An inductor (L = 20 mH), a resistor (R = 100 Ω) and a battery (ε = 10 V) are connected in series.Â
Find
i. The time constant
ii. The maximum current
iii. The time elapsed before the current reaches 99% of the maximum value.Â
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)"
Comments
Leave a comment