Question

A coil of resistance 12Ω and self-inductance 2mH is connected across an ideal 12V battery.

Answer 1

Answer:

T= 0.167 millisecs (T is time constant)

$I = 1- e^{-6000t}$

T₉₉ = 0.768 millisecs (T₉₉ is time taken to reach 99% current value)

E = 10⁻³ Joules or 1 milli Joule

Explanation:

Here the expression of Time Constant for a series L-R circuit is$T =\frac{L}{R}$

Thus $T = \frac{2* 10^{-3} }{12}$ = 0.167 millisecs

The current flowing in series L-R circuit is given by the expression

$I = \frac{V}{R} (1 - e^{-\frac{t}{T} } )$

Substituting required values we get

$I = 1- e^{-6000t}$

Final current is when t -> ∞ I(final) -> 1

Thus when I = 0.99 then t = T₉₉

$0.99 = 1- e^{-6000T_{99} }$

Solving we get

T₉₉ = 0.768 millisecs

Energy stored in inductor is $\frac{1}{2} Li^{2}$

Here final current is I(final)->1

$E = \frac{1}{2}* 2*10^{-3} *1^{2}$

Thus,

E = 10⁻³ Joules or 1 milli Joule