# source free RC circuit

#### PG1995

Joined Apr 15, 2011
832
Hi

Please have a look on the attachments. You can find my questions there. Please help me with them. Thank you.

Regards
PG

#### steveb

Joined Jul 3, 2008
2,436
Hi

Please have a look on the attachments. You can find my questions there. Please help me with them. Thank you.

Regards
PG
Q1: One is free to define current in either direction, but you have to be consistent and write the wformulas correctly based on the way you define it. Your way if fine, and the author's way is fine. Current does go through the capacitor in the form of displacement current. Later you will learn what this means, but for now just accept that current does flow through a capacitor in a way that is consistent with the equations.

Q2: The author is deriving the equation without calculus, but it is more elegant and proper to use calculus to derive it. The definition of capacitance is given by the formula Q=CV.

Now take the derivative of both sides, assuming C is constant.

dQ/dt=C dV/dt

and since current is defined to be the rate of change of charge I=dQ/dt.

I=C dV/dt

That's all there is to it.

Q3: Both A and ln(A) are correct because the value is an integration constant. Integration constants are arbitrary, so it doesn't matter what you call the constant. It can be A, ln(A), sin(A) or any other arbitrary constant. I think you are confused because the A was used in both cases, but A is arbitrary. Instead, think of it as A1=ln(A2) where A1 and A2 are different arbitrary constants. In other words, A and ln(A) are not equal to each other, but since you haven't assigned a value to A yet, you don't really care.

Note that integration constants can play many tricks with your mind, as in this case. Sometimes you can derive two different formulas for the same problem, by two different methods, but they look very different in form. But they end up being the same because the integration constant looks very different in form, but has the same numerical value. Keep an eye out for this in your studies.

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• PG1995

#### PG1995

Joined Apr 15, 2011
832
Thank you very much, Steve.

On this page the author derived equation 7.5 from 7.4b by rearranging the terms, otherwise I don't think he would have solved it. Is there general guide how to rearrange the terms so that it simple differential equation can be solved? Please let me know. Thanks.

Best wishes
PG

#### steveb

Joined Jul 3, 2008
2,436
Thank you very much, Steve.

On this page, the author derived equation 7.5 from 7.4b by rearranging the terms, otherwise I don't think he would have solved it. Is there general guide how to rearrange the terms so that it simple differential equation can be solved? Please let me know. Thanks.

Best wishes
PG
Solving differential equations can be tricky in general, especially if you try to use simple techniques like that. The simple approaches don't always work. One accepted practice in solving differential equation is the trial and error approach. You guess a form and try it to see if it works. That should give you an idea of just how tricky this subject is. When teachers tell you guessing is ok, you know you are in the deep stuff!

In the future, you will be learning more general methods, such as the Laplace transform approach, which lets you solve some linear differential equations in the frequency domain, and the state transition matrix approach, which lets you solve them in the time domain.

I can't really offer a general guide. In your studies you will gradually encounter more and more differential equations and you will learn more and more techniques to solve them. Below is a link to give you a flavor of some more general methods, but they will look very cryptic to you now. In time you will find all this straightforward, but you have to walk before you run, so be patient. Note that even the more general methods are still limited to very particular types of differential equations, hence this subject is quite vast in scope.

http://www.ece.ucsb.edu/~roy/classnotes/147b/lecture9_small.pdf

• PG1995

#### PG1995

Joined Apr 15, 2011
832
Hi

Please have a look on the attachments, you can find my questions there. These two pages follow directly the attached pages in my first post above. Please help me with the queries. Thanks a lot.

Regards
PG

PS: Please excuse Q2.

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Please have a look on the attachments, you can find my questions there. These two pages follow directly the attached pages in my first post above. Please help me with the queries. Thanks a lot.

Regards
PG

PS: Please excuse Q2.
For question Q1, you are correct that the author should define tau=RC beforehand, but it is a slight oversight that can be forgiven since the relation is given immediately afterwards.

You said to excuse Q2. Does this mean you figured it out already? I'm not sure, so I'll answer it along with questions Q3 which is related.

The derivative of exp(ax) with respect to x is a*exp(ax). Similarly, the antiderivative of exp(ax) with respect to x is exp(ax)/a. These are just basic relations for exponential functions and you typically derive them in calculus class. There are many ways to figure these relations out, but I don't know the one that will appeal to you at your level of study. One can use a power series expansion for exponential and take anti-derivatives or derivatives term by term. Or, one can use the chain rule and substitution as follows.

$${{d f({u})}\over{dx}} = {{d f({u})}\over{du}} {{d {u}}\over{dx}}$$

• PG1995