# Kirchoff's Law and Critical Points

1. Mar 11, 2009

### JJBladester

1. The problem statement, all variables and given/known data[/b]

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

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

3. The attempt at a solution

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

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

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?

2. Mar 11, 2009

### Staff: Mentor

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?

3. Mar 12, 2009

### JJBladester

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?

4. Mar 12, 2009

### Staff: Mentor

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

5. Mar 12, 2009

### 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.

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