Question

An inductor (L = 20 mH), a resistor (R = 100 Ω) and a battery (ε = 10 V) are connected in series. 

Find

i. The time constant

ii. The maximum current

iii. The time elapsed before the current reaches 99% of the maximum value. 

Answer 1

i.

$\tau=L/R=0.02/100=0.0002\ (s)$

ii.

$I_{max}=10/100=0.1\ (A)$

iii.

$0.99I_{max}=I_{max}(1-e^{-t/\tau})\to t=-\tau\ln0.01=$

$=-0.0002\cdot\ln0.01=9.2\cdot10^{-4}\ (s)$