CPL.Luke Posted April 28, 2006 Share Posted April 28, 2006 Hmmm... so my AP book has one equation in the book for the current through an RL (resistor, inductor) circuit but I can;t seem to get the same result. I went through solving the differential equation with respect to I however I end up with the result 1/R(V-e^(-tR/L)) instead of the standard I_max (1-e^(-tR/L)) by any chance could someone oblige me and derive the solution for the current in an RL circuit? Link to comment Share on other sites More sharing options...
ydoaPs Posted April 28, 2006 Share Posted April 28, 2006 solving what differential equation? Link to comment Share on other sites More sharing options...
[Tycho?] Posted April 28, 2006 Share Posted April 28, 2006 Isn't that the solution to an integral? Link to comment Share on other sites More sharing options...
ydoaPs Posted April 28, 2006 Share Posted April 28, 2006 no, but you do integrate to get the function. Link to comment Share on other sites More sharing options...
CPL.Luke Posted April 28, 2006 Author Share Posted April 28, 2006 I did end up integrating, but I started with kirkoffs loop rule V=V_L+V_R then substituing in V_L with L dI/dt and V_R with IR I got V=L dI/dt +IR L dI/dt=V-IR L/(V-IR) dI=dt then integrating both sides you get a ln function then solving for t I get the solution I mentioned in the opening post which I don't believe it can be right as it doesn't seem to fit with dimensional analasys, unless something funky is happening in the exponent (I'm not sure what a unit in the exponent translates to), but still the book has a different solution which it gives without deriving. Link to comment Share on other sites More sharing options...
YT2095 Posted April 28, 2006 Share Posted April 28, 2006 I=(Vo/R)(1-e^-Rt/L) growth of constant current Vo. I=Ioe^-Rt/L = Ioe^-t/t decay of current, Initial Io. I not sure, but Maybe this might be of some help? Link to comment Share on other sites More sharing options...
CPL.Luke Posted April 28, 2006 Author Share Posted April 28, 2006 it is, but how would you derive that equation? from the basics my main problem is that I don't like memerizing equations I don't understand (partly because its alot more difficult for me to remember) and in order to understand it I need to know how to derive it from the basics on up. I also think it makes me a better student as I'll know how to tackle these problems when the come up in the future. Link to comment Share on other sites More sharing options...
s pepperchin Posted April 29, 2006 Share Posted April 29, 2006 This is how you derive that eqtn: start with your given eqtns 1 [math]V=V_L + V_R[/math] 2 [math]I_{max}=\frac{V_0}{R}[/math] 3 [math]V_L=L\frac{dI}{dt}[/math] 4 [math]V_R=IR[/math] Starting from eqtn 1 insert eqtns 3 and 4 [math]V=L\frac{dI}{dt} + IR[/math] the next set of steps is just some algebra to get it in the form we want before we integrate [math]V-IR=L\frac{dI}{dt}[/math] [math]V(1-I\frac{R}{V})=L\frac{dI}{dt}[/math] [math]V(1-I\frac{R}{V}) dt=L dI[/math] [math]\frac{V}{R}(1-I\frac{R}{V}) dt=\frac{L}{R} dI[/math] [math]\frac{R}{L}dt=\frac{R}{V(1-I\frac{R}{V})}dI[/math] [math]\frac{R}{L}\int dt=\frac{R}{V}\int\frac{1}{(1-I\frac{R}{V})}dI[/math] at this point we can use the follwing integral that is in most integral tables. [math]\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+c[/math] in this case a = 1 and b = -R/V [math]\frac{Rt}{L}=\frac{R}{V}(\frac{-V}{R}ln(1-I\frac{R}{V}))[/math] [math]\frac{Rt}{L}=-ln(1-I\frac{R}{V})[/math] [math]\frac{-Rt}{L}=ln(1-I\frac{R}{V})[/math] [math]e^{\frac{-Rt}{L}}=1-I\frac{R}{V}[/math] [math]e^{\frac{-Rt}{L}}-1=-I\frac{R}{V}[/math] [math]1-e^{\frac{-Rt}{L}}=I\frac{R}{V}[/math] [math]\frac{V}{R}(1-e^{\frac{-Rt}{L}})=I[/math] [math]I=\frac{V}{R}(1-e^{\frac{-Rt}{L}})[/math] [math]I=I_{max}(1-e^{\frac{-Rt}{L}})[/math] I hope this helps you Link to comment Share on other sites More sharing options...
CPL.Luke Posted April 30, 2006 Author Share Posted April 30, 2006 Absolutely! thankyou a couple of questions though if you look up a couple of posts I wrote out how I worked the algebra to get the integral, it looks like you worked it a bit differently by factoring out a V and dividing by R, what was the logic behind doing this and why did the problem not work out when I did it the other way? Link to comment Share on other sites More sharing options...
s pepperchin Posted April 30, 2006 Share Posted April 30, 2006 When I worked through it the first time I got the same thing you did but when I looked at the steps the units didn't work out the whole way through. I wasn't sur ehow exactly I had messed up but I plugged the integral into Mathematica and it gave me a weird answer that didn't even seem similar to what I had imput. so at that point I loked at a couple of different factors for the same integral and did some backwards engineering from the actual eqtn. Link to comment Share on other sites More sharing options...
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