RL circuits

[CPL.Luke]

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?

[ydoaPs]

solving what differential equation?

[[Tycho?]]

Isn't that the solution to an integral?

[ydoaPs]

no, but you do integrate to get the function.

[CPL.Luke]

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.

[YT2095]

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?

[CPL.Luke]

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.

[s pepperchin]

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

[CPL.Luke]

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?

[s pepperchin]

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.