Question

A coil of inductance 1 H and resistance 100 Ω 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 \(ln2=0.693\), e^–3/2 = 0.25)

• A. t = 10 ms; U = 2 mJ
• B. t = 10 ms; U = 1 mJ
• C. t = 7 ms; U = 1 mJ
• D. t = 7 ms; U = 2 mJ

Answer 1

The Correct Option is C

\(i(t) = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right)\) \(⋯(1)\)
\(\frac{L}{R} = \frac{1}{100s}\)

\(⇒\) \(\frac{L}{R} = 10 \ \text{ms} \quad \dots (2)\)

\(\frac{V}{2R} = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right)\)

\(⇒\) \(e^{-\frac{Rt}{L}} = \frac{1}{2}\)

\(⇒\)\(t = \frac{L}{R} \ln 2 = 6.93 \ \text{ms}\)

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

\(=\frac{1}{2} \left[1 - e^{-\frac{15}{10}}\right]^2 \left(\frac{6}{100}\right)^2\)
\(=\frac{1}{2}[1 - 0.25]^2 \times 36 \times 10^{-4}\)
\(= 1 \)mJ
So, the correct option is (C): t = 7 ms; U = 1 mJ