# Thread: RC Circuit Differential Equation

1. ## RC Circuit Differential Equation

I was trying to do this practice problem, but the constants of the equation are killing me!
The question is:

-The differential equation for the voltage vc across the capacitor is dvc/dt=(V(t)-vc)/RC.
Find the formula for the general solution of the RC circuit equation above if the voltage source is contant for all time, i.e. V(t)=K for all t.
-Find the solution for the voltage vc(t) with initial-value vc(0)=1 in the RC circuit with constant voltage source V(t)=K

The first question I get stuck because once I replace V(t)=K, it becomes dvc/dt=(K-vc)/RC, now I could isolate RC by dvc/dt=1/RC(K-vc). The problem is after that I'm having problems differentiating.
The second question is easy. Once the formula for the general solution of the RC circuit is found all I have to do is plug it in vc(t)=1 and solve for contant right?

2. Your first question is separable. That is, you have

$\displaystyle \dfrac{dv_{c}}{K-v_{c}}=\dfrac{dt}{RC}.$

Now integrate.

Your intuition about the second part is correct, just make sure to do $\displaystyle v_{c}(0)=1,$ not $\displaystyle v_{c}(t)=1.$

Make sense?

3. dvc/(k-vc)= dt/RC,
Then I integrate both sides but first taking 1/RC out of the integral, since its constant. then
integral of (dvc/(k-vc))= (1/RC) integral of dt

that will equal to

ln|K-vc|= (1/RC)t+c,

solving for vc I have to exponentiate both sides which will equal to

e^(ln|K-vc|)= e^(t/RC+c), in which is the same as K-vc=e^(t/RC)e^(c),

solving for vc the equation will be,

vc=-e^(t/RC)e^(c)+K

Is that correct? Are the signs correct (-,+)?
Now do I have to replace e^(c) with another letter. My professor usually subs with k, but I feel like I can't since there is a K on the equation already. Does it matter?

As you said, my intuition on the second question is right, however when I plug 0 for all ts and equal the equation to 1
i get e^c=K-1

what I did was since vc(0)=1, so
vc(0)=-e^(t/RC)e^(c)+K
1=-e^(0/RC)e^(c)+K, aince e^(0)=1 then
1=(-1)e^(c)+K , solving for e^(c)
1-K=(-1)e^(c)
K-1=e^(c)

Now that i know that K-1=e^(c), I can go back to my first equation and plug (K-1) to e^(c)
by doing that i get
vc=-e^(t/RC)e^(c)+K
vc=-e^(t/RC)(K+1)+K

Would these be the correct answer or they are asking something else? I don't know the values for R nor C, neither the values for V(t)= K. Did the answers I gave satisfy what both question are asking? I feel like I did something is wrong and the solution is supposed to be something simpler.

4. Originally Posted by hansbahia
dvc/(k-vc)= dt/RC,
Then I integrate both sides but first taking 1/RC out of the integral, since its constant. then
integral of (dvc/(k-vc))= (1/RC) integral of dt

that will equal to

ln|K-vc|= (1/RC)t+c,

1. The integration is incorrect. It should be

$\displaystyle -\ln|K-v_{c}|=\dfrac{t}{RC}+B.$

2. As you have already seen, I used a symbol other than $\displaystyle C$ for my constant of integration. You don't want to confuse the constant of integration with the capacitance!

If you carry these corrections through, what does that give you?

5. Oh yes! I forgot to put the "-" sign when using substitution, with that being fixed all i have to change is

-ln|K-vc|= (1/RC)t+B
vc=+e^(t/RC)e^(B)+K

and the second part solving for e^(B)

1-K=(1)e^(c)
e^(c)=1-K

Now that i know that e^(c)=1-K, I can go back to my first equation and plug (1-K) to e^(c)
by doing that i get

vc=e^(t/RC)e^(c)+K

vc=e^(t/RC)(1-K)+K

Now everything should be right. However, does the these two answer satisfy what its being asked? Should I change K to V(t)?

vc(t)=e^(t/RC)(1-K)+K

6. The cool thing about differential equations is that it's easy to check your own answer: differentiation is easier than integration. Just differentiate your solution, plug it into the DE and the initial conditions, and see if they're all satisfied. Are they?

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