《电路》（英文版）5-2 The source-free RL circuit

85-2 The source-free RL circuit i(t)--time var ying current R UR( L)ULO at t=0, i(t) KVL: VR+V,=0 or i+l-=0 dt di R di R R +--dt=0 or di t→ dt R In i(t)-Inl=--t .i(t=le l L

§5-2 The source-free RL circuit 0 0 0 0 0 I i( ) I t at t , i(t ) i(t ) time var ying current = = = = − − t L R t i t I e L R i t I − − 0 = −  = 0 ln ( ) ln ( ) : + = 0 + = 0 dt di KVL vR vL or Ri L   + = = −  = − i( t ) t I dt L R i di dt L R i di dt or L R i di 0 0 0 (t)  L + − (t)  R − + R i(t) L

Let us assume a solution i(t=Ae, where a and S, are constant to be determined d(Ae R(Ae”)+L 0 R RAel+ lasel=o or A(s,+el=0 R the solution are S, L ∴i(t)=AeL if i(t) I。→A=I, 0 ..i( t)=1o 2R Pr=Ri= RIoe l It =RI R PR dt 0 L LIo)

t L R R p Ri RI e 2 2 0 2 − = = Let us assume a solution i(t)=A , where A and s1 are constant to be determined. s t e 1 0 ( ) ( ) 1 1 + = dt d Ae R Ae L s t s t 0 ( ) 0 1 1 1 + 1 = 1 + = s t s t s t e L R RAe LAs e or A s L R the solutionare s1 = − t L R i(t ) Ae −  = t L R i t I e −  = 0 ( ) ) 2 1 ( 2 1 2 0 2 0 2 0 2 0 0 w p dt R I e dt LI w LI L t L R R = R = = = =   −   0 0 I A I t 0 i f i( t ) = → = =

)L1 R1 10mH 1k 0 -- 10u 20us 40us 50us I(L1)

Time 0s 10us 20us 30us 40us 50us I(L1) 0A 1.0A 2.0A 0 I L1 10mH 1 2 R 1 1k

The series RL circuit: i(t)=loe At zero time. the current is the o And as time increases. the current decreases and approaches zero The initial rate of decay is found by evaluating the derivative at zero time i/I R dt t RL

The series RL circuit: t L R i t I e − = 0 ( ) At zero time, the current is the I0 . And as time increases, the current decreases and approaches zero. i t 0 0 I The initial rate of decay is found by evaluating the derivative at zero time. L R t e L R dt t I i d t L R = − = = − = − 0 0 ( ) 0 1 t 0 0 i / I 

We designated the value of time it takes for io to drop from unity to zero, assuming a constant rate of decay, by the greek letterτ R f-t=l or T =L/R--time-constant i(t=loe t ori(z)=0.36810=36810% The value of io at tT =er=e=0.368 i/I In one time constant the response has dropped to 36.8 0.368 percent of its initial value. 0.135 T 2t 3T

We designated the value of time it takes for i/I0 to drop from unity to zero, assuming a constant rate of decay, by the Greek letter .  The value of i/I0 at t=  0.368 ( ) 1 0 = = = − − e e I i    In one time constant the response has dropped to 36.8 percent of its initial value. If =1 or =  L/R--time-constant L R   t i t I e −  = 0 ( ) 0 0 0 8 0 or i( ) = 0.368I = 36. I 0.368 t 0 0 i / I  2 3 0.135 0.051

Why does a larger value of the time constant L/R produce response curves, which decays more slowly (↑τ↑w=L2/2↑t↑orR↓↑p=R2↓t↑) If a circuit contains any number of resistors and one inductor. we fix our attention on the two terminals of the inductor and determine the equivalent resistance across these terminals. The circuit is reduced to simple series case

If a circuit contains any number of resistors and one inductor, we fix our attention on the two terminals of the inductor and determine the equivalent resistance across these terminals. The circuit is reduced to simple series case. Why does a larger value of the time constant L/R produce response curves, which decays more slowly? (L↑ ↑ wL=Li 2/2↑ t↑or R ↓ ↑ p=Ri   2↓ t↑)

L L R3 R R R4 R R=R3+R4R1R2→zs →=i(0)ex R1+R2 R eq R R +r iL(oet and e + r

  t L L eq eq i i e R L R R R R R R R − → = → = + = + + (0) 1 2 1 2 3 4   t L t L i ( )e R R R i ( )e and i R R R i − − + = − + = − 0 0 1 2 1 2 1 2 2 1

If we are given the initial value have i as i, (0 then i2(0)= R1i(0+) R i1(0)=i1(0+) R3 /i(0+)+i2(0+ R4 R2 R4 R 2 R1+R2i(0) R i1(t)=i1(0+)e 2()R i1(0+)e ()=-R+R, i1(0+)e (t=L/Rea) R Since the inductor current decays exponentially as e t, then every current and voltage in the resistive network must have the same function behavior

( ) (0 ) ( / ) 1 2 1 2 eq t L i e L R R R R and i t = + = − − +   2 1 1 2 (0 ) (0 ) R R i then i + + = i ( ) R R R [i ( ) i ( )] i ( ) i ( ) L L + + + + + = − = − + = 0 0 0 0 0 1 2 1 2 1 2   t t i e R R i t i e i t − + − +  ( ) = (0 ) ( ) = (0 ) 1 2 1 1 1 2 If we are given the initial value have i1 as (0 ). 1 + i Since the inductor current decays exponentially as , then every current and voltage in the resistive network must have the same function behavior.  t e −

Example 2: Find i, and i, for t>0. solution 120 2×3 eq +1=2.2mH 2+3 ImH 18V 2mwmv2=50(120+60)×90 120+60+90 110 ea 2.2×10 0.36 =2×10s 0.2 R 11o ea i(0)=0.364→i(0+)=i(0-)=0.364 180 0.24 i1(0)=18/90=0.2 A and i1(0)=-i2(0) =-0.24A 180+90 i1(t)=0.36e4andi1(t)=-0.24e04

Example 2: Find iL and i1 for t>0. 110 120 60 90 (120 60) 90 50 = + + +  Req = + i (t ) . e A and i (t ) . e A t t L 50000 1 50000 0 36 0 24 − −  = = − i L(0 ) = 0.36A → i L(0 ) = i L(0 ) = 0.36A − + − i ( ) / . A and i ( ) i L ( ) 0.24A 180 90 180 0 18 90 0 2 0 0 1 1 = − + = = = − − + + solution: s R L eq eq 5 3 2 10 110 2 2 10 − − =   = = .  L i R i 0.36 −0.24 0.2 Leq 1 2.2mH 2 3 2 3 + = +  =

R4 120 R3 TOPEN =0 60 R1 1mH 50 R2 L2 18V 90 1.2mH 0

I R 1 50 R 2 90 L1 1mH 1 2 R 4 120 R 3 60 18V L2 1.2mH 1 2 U 1 TOPEN = 0 1 2 0