91Ó°ÊÓ

It is given that$I=\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$R$}\right.\left[1-e^{-\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$L$}\right.\right)t}\right].$If$E=12V,R={10}^{-2},L=5$henrys. We need to write how long it takes to obtain a current of$0.5$ampere and$1.0$ampere and also we need to graph the equation.

Substitute $12$to $E,10$to $R,5$to $L$and $0.5$to $I$in role="math" localid="1647187235340" $I=\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$R$}\right.\left[1-e^{-\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$L$}\right.\right)t}\right].$

role="math" localid="1647187295650" $0.5=\frac{12}{10}\left[1-e^{-\left(\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)t}\right].$

$0.5=1.2I=\left[1-e^{-\left(\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)t}\right].$

$\frac{0.5}{1.2}=1-e^{-2t}.$

$\frac{5}{12}-1=-e^{-2t}.$

$-\frac{7}{12}=-e^{-2t}.$

$e^{-2t}=\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$12$}\right..$

$-2t=\ln \left(\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\right).$

$t=\frac{\ln \left(\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\right)}{-2}.$

$t=0.2694982504.$

$≈0.27sec.$

It will take around$0.27$seconds to obtain a current of$0.5$ampere.

Substitute $12$to $E,10$to $R,5$to $L,1.0$to $I$in,

$I=\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$R$}\right.\left[1-e^{-\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$L$}\right.\right)t}\right].$

$1.0=\frac{12}{10}\left[1-e^{-\left(\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)t}\right].$

$1=1.2\left[1-e^{-2t}\right].$

$\frac{1}{1.2}=1-e^{-2t}.$

$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=1-e^{-2t}.$

$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.-1=-e^{-2t}.$

role="math" localid="1647187447064" $-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=-e^{-2t}.$

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=e^{-2t}.$

$-2t=\ln \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right).$

$t=\frac{n\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)}{-2}.$

$t≈0.90.$

It takes around $0.90$seconds to obtain a current$1.0$ampere.

$I=1.2\left(1-e^{-2t}\right).$