MHB What is meant by find the time constant?

[Dustinsfl]

I solved a DE for an electrical circuit where the input was a step input.
\[
\mathcal{U}(t) =
\begin{cases}
0, & \text{if } t <0\\
V, & \text{otherwise}
\end{cases}
\]
So the solved DE for \(t > 0\) is
\[
q(t) = VC + Ae^{\frac{-R}{C}t}.
\]

1. How do I find the time constant?
2. Also, \(q(t)\) is the charge. How can I go from \(q(t)\) to the current with respect to time?

 

[Ackbach]

If the circuit is a series $RC$ circuit, then I don't buy your solution. Can you post the circuit? To get from charge to current, you simply use the definition of current:
$$i= \frac{dq}{dt}.$$
Incidentally, electrical engineers solve circuits for $i$, and almost never bother with charge, because current is so much easier to measure in a lab.

 

[Dustinsfl]

If the circuit is a series $RC$ circuit, then I don't buy your solution. Can you post the circuit? To get from charge to current, you simply use the definition of current:
$$i= \frac{dq}{dt}.$$
Incidentally, electrical engineers solve circuits for $i$, and almost never bother with charge, because current is so much easier to measure in a lab.

The circuit is \(\mathcal{U}(t) = iR + \frac{1}{C}\int i(t)dt\)

 

[Ackbach]

The circuit is \(\mathcal{U}(t) = iR + \frac{1}{C}\int i(t)dt\)

So, differentiating once yields
\begin{align*}
0&=R \frac{di}{dt}+\frac{i}{C} \\
R \frac{di}{dt}&=- \frac{i}{C} \\
\frac{di}{dt}&=- \frac{1}{RC} \, i.
\end{align*}
What is the solution to this DE?

 

[Dustinsfl]

So, differentiating once yields
\begin{align*}
0&=R \frac{di}{dt}+\frac{i}{C} \\
R \frac{di}{dt}&=- \frac{i}{C} \\
\frac{di}{dt}&=- \frac{1}{RC} \, i.
\end{align*}
What is the solution to this DE?

So that is trivial to solve. One question then. Do we always differentiate the DE to begin with after it is written?

 

[Ackbach]

So that is trivial to solve. One question then. Do we always differentiate the DE to begin with after it is written?

If you want to solve for $i$, then yes, essentially you differentiate once to get the DE in terms of $i$. You can do this if you have inductors in the circuit as well.

Now if some bozo (typically a physics professor like myself) wants to see the charge, you can just integrate the current and find the right integration constant.