A35.0-v battery with negligible internal resistance, a 50.0 ? resistor, and a 1.25-mh inductor with negligible resistance are all connected in series with an open switch. the switch is suddenly closed.

(a) how long after closing the switch will the current through the inductor reach half of its maximum value?
(b) how long after closing the switch will the energy stored in the inductor reach half of its maximum value?

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:

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

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

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