# RC / RL Differential Equations - beginner

Discussion in 'Homework Help' started by kvnsmith.at.uw, May 16, 2012.

1. ### kvnsmith.at.uw Thread Starter New Member

May 16, 2012
4
0
First, I hope this topic doesn't already exist. I've searched and searched and Google'd and compiled everything that seemed to make sense.

Just getting started on these, I don't know differential equations but I don't think first-order are that hard so I'm not too worried. What I do want to know is that the work and figuring that I have done is not in error.

My goal was to get a GENERAL equation. I don't know why it was so hard. My solution ended up as this:

y(t) = (y(0-) - y(∞))e^-at + y(∞)

Where y can be either current or voltage and a can be either the time constant for an RC or for an RL. From what I've found, this should work BUT I really don't know what I'm doing and maybe it's pure falsehood.

The only real problem I have "solving" these after I get this form set up is what to do with y(∞). There's a table in the textbook where if that value is a constant you just replace it with a different constant and a couple other if-else situations. The ones we did in class were super confusing with 800 more letters in them, so I'm not confident I should say (for a small circuit with a switch, a Vs, an R(4) and a C(.25) where Vc(0-) is 2) that the solution is:

Vc(t) = (2 - K)e^-t + K

I know that's probably hard to follow, but the general equation is what I'm really after for now.

2. ### panic mode Senior Member

Oct 10, 2011
1,752
524
if you are only using first order circuit you can use

y(t)=N+M*e^-t/τ

where
τ (tau) is time constant

if you know initial condition (t=0) then you know N+M because

e^0=1 so

y(0)=N+M

if you know t=∞, then e^-∞=0 so y(∞)=N

in between you get exponential function. N and M are constants (which you find using evaluation of t=0 and t=∞)

for example you have series RC circuit that will be connected to 5VDC, and you are measuring voltage across capacitor as output variable. initially circuit is not powered so C is fully discharged. therefore at t=0 we get y(0)=0 and at infinity we get capacitor fully charged so y(∞)=5V

y(0)=0=N+M*e^0
0=N+M
N=-M

y(∞)=5 so

5=N+M*e^-∞=N+M*0
therefore N=5
and so it must be that M=-5

now we just need RC constant

τ=RC so our result is equation

y(t)=5(1-e^(-t/RC))

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