QJEEM19Phaf57: JEE Main 2019 Physics - Class 12 JEEM19Phaf57 Solution: A coil of self inductance 10 mH and resistance 0.1 Ω is…

Question

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A coil of self inductance 10 mH and resistance 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: [take ln 5 = 1.6]

Accepted Answer

L = 10 × 10^–3 H, r₁ = 0.1

Ω

i = ε {1 - e^{- 1 / 2}}

i_{s a t u r a t i o n} = ε

80 % i_{s a t u r a t i o n} = 0.8 ε

0.8 = 1 - e^-t/2 ; e^-t/2 = 0.2

e^t/L = 5
t = L ln 5 = 10 × 10^–3 × 1.6 = 16 × 10^–3

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