Question 167: AC fundamentals

Question

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 $$\ln 2=0.693, \mathrm{e}^{-3 / 2}=0.25$$)

Accepted Answer

$$i(t) = {V \over R}(1 - {e^{ - Rt/L}})$$ ...... (1)
$${L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms$$ ...... (2)
$${V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})$$
$$ \Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms$$
$$U = {1 \over 2}L{i^2} = {1 \over 2}{[1 - {e^{ - 15/10}}]^2}{\left[ {{6 \over {100}}} ight]^2}$$
$$ = {1 \over 2}{[1 - 0.25]^2} 	imes 36 	imes {10^{ - 4}}$$
$$ = 1\,mJ$$

Answer

t = 10 ms; U = 2 mJ

Answer

t = 10 ms; U = 1 mJ

Answer

t = 7 ms; U = 1 mJ

Answer

t = 7 ms; U = 2 mJ

AC fundamentals

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