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 values is: [take $\ln 5=1.6$]

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Electromagnetic Induction

A coil of self inductance 10 mH and resistance

0.1Ω

is connected through a switch to a battery of internal resistance

0.9Ω

. After the switch is closed the time taken for the current to attain

80%

of the saturation values is: [take

ln 5 = 1.6

]

(A)0.016 s✓ Correct
(B)0.324 s
(C)0.002 s
(D)0.103 s

Answer: (A)

Solution

In an

L R

circuit the current climbs as

i = i_{0} (1 - e^{- t / τ})

with time constant

τ = \frac{L}{R _{total}} = \frac{10 \times 1 0 ^{- 3}}{0.1 + 0.9} = 0.01 s

. Reaching

80%

means

e^{- t / τ} = 0.2

, i.e.

e^{t / τ} = 5

, so

t = τ ln 5 = 0.01 \times 1.6 = 0.016 s

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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