Question

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

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

Answer 1

The correct option is A $1.06$

The growth of current $\left(I\right)$ in $LR$ series circuit is given by

$I=\frac{V}{R}(1-e^{-\frac{tR}{L}})$

$\text{At t =}\infty ,$

$I_{\infty }=\frac{20}{5}(1-e^{\frac{-\infty }{L/R}})=4\dots \left(i\right)$

$\text{At t = 40s}$

$I_{40}=\frac{20}{5}(1-e^{\left(\frac{-40\times 5}{10\times {10}^{-3}}\right)})$

$I_{40}=4(1-e^{-20,000})\dots \left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$ we get

$\Rightarrow \frac{I_{\infty }}{I_{40}}=\frac{1}{1-e^{-20,000}}$

Which is slightly greater than $1$

$\Rightarrow \frac{I_{\infty }}{I_{40}}\approx \left(1.06\right)$

Hence, option $\left(\text{A}\right)$ is correct.