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

Option 1 - 1.06

Option 2 -  0.84

Option 3 - 1.46

Option 4 - 1.15

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Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

$\begin{array}{r} i & = \frac{V}{r} (1 - e^{- R t / L}) \\ = 4 (1 - e^{- 500 t}) \end{array}$

At $t = \infty$

$i_{1} = 4 A$

at $t = 40 s$

$i_{2} = 4 (1 - e^{- 20000})$

= 4 [1 - \frac{1}{{(e^{2})}^{10000}}]

= 4 [1 - \frac{1}{(7.389)^{10000}}]

\frac{i_{1}}{i_{2}}

is slightly greater than 1 .

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An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 m H inductor and 5 Ω resistor. The ratio of the currents at time t = ∞ and at t = 40 s is close to: (take e 2 = 7.389

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