# Time for current in solenoid to be 10% less than the steady state current

lorenz0
Homework Statement:
A solenoid (##L=230mH##) is connected to a constant voltage source via a resistive wire with resistance ##R=0.15\Omega##. Initially, the circuit is open.
How much time after closing the circuit is the current in the solenoid 10% less than the steady-state current?
Relevant Equations:
##V-iR-L\frac{di}{dt}=0##
I set up the equation ##V-iR-L\frac{di}{dt}=0##, with ##i(0)## and by solving it I got ##i(t)=\frac{V}{R}(1-e^{-\frac{R}{L}t})##.
Then, since the steady state current is ##i_s=\frac{V}{R}## I imposed the condition ##i(t_1)=\frac{9}{10}\frac{V}{R}\Leftrightarrow \frac{9}{10}\frac{V}{R}=\frac{V}{R}(1-e^{-\frac{R}{L}t_1})\Leftrightarrow t_1=\frac{L}{R}\ln(10)\approx 3.53 s##, but this answer is different from the solution proposed in the book I took the problem from.

I don't see what I am doing wrong here, so I would appreciate if someone would point me in the right direction, thanks.

Last edited:

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\frac{9}{10}\frac{V}{R}\Leftrightarrow \frac{9}{10}\frac{V}{R}=\frac{V}{R}(1-e^{-\frac{R}{L}}t_1)\Leftrightarrow t_1=\frac{L}{R}\ln(10)\approx 3.53 s##, but this answer is different from the solution proposed in the book I took the problem from.
##\frac{9}{10}\frac{V}{R}=\frac{V}{R}(1-e^{-\frac{R}{L}}t_1)##
looks OK but the next bit
##t_1=\frac{L}{R}\ln(10)##

Whoops - sorry, see Post #4,

Last edited:
• hutchphd
lorenz0
##\frac{9}{10}\frac{V}{R}=\frac{V}{R}(1-e^{-\frac{R}{L}}t_1)##
looks OK but the next bit
##t_1=\frac{L}{R}\ln(10)##
there was a typo in my answer (now corrected): the ##t## was part of the exponent and so the formula I had obtained should be correct.

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there was a typo in my answer: the ##t## was part of the exponent and so the formula I had obtained should be correct.

• lorenz0
lorenz0
The mistake was purely mine, thanks. Nonetheless the book claims the answer should be ##0.16s##.

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The mistake was purely mine, thanks. Nonetheless the book claims the answer should be ##0.16s##.
Why don't you put your value and the book's value in the equation for ##I(t)## and see which one gives you 10% less than the steady state current? Then you will have to decide if you believe the book or your own work.

• lorenz0
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The mistake was purely mine, thanks. Nonetheless the book claims the answer should be ##0.16s##.
0.16s is the time to reach 10% of the steady-state current. Probably not a coincidence!

Looks like the wording in the question is wrong/misleading or the person who worked out the ‘official’ answer misinterpreted the question.

• berkeman and lorenz0
lorenz0
0.16s is the time to reach 10% of the steady-state current. Probably not a coincidence!

Looks like the wording in the question is wrong/misleading or the person who worked out the ‘official’ answer misinterpreted the question.
Thank you very much, no wonder I couldn't see what I was doing wrong.

• berkeman