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Question
A coil of resistance 40 Ω is connected across a 4.0 V battery. 0.10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil.
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Solution
Given:-
Resistance, R = 40 Ω
Emf of the battery, E = 4 V
Now,
The steady-state current in the LR circuit is given by
\[i_0 = \frac{4}{40} = 0 . 1 A\]
At time, t = 0.1 s, the value of current i is 63 mA = 0.063 A
The current at time t is given by
i = i0(1 − e−t/τ)
⇒ 0.063 = 0.1(1 − e−tR/L)
⇒ 63 =100(1 − e−4/L)
⇒ 63 = 100(1 − e−4/L)
⇒ 1 − 0.63 = e−4/L
⇒ e−4/L = 0.37
\[\Rightarrow- \frac{4}{L}=\ln(0.37)\]
\[\Rightarrow L=\frac{- 4}{- 0 . 994}\]
= 4.024 H
= 4 H
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