Question 1.SGPYQ.28: For the value of obtained in the above question, the time ta......

$E_1$ is the stored energy and $E_2$ is the dissipated energy. So,

$\frac{i_2^2}{i_1^2}=\frac{E_2}{E_1}=0.95$

Therefore, $i_2^2=0.95 i_1^2 \Rightarrow i_2=0.9746 i_1$

When the switch is in position 2,

$i(t)=i_1 e^{-(R / L) t}=2 e^{-(60 / 10) t}=2 e^{-6 t}$          (i)

After 95% of energy dissipation,

i(t) = 2 – 2 × 0.97 = 0.05 A        (ii)

Substituting Eq. (ii) in Eq. (i), we get

$0.05=2 e^{-6 t} \Rightarrow t=0.5  s$