082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

19.3 First-order equations: simple equations and separation of variables 545
so that
L
R ln E
E−iR =t
Then
E
ln
E−iR = Rt
L
hence
E
E−iR = eRt/L
Rearranging toobtainE gives
hence
and so
so
E =(E−iR)e Rt/L
E =Ee Rt/L −iRe Rt/L
iRe Rt/L =E ( e Rt/L −1 )
iR=E ( 1 −e −Rt/L)
Finally we have
i = E ( )
1 −e
−Rt/L
R
ThegraphofthiscurrentagainsttimeisshowninFigure19.2.Wenotethatast → ∞,
i → E R .Therateatwhichthecurrentincreasestowardsitsfinalvaluedependsupon
thevaluesofthecomponentsRandL.Itiscommontodefineatimeconstant, τ,for
the circuit. Inthis case τ = L and the equation forthe current can be written as
R
i = E R
(
1 −e
−t/τ )
The smaller the value of τ, the more rapidly the current reaches its final value. It is
possibletoestimateavalueof τ fromalaboratorytestcurvebynotingthatafterone
time constant, thatist = τ,ihas reached 1 −e −1 ≈ 0.63 of its final value.
EXERCISES19.3
1 Find the general solution ofthe followingequations:
(a)
(c)
dy
dx =3
dy
dx = 2x
(b) dx
dt = 5
(d) dy
dt = 6t
(e)
(g)
(i)
dy
dx = 8x2
dy
dx = x2
y
dx
dt = et
x
(f)
(h)
(j)
dx
dt =3t3
dx
dt = t3
x 2
dy
dx = e−2x
y 2