- #1

lorenz0

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- 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.

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.

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