# Help Needed for Solving Electrical Circuit Problem

• lorenz0
In summary, the correct expression for ##I(t)## is ##i(t)=0.06e^{-(R/L)t}## and by setting ##i(t^*)=35\cdot 10^{-3} A##, the correct value can be obtained. The mistake was using the wrong formula and not accounting for the absence of a battery in the "new" circuit. f

#### lorenz0

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.

#### Attachments

• RL_circuit.png
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• Delta2
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)##.

Last edited:
• lorenz0, berkeman, Steve4Physics and 1 other person
(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!

• lorenz0, berkeman and kuruman
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.

• berkeman, kuruman and Steve4Physics