Help Needed for Solving Electrical Circuit Problem

AI Thread Summary
The discussion revolves around solving an electrical circuit problem involving current calculations. The initial calculations were based on an incorrect expression for current, leading to an erroneous result. Participants pointed out the need for the correct formula, emphasizing the importance of proper LaTeX formatting for clarity. The correct expression for current should be i(t) = 0.06e^{-(R/L)t}, reflecting the circuit's conditions. The user ultimately corrected their approach and arrived at the right solution.
lorenz0
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Homework Statement
At the beginning the switch c of the circuit is in position 1. At t=0, it switches to position 2.
(1) Find ##I(0).## (2) Find ##t*## such that ##I(t*)=35mA.##
Relevant Equations
##I(t)=I(0)(1-e^{(-R/L)t)},\ I(0)=\frac{V}{R}##
What I have done:

(1) ##I(0)=\frac{V}{R}=\frac{1.5}{25}A=0.06 A.##

(2) By setting ##I(t*)=0.06(1-e^{-(35/0.4)t*})=35 mA## we get ##t*\approx 0.01 s##

What I have done seems correct to me, but the result for part (2) should be different.
I would be grateful if someone could point out to me where I have made a mistake.
 

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You have the wrong expression for ##I(t)##. It does not predict that ##I(0) =0.06## A. What is the correct expression to use?

Also, please enter the exponent correctly in LaTeX as e^{-Rt/L} to render as ##e^{-Rt/L}##. Or you could write it as ##\exp(-R t/L)##.
 
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lorenz0 said:
(2) By setting ##I(t*)=0.06(1-e^((-35/0.4)t*))=35 mA## we get ##t*=\approx 0.01 s##
You have used the wrong formula! Can you see why? If you can't, click the spoiler for a hint:
When the switch has been at position 2 for a long time, do you expect the current through the inductor to be 0.06A?

Also, here's a LaTeX formatting hint:
e^(-R/L)t gives: ##e^(-R/L)t## (yuchy)
but
e^{(-R/L)t} gives: ##e^{(-R/L)t}## (slightly yuchy)
Another alternative is:
e^{-(R/L)t} gives: ##e^{-(R/L)t}## (almost not yuchy)

For zero-yuchiness, I would define the time-constant as ##\tau = \frac L R##. Then
e^{-\frac t {\tau}} gives: ##e^{-\frac t {\tau}}##.

Edit. @kuruman beat me to it or I would not have replied!
 
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kuruman said:
You have the wrong expression for ##I(t)##. It does not predict that ##I(0) =0.06## A. What is the correct expression to use?

Also, please enter the exponent correctly in LaTeX as e^{-Rt/L} to render as ##e^{-Rt/L}##. Or you could write it as ##\exp(-R t/L)##.
I understand now, thanks. It should have been ##i(t)=0.06e^{-(R/L)t}## since, from time ##t=0## onwards, there isn't a battery connected to the "new" circuit anymore. By setting ##i(t^*)=35\cdot 10^{-3} A## I now get the correct value, thanks.
 
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