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

Answer: (C)

t = 7 ms; U = 1 mJ

Answer 2

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

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

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

$⇒$ $e^{-\frac{Rt}{L}} = \frac{1}{2}$

⇒$$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