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