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(a) Current througn ' L ' at t=0i_{0}=\frac{E}{R}=\frac{24}{3}=8 \mathrm{~A}(b)\begin{array}{l}Y=\frac{L}{R}=\frac{8}{3} \times 10^{-3} \mathrm{~s} \\i=e_{0}^{-\frac{t}{r}} \\i=8 e^{-\frac{3.5 \times 10^{-4}}{\frac{8}{3} \times 10^{-3}}} \\i=8 e^{-0.13125}=2.15 \mathrm{~A}\end{array}(c) voltage across ' R ' at t=3.5 \times 10^{-4} \mathrm{~s}V=i R=2.15 \times 3=6.46 \mathrm{~V}(d)\begin{array}{l}i=i_{0} e^{-\frac{t}{r}} \\\frac{d i}{d t}=i_{0}\left[\frac{-1}{r}\right] e^{-\frac{t}{r}} \\\frac{d i}{d t}=8\left[-\frac{3}{8 \times 10^{-3}}\right] e^{-\frac{3.5 \times 10^{-4}}{\frac{8}{3} \times 10^{3}}} \\\frac{d i}{d t}=-3 \times 10^{3} e^{-0.13125} \\\frac{d i}{d t}=-2.63 \times 10^{3} \mathrm{~A} / \mathrm{s}\end{array}negatine \operatorname{sig} x reprenent decreaxe.\frac{d i}{d t}=2.6 .3 \times 10^{3} \mathrm{~A} / \mathrm{s}current in decreasine at the nate of 2.63 \times 10^{3} \mathrm{~A} / \mathrm{s} ...