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Solve the given problems.The rate of change of current $i$ (in $m A$ ) in a circuit with a variable inductance is given by $d i / d t=300(5.0-t)^{-2},$ where $t$ (in ms) is the time since the circuit is closed. Find $i$ as a function of $t$ if $i=300 \mathrm{mA}$ for $t-2.0 \mathrm{ms}$.
Calculus 1 / AB
Chapter 25
Integration
Section 2
The Indefinite Integral
Integrals
Campbell University
Harvey Mudd College
University of Nottingham
Lectures
03:09
In mathematics, precalculu…
31:55
In mathematics, a function…
01:54
A series $R L$ circuit lik…
00:48
In the circuit, switch $S$…
03:07
An electric circuit contai…
03:17
Electric Circuits An elect…
06:34
An (open) electric circuit…
03:26
At time $t=0,$ a a 12.0 $\…
02:44
The current in a 90.0 -mH …
03:57
An inductor that has an in…
05:37
Consider the $R L$ circuit…
05:21
Consider a series circuit …
for a series R L circuit. The current at Time T is given by the IMF over resistance times one minus me to the minus our TV over Inducted its And so the rate of change of current is just the first derivative of this with respect the time and so this will be negative. Negative or positive? I m f over Anil because you have one over r uh, times are over. L um, times e to the minus party Overall, you differentiate us and so a t equals four seconds are zero seconds. Um d i d t rate of change Current is just eso This term is one that's just I m f over l and so that's 60 votes over 1.5 Henry. And that gives you 40 amperes per second. Um, and so part B A T equals 0.1 seconds. Uh, the rate of change of current is now he cool to 60 year old 600.5 again. So 40 I appear per second times e to the minus 22 times 220.1. This is not there anymore. So this term, it's not one anymore. I was 22 times point 1/1 220.0.5. Um And so? So this gives you rate of change off the 9.23 up yours per second.
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