# RC Circuit Differential Equation

• March 9th 2011, 11:46 AM
hansbahia
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?
• March 9th 2011, 12:01 PM
Ackbeet
Your first question is separable. That is, you have

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

Now integrate.

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

Make sense?
• March 9th 2011, 12:56 PM
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,

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.
• March 9th 2011, 04:00 PM
Ackbeet
Quote:

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

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

2. As you have already seen, I used a symbol other than $C$ 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?
• March 10th 2011, 06:27 AM
hansbahia
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
• March 10th 2011, 06:38 AM
Ackbeet
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?