(t): v(t)=-160 e -5 t p(t)=v(t) 2 / 10 p(t)=160 * 160 / 10e -2( 5 t ) p(t)=2560 e -10 t -- Units of Watt [W]

39 f.) Example 2 – Energy w 10 in 10 Ω Energy is integral of power over time Power in 10 Ω resistor, p(t) = 2560 e -10 t Time runs from 0 to ∞ w 10 = 2560 e -10 t dt -- from 0 to ∞ w 10 = 2560 / -10 (e -10 ∞ - e -10 *0 ) w 10 = 2560 / -10 ( ) w 10 = units of Joule [J]

40 g.) Example 2 – Energy w L in L Energy is integral of power over time Integral starting at 0 yields 1 for t = 0 w(t)=½ L i 0 2 ( 1 - e -2*(R/L) t ) e -2*(R/L) t = 0 -- for t = ∞ w L =½ L i 0 2 ( ) w L =½ L i 0 2 w L =½ L w L =½ * 2 * 20 2 w L = units of Joule [J]

41 h.) Example 2 – Percent of w in 10 Ω we know w L = 400 J And w 10 = 256 J Hence the ratio r is r=w 10 / w L r=25 / 400 r=0.64= 64%

42 Bibliography  Electric Circuits, 10 nd edition, Nilsson and Riedel, Pearsons Publishers, © 2015 ISBN-13:  Table of integrals: table.com/downloads/single-page-integral-table.pdf