Question

The current in an RL circuit increases to 98 % of its final value 2.22 safter the switch is closed.

PART A)

What is the time constant for this circuit?

PART B)

If the inductance in the circuit is 0.298 H , what is the resistance?

Answer 1

I = I₀( 1-e ^{- t/$\dpi{100} \tau$} )

$\dpi{100} \tau$, time constant = L/R , t = time in second =2.22s

I_o is full current

0.98 =1(1-e^{-2.22/$\dpi{100} \tau$})

e^{-2.22/$\dpi{100} \tau$} = 0.02

taking log_e

-2.22/$\dpi{100} \tau$ = log_e 0.02

$\dpi{100} \tau$ = 0.56748 sec ^{part A answere}

part B

$\dpi{100} \tau$ =L/R

0.56748 = 0.2986/R

R= 0.52618 $\dpi{100} \Omega$