An emf of 20 V is applied at time t=0


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 $e^{2}=7.389$ )

  1. (1) $1.06$

  2. (2) $1.15$

  3. (3) $1.46$

  4. (4) $0.84$

Correct Option: 1


(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 $1 \Rightarrow(1.06)$

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