Question

A coil of inductance $1\text{H}$ and resistance $100\Omega$ is connected to a battery of $6\text{V}$. 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 $15\text{ms}$ after the circuit is switched on. (Given $\ln 2=0.693,e^{-3/2}=0.25$)

Answer 1

$I=\frac{\in }{R}(1-e^{-\frac{Rt}{L}}),I_0=\frac{\in }{R}$

$I=\frac{I_0}{2}=\frac{\in }{2R}=\frac{\in }{R}(1-e^{-Rt/L})$

$\frac{1}{2}=e^{-Rt/L}$

$\frac{Rt}{L}=ln2$

$t=\frac{L}{R}ln2=\frac{1\times 0.693}{100}$

$t=7\text{ms}$

$U=\frac{1}{2}L{I}^2$

$=\frac{1}{2}\times 1\times \left[\frac{6}{100}\right(1-e^{-3/2}{)}^2]$

$=\frac{1}{2}{[\frac{6}{100}\times \left(\frac{3}{4}\right)]}^2=1\text{mJ}$