A coil of self inductance 10 m H and resistance of 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω . After the switch is closed, the time taken for the current to attain

Question

A coil of self inductance 10 m H and resistance of 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω . After the switch is closed, the time taken for the current to attain 80 % of the saturation value is: [ ln 5 = 1.6 ]

TestHub · Accepted Answer

The instantaneous current in an LR circuit is given by the equation :
(instantaneous current) $I = I_{s} (1 - e^{- \frac{R t}{L}})$

Where $I, I_{s}, R, t, L$ are the instantaneous current, saturation current, resistance, time and inductance respectively.

Given that the current becomes so percent of saturation current.
$\Rightarrow (0.8) I_{s} = I_{s} (1 - e^{- \frac{R t}{L}})$

$R = 0.1 Ω + 0.9 Ω = 1 Ω$
$\Rightarrow 0.8 = (1 - e^{- \frac{1 \times t}{10 \times 10^{- 3}}})$
$\Rightarrow 0.8 = 1 - e^{- 100 t}$
Therefore, the time is:
$t = \frac{l n (\frac{1}{1 - 0.8})}{100} = \frac{\ln (5)}{100} = 0.016 s$

Physics - EMI/AC Question with Solution | TestHub

Options:

Doubts & Discussion