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Time required store half the maximum energy in an inductor

  • Thread starter prodo123
  • Start date
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Homework Statement


A 35.0 V battery with negligible internal resistance, a 50 Ω resistor and a 1.25 mH inductor forms a RL circuit. How long will it take for the energy stored in the inductor to reach one-half of its maximum value?

Homework Equations


##i(t)=I_0 e^{\frac{-tR}{L}}##
##U(t)=\frac{1}{2}Li(t)^2##

The Attempt at a Solution


##U=\frac{1}{2}Li^2=\frac{1}{2}U_0##
##\frac{1}{2}Li^2=\frac{1}{4}LI_0^2##
##i^2=\frac{1}{2}I_0^2##
##i=I_0 e^{\frac{-tR}{L}}=\frac{1}{\sqrt{2}}I_0##
##\frac{-tR}{L}=\text{ln}(\frac{1}{\sqrt{2}})=-\frac{1}{2}\text{ln}(2)##
##t=\frac{L}{2R}\text{ln}(2)##
##t=8.66\text{ µs}##

Textbook has the following answer:
##t=30.7\text{ µs}##

Am I doing something wrong? The same method worked for finding how long it takes for the current to reach half the maximum value ##I_0=\frac{\mathcal{E}}{R}##.
 

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