Answer: a) 43.32 ms; b) 77.37 ms
Explanation: In order to explain this problem we have to take into account the expression for the current for a RL circuit, which is given by:
[tex]i(t)=if (1-e^{-t/RL} )[/tex]
when i(t) is half of the final currrent we have:
1/2=1-e^(-t/RL) then 1-1/2=-e^(-t/RL)
1/2=e^(-t/RL)
ln(1/2)=-t/RL
then t= 0.69*50*1.23*10^-3=43.32 ms
On the other hand to calculate the time when the stored magnetic energy get the half value we also know that;
The magnetic energy is given by: Um=1/2*L*if^2
then we need to calculate the time
when i(t)=if/(2)^1/2[tex]i(t)=\frac{if}{\sqrt{2}}[/tex]
then we have
1/(2)^1/2=1-e^(-t/RL)
1-(1/(2)^1/2)=e^(-t/RL)
ln (0.29)= -t/RL
t= ln(0,29)*50*1.25*10^-3=77.37 ms
