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.

 

[Steve4Physics]

\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
$t_1=\frac{L}{R}\ln(10)$
doesn't follow. Check your algebra/arithmetic.

Whoops - sorry, see Post #4,

 

[lorenz0]

$\frac{9}{10}\frac{V}{R}=\frac{V}{R}(1-e^{-\frac{R}{L}}t_1)$
looks OK but the next bit
$t_1=\frac{L}{R}\ln(10)$
doesn't follow. Check your algebra/arithmetic.

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.

 

[Steve4Physics]

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

Apologies - I think your answer is correct.

 

[lorenz0]

Apologies - I think your answer is correct.

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

 

[kuruman]

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.

 

[Steve4Physics]

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.

 

[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.