An inductor `(L = 20 mH)`, a resistor `(R = 100 Omega)` and a battery`(varepsilon = 10 V) ` are connected in series. Find (a) the time constant, (b) the maximum current and (c ) the time elapsed before the current reaches 99% of the maximum value.

Explanation

Here, `L = 20 mH = 20 xx 10^(-3) H, R = 100 ohm`.
`E = 10 V`
Time constant, `tau = (L)/(R ) = (20 xx 10^(-3))/(100) = 2 xx 10^(-4) s`
Maximum current, `I_(0) = (E)/(R ) = (10)/(100) = 10^(-1) A`
Using `I = I_(0) (1 - e^(- t//tau))`
`0.99 I_(0) = I_(0) (1 - e^(-t//tau))`
`:. e^(-t//tau) = 0.01`.
Solve to get `t = 0.92 xx 10^(-3) s`