Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
(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} ...