What is meant by find the time constant?

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SUMMARY

The discussion centers on finding the time constant in a series RC circuit and transitioning from charge \(q(t)\) to current \(i(t)\). The differential equation governing the current is derived as \(\frac{di}{dt} = -\frac{1}{RC} i\), indicating that the time constant \(\tau\) is equal to \(RC\). Electrical engineers typically focus on current measurements rather than charge due to practical considerations. The solution to the differential equation is straightforward, and differentiation is a common initial step when solving such equations.

PREREQUISITES
  • Understanding of differential equations in electrical circuits
  • Familiarity with series RC circuit behavior
  • Knowledge of the relationship between charge and current
  • Basic calculus for differentiation and integration
NEXT STEPS
  • Study the derivation and application of the time constant in RC circuits
  • Learn about the Laplace transform for solving differential equations in circuits
  • Explore the impact of varying resistance and capacitance on circuit behavior
  • Investigate the role of inductors in circuit analysis and their differential equations
USEFUL FOR

Electrical engineers, students studying circuit theory, and anyone involved in analyzing or designing RC circuits will benefit from this discussion.

Dustinsfl
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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?
 
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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.
 
Ackbach said:
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 \(\mathcal{U}(t) = iR + \frac{1}{C}\int i(t)dt\)
 
dwsmith said:
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?
 
Ackbach said:
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?
 
dwsmith said:
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.
 

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