Question:
An emf of $20 \mathrm{~V}$ is applied at time $t=0$ to a circuit containing in series $10 \mathrm{mH}$ inductor and $5 \Omega$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40 \mathrm{~s}$ is close to: (Take $e^{2}=7.389$ )
Correct Option: 1
Solution:
(1) The current (I) in LR series circuit is given by
$I=\frac{V}{R}\left(1-e^{-\frac{t R}{L}}\right)$
At $t=\infty$,
$I_{\infty}=\frac{20}{5}\left(I-e^{\frac{-\infty}{L / R}}\right)=4$ ...(i)
At $t=40 \mathrm{~s}$,
$I_{40}=\frac{20}{5}\left(1-e \frac{-40 \times 5}{10 \times 10^{-3}}\right)=4\left(1-e^{-20,000}\right)$ ...(ii)
Dividing (i) by (ii) we get
$\Rightarrow \quad \frac{I_{\infty}}{I_{40}}=\frac{1}{1-e^{-20,000}}$
which is slightly greater than $1 \Rightarrow(1.06)$