A coil of inductance \(1\,H\) and resistance \(100\,\Omega\) 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 In \(2=0.693\), \(\left.e ^{-3 / 2}=0.25\right)\)

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Answer & Solution

Correct Answer

(C) \(t =7\,ms ; U =1\,mJ\)

Detailed explanation

\(i =\frac{ E }{ R }\left(1- e ^{- t / \tau}\right)\)

\(i =\frac{ E }{2 R }=\frac{ E }{ R }\left(1- e ^{- t / \tau)}\right)\)

Solving \(t =\tau \ln 2\)

\(t =\frac{1}{ R } \ln 2=\frac{1}{100} 0.693=0.00693\)

\(=7\,ms\)

\(i (15\,ms )=\frac{ E }{ R }\left(1- e ^{-\frac{15}{10}}\right)\)

\(i =\frac{6}{100}(1-1 / 4)=\frac{3}{4} \times \frac{6}{100}\)

\(U =\frac{1}{2} LI ^{2}\)

by solving we get \(U=1\,mJ\).

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