Derivation of Source-Free RC circuit

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  • Thread starter Brute
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  • #1
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Homework Statement


Hi,
If there is a series combination of a resistor and an initially charged capacitor, I know that the decay of the voltage is given by the equation v(t) = Ae^(-t/RC) where V(0) = A = V0. But i am unsure of how to get to this equations.

Homework Equations


If I assume Ir = current flowing through resistor
and Ic = current flowing through Capacitor and assume there both flowing out of the node.


The Attempt at a Solution


Then:
Ic + Ir = 0
C(dv/dt) + V/R = 0
V/R = -C(dv/dt)
1/V(dv) = -1/RCdt
integrate both sides
ln v = -t/RC + A
A= integration constant

Here is where i cant go no more I saw somewhere they got: ln v = -t/RC + ln A, from there I know how to get the solution but how to arrive at ln A?, can we just do this automatically because it is a constant to make life easier or is there some logic behind it?
 

Answers and Replies

  • #2
64
6
Two things you can do here. You're answer is already correct even though it says "+A" and not "+ln(A)".

Let me explain.

When you get a constant of integration, it is representative of a single value determined from initial conditions. If no conditions are given, it remains a letter; in your case, "A". Now think what ln(A) is. The natural log of a constant is what? A constant! So leaving it as "A" is totally fine, because "A" just represents a constant anyway.

The reason the solution has "+ln(A)" is because when they evaluated the left hand integral of dV/V, then wrote it as ln(V) - ln(A). Personally, I write mine like this so I can evaluate the integral as ln(V/A). It's all preference. What really matters is how you implement the initial conditions.
 
Last edited:
  • #3
64
6
Now, to elaborate on the above (sorry for separate post) equation of v(t) = Ae(-t/RC) where V(0) = A = V0.

Your equation is ln(V) = -t/RC + A.
So let's check this out with the info I gave you in my previous post:

Remove the natural log: eln(V) = e-t/RC + A
Now rewrite: V = e-t/RCeA
Rearrange: V = eAe-t/RC

Now above I told you ln(A) is a constant, as is A..so it's literally the same thing because it just represents a constant. So why don't we look at eA. That's a constant raised to a constant power, right? So it's still just a constant. Just call it A (or C, or whatever, it doesn't matter, as long as you know it's just a single value).

Now we have: V = Ae-t/RC.

Now just implement your initial conditions!
 
  • #4
rude man
Homework Helper
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Homework Statement


Hi,
If there is a series combination of a resistor and an initially charged capacitor, I know that the decay of the voltage is given by the equation v(t) = Ae^(-t/RC) where V(0) = A = V0. But i am unsure of how to get to this equations.

Homework Equations


If I assume Ir = current flowing through resistor
and Ic = current flowing through Capacitor and assume there both flowing out of the node.


The Attempt at a Solution


Then:
Ic + Ir = 0
C(dv/dt) + V/R = 0
V/R = -C(dv/dt)
1/V(dv) = -1/RCdt
integrate both sides
ln v = -t/RC + A
A= integration constant

ln v = -t/RC + A
ln v(0+) = -t/RC|t=0 + A
So A = ln v(0+)
and ln v = ln v(0+) - t/RC
so take exp of both sides to get your answer. Remember exp(a + b) = exp(a)exp(b).
 
  • #5
8
0
I would like to thank both of you and taking the time out to answer my question. It has made me understand the derivation thoroughly.
 

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