Question

In the circuit L=2.o H, R=10.0 Ohms and there is 110 V
(A)After the switch is first thrown to a, what time interval clapses before the current reaches 60% of its final steady-state value?
(b) After a long while, the switch is quickly thrown from a to b. What time interval clapses before the current in the inductor falls to 60% of its steady-state value?


*NEEDED AS SOON AS POSSIBLE

Answer 1

a)

$$0.6I=I(1-e^{-\frac{R}{L}t})\\0.6=1-e^{-\frac{10}{2}t}\\t=0.183\ s$$

b)

$$0.6I=Ie^{-\frac{R}{L}t}\\0.6=e^{-\frac{10}{2}t}\\t=0.102\ s$$