Question

Consider the RL circuit shown in the figure (Figure 1). When the switch is closed, the current in the circuit is observed to increase from 0 to 0.33 A in 0.14 s .

What is the inductance L?

Express your answer using two significant figures

How long after the switch is closed does the current have the value 0.51 A ?

Express your answer using two significant figures.

Answer 1

As you have not provided figure, I am considering voltage of battery = 9.0 volts and resistance = 5.5ohms

hence,

we know , Time constant T_c = L/R - --------------->> (1)
also from question I₀ = 0, I₁ = 0.33 A, t₁= 0.14 s

Final current, I_f = V/R = 9/5.5 = 1.36A

also T_c = -t₁/ln((I₁-I_f)/(I₀-I_f)) ----------------->> (2)

T_c =-0.14/ ln(1.03/1.36) = 0.50sec
from 1 we know inductance,

L = T_c*R = 0.50*5.5 = 2.78 Henry

now when current I₂ = 0.51 Amps, time t₂ will be equal to, using formula 2
t₂= -T_c*ln((I₂-I_f)/(I₀-I_f)) = 0.236 s

t₂=  0.236 s