Question

A 12.0 V ideal battery is connected in series to a 20 ohm resistor and an inductor. The switch is closed and the circuit reaches a steady state, then the switch is opened. At what rate is the battery transferring energy to the inductors field at a time equal to 2.0 times the inductive time constant?

Answer 1

Solution: Given that V=12 V R-2002 The switch is closed and the circuit reaches a steady state, then the switch is opened. This means it is the case of circuit without battery Hence the equation of state is =1.e where τ is the time constant of RL circuit The rate at which energy is being transferred to the inductors field is dI dt dt Given t = 2 τ R. 12 x( e-2 )-x (-20) Rate=-0.1318 w The negative si gn shows that energy is reducing