Solving Equation to Analyze Steady State Current

In summary: I will try to do that and see if the answer comes out the same. Thank you very much, no wonder I couldn't see what I was doing wrong. I will try to do that and see if the answer comes out the same.
  • #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.
 
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  • #2
lorenz0 said:
\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,
 
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  • #3
Steve4Physics said:
##\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.
 
  • #4
lorenz0 said:
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.
 
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  • #5
Steve4Physics said:
Apologies - I think your answer is correct.
The mistake was purely mine, thanks. Nonetheless the book claims the answer should be ##0.16s##.
 
  • #6
lorenz0 said:
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.
 
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  • #7
lorenz0 said:
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.
 
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  • #8
Steve4Physics said:
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.
 
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Related to Solving Equation to Analyze Steady State Current

What is the equation for analyzing steady state current?

The equation for analyzing steady state current is I = V/R, where I is the current in amperes, V is the voltage in volts, and R is the resistance in ohms.

How do you solve for steady state current in a circuit?

To solve for steady state current in a circuit, you need to know the voltage and resistance values. Then, you can plug these values into the equation I = V/R and solve for I. The resulting value will be the steady state current in the circuit.

What does steady state mean in terms of current?

In terms of current, steady state refers to a condition where the current in a circuit remains constant over time. This means that there are no changes in the voltage or resistance, and the current is not affected by external factors.

How is steady state current different from transient current?

Steady state current is the current that exists when a circuit is in a stable, unchanging state. Transient current, on the other hand, refers to the current that occurs during a period of change in the circuit, such as when the circuit is turned on or off.

Why is it important to analyze steady state current in a circuit?

Analyzing steady state current in a circuit is important because it allows us to understand how the circuit behaves under normal conditions. This information is crucial for designing and troubleshooting circuits, as well as predicting the performance of electronic devices.

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