Question

https://prnt.sc/su2cfr

Answer 1

With reference to the image :

For the given circuit,$\tau=\frac{L}{R_1}=\frac{L}{15}=\frac{1}{2}\implies L=7.5$

And at $t=0,W=\frac{1}{2}Li_o^2=2.22\ J\implies i_o=\sqrt{\frac{4.44}{L}}=$ $\sqrt{\frac{4.44}{7.5}}=0.77A$

At $t=0,$ inductance offer infinite resistance,So,no current pass through it.

$i_o=\frac{E}{R_2}\implies R_2=\frac{E}{i_o}=\frac{10}{0.77}=13 \Omega$

And $i=i_o(1-e^{t/\tau})=0.77(1-e^{2t})$