A) 6.7 Ma
B) 0.67 mA
C) 100 mA
D) 67 mA
Correct Answer: B
Solution :
[b] \[I\,(0)=\frac{15\times 100}{0.15\times {{10}^{3}}}=0.1A\] \[I\,(\infty )\,=0\] \[I\,(t)\,={{[I\,(0)-I(\infty )]}_{{{e}^{^{\frac{-t}{L/R}+i(\infty )}}}}}\] \[I\,(t)\,=0.1\,{{e}^{^{\frac{-t}{L/R}+i(\infty )}}}=0.1\,{{e}^{\frac{R}{L}}}\] \[I(t)\,=0.1\,e\frac{0.15\times 1000}{0.03}\,=067\,mA\]You need to login to perform this action.
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