Question 130 of 190

An emf of $20V$ is applied at time $t=0$ to a circuit containing in series $10mH$ inductor and $5\text{Ω}$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40\text{s}$ is close to: (Take ${e}^{2}=7.389$ )

A$1.06$B$1.15$C$1.46$D$0.84$

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Answer: (A) $1.06$

$i={i}_{0}\left(1-{e}^{\dfrac{-t}{L/R}}\right)$

$=\dfrac{20}{5}\left(1-{e}^{\dfrac{-t}{0.01/5}}\right)$
$=4\left(1-{e}^{-500t}\right)$now at $t=\infty$
${i}_{\infty }=4$at $t=40\text{s}$
${i}_{40}=4\left(1-{e}^{-500\times 40}\right)$
$=4\left(1-\dfrac{1}{{\left({e}^{2}\right)}^{10000}}\right)=4\left(1-\dfrac{1}{{7.29}^{10000}}\right)$
$\dfrac{{i}_{\infty }}{{i}_{40}}\approx 1$ Slightly greater then one.