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] (a) 0.324 s (b) 0.002 s (c) 0.103 s (d) 0.016 s

Answer: (d). Total resistance R = 0.1 + 0.9 = 1 Ω, so τ = L/R = 10×10⁻³1 = 0.01 s 0.8 = 1 - e-t/τ ⇒ e-t/τ = 0.2 = 1/5 t = τln 5 = 0.01 × 1.6 = 0.016 s So the answer is (d) 0.016 s.

A coil of self inductance 10 mH and resistance 0.1 Ω is connected… | JEE Main 2019 (10 Apr, Shift 2)

Q18JEE Main 2019 · 10 APR · SHIFT 2Medium

Inductors & DC Transients (Mixed)

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

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Answer: (d)

Total resistance $R = 0.1 + 0.9 = 1\ \Omega$ , so $\tau = \frac{L}{R} = \frac{10\times10^{-3}}{1} = 0.01\text{ s}$
$0.8 = 1 - e^{-t/\tau} \Rightarrow e^{-t/\tau} = 0.2 = \frac{1}{5}$
$t = \tau\ln 5 = 0.01 \times 1.6 = 0.016\text{ s}$
So the answer is (d) 0.016 s.

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