An electric circuit contains a battery that produces a volta

An electric circuit contains a battery that produces a voltage of 60 volts (V), a resistor with a resistance of 11 ohms (), and an inductor with an inductance of 3 henrys (H), as shown in the figure. Using calculus, it can be shown that the current I = I(t) (in amperes, A) t seconds after the switch is closed is

= 60/11(1-e^-11t/3)

Use this equation to express the timet as a function of the current I.

After how many seconds is the current3 A? (Round your answer to three decimal places.)

Solution

at t = 0+ ( on switching) l behaves as short circuit = V/R = 60/11 amp

Write KVL equation cosnideruing series circuit : V = Ir + LdI/dt

taking Laplce transform :

V = Ir + sLI(s)

V = RI(s) + sLI(s)

I(s) = V/(R + sL)

taking inverse laplace transform we get:

I(t) = (V/R)[ 1- e^-(R/L)t ]

Plugging all values : V = 60 v ; R = 11 ohms ; L = 3 henry

I(t) = (60/11)[ 1- e^-(11/3)t ]

11I/60 = 1- e^(-11/3t)

e^(-11/3)t = 1- 11*I/60

take natural log on both sides:

-11t/3 = ln[1- 11*I/60]

t = - 3ln[1 -11*I/60]/11

I(t) = 3 amp

3/5.45 = 1 - e^-(11/3)t

e^(-11/3)t = 0.449

take natural log on both sides:

-11/3t = ln(0.449)

t = 0.218 sec