Kirchoff's Law and Critical Points

[JJBladester]

1. Homework Statement [/b]

What is/are the critical points of Kirchoff's Law:

L\left(\frac{di}{dt}\right) + Ri = E

The Attempt at a Solution

I solved the differential equation above and got the following solution (which I verified to be correct):

i = \left(\frac{E}{R}\right) + Ce^{-\left(\frac{R}{L}\right)t}

If I remember correctly, the critical points would be when \left(\frac{di}{dt}\right) = 0.

\left(\frac{di}{dt}\right) = \left(\frac{E}{L}\right) - \left(\frac{R}{L}\right)i so you have a critical point when

\left(\frac{E}{L}\right) = \left(\frac{R}{L}\right)i

Is this correct or am I on the wrong path?

 

[berkeman]

Sorry, what is the definition of a critical point in this context? Do they mean critical damping of this series RL circuit, or something else?

 

[JJBladester]

Sorry, what is the definition of a critical point in this context? Do they mean critical damping of this series RL circuit, or something else?

I asked my professor about the definition of a critical point in this context. She wrote back:

The critical point of a first order DEQ is the value of the dependent variable found by setting its derivative to zero.

So, my crack at an answer is:

L(di/dt) + Ri = E

di/dt + (R/L)i = E/L

0 + (R/L)i = E/L

i = E/R <----- since "i" is the dependent variable, by setting di/dt = 0, we have a critical point at E/R, or voltage/resistance.

How does this sound?

 

[berkeman]

Seems like you did the math right, I'm just not able to intuit what it means physically.

 

[JJBladester]

Current equals voltage over resistance. This is Ohm's law, if I'm not mistaken. If di/dt = 0 then the change in current with respect to time is zero, which means if you have a circuit running at constant current, that current can be measured as voltage/resistance.