Question

A 12.0 VV dc battery having no appreciable internal resistance, a 154.0 ΩΩ resistor, an 11.2 mH inductor, and an open switch are all connected in series.

1.After the switch is closed, what is the time constant for this circuit ?

2.After the switch is closed, what is the maximum current that flows through it ?

3.What is the current 73.1 μs after the switch is closed ?

4.After the switch is closed, what is the maximum energy stored in the inductor?

Answer 1

| 0 Tue Constant = % T= 11.2 x 103 154 7.273 x 10 seconds we know (t) = (1-2 ) at t=0 It) = ☆ (1-2) = 0 Hence, current - 0 A mi At t-73.1 us 4 = 12 (1-12 43) > Ile = 0.0494 A At any time energy in the inductor E = 2 2 2 1 ² So, & is maximum when I is maximum and I = Imax = ě So, Emax= 2 x 11.2 x ( 12 ) ² = 3.481035 Hence, Emax = 3.4 x 10-5 J 1000