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SalutionGiven\begin{array}{l}L=2 H, R=100 \Omega \\V=9 V\end{array}a) At t=0, the switch S is closedand the circuit is completed.From kirchoft"s Voltage rull,\begin{aligned}V_{L}+V_{R} & =V \\L \frac{d i}{d t} & +i R=V \\L \frac{d i}{d t}=V-i R \Rightarrow & \int_{0}^{i} \frac{d i}{V-i R}=\int_{0}^{t} \frac{d t}{L}\end{aligned}-) on integrating weget\begin{array}{l} i(t)=i_{0}\left(1-e^{-t / \tau}\right), \text { where } i_{0}=\frac{V}{R} \\\tau=L / R \\\Rightarrow i(t=5.8 \mathrm{~ms})=\frac{9}{100}\left[1-e^{-\frac{5.8 \times 10^{-3}}{\alpha} \times 100}\right] \\\Rightarrow i=9 \times 10^{-2}\left[1-e^{-0.29}\right], 9 \times 10^{-2}[1-0.748]\end{array}\begin{array}{c}\Rightarrow i(5.8 \mathrm{~ms})=0.02268 \mathrm{~A} \\i=22.68 \mathrm{~mA}\end{array}B) Maximum Power delivered by the Rower suthby\begin{aligned}P & =i_{\max }^{2} R=\frac{v^{2}}{R} \quad\left[\because i_{\text {max }}=\frac{v}{R}\right] \\\Rightarrow P=\frac{(q)^{2}}{100} & =0.81 \text { Watt }\end{aligned}please upvote ud83dudc4d Comment for any doubt ...