A coil of inductance 1 H and resistance 100Ω is connected to a battery of 6 V. Determine approximately:
(a) The time elapsed before the current acquires half of its steady – state value.
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on.
(Given In2 = 0.693, e ^-3/2= 0.25)

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t = 7 ms; U = 2 mJ

t = 10 ms; U = 2 mJ

t = 10 ms; U = 1 mJ

t = 7 ms; U = 1 mJ

answer is C.

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Detailed Solution

Given circuit is R – L growth circuit
Question Image

$\begin{array}{l} i = \frac{E}{R} (1 - e^{- t / τ}) \\ i = \frac{E}{2 R} = \frac{E}{R} (1 - e^{- t / τ}) \\ Solving t = τln 2 \\ t = \frac{L}{R} \ln 2 = \frac{1}{100} 0 .693 = 0 .00693 \\ = 7 ms \\ i (15 ms) = \frac{E}{R} (1 - e^{- \frac{15}{10}}) \\ i = \frac{6}{100} (1 - \frac{1}{4}) = \frac{3}{4} \times \frac{6}{100} \end{array}$
$U = \frac{1}{2} {LI}^{2}$
By solving we get U = 1 mJ

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