University of Florida/Egm4313/s12.team11.perez.gp/R1.4

Problem: Derive (3) and (4) from (2).

Given:

$$ V = LC\frac{dV^2_c}{dt^2} + RC\frac{dV_c}{dt} + V_c $$ (2)

$$ V' = LI'' + RI' + \frac{1}{C}I $$                     (3)

$$ V = LQ'' + RQ' + \frac{1}{C}Q $$                      (4)

Solution:

First, let's solve for (3).

Recall that:

$$ Q = CV_c = \int Idt $$,

and

$$ I = C\frac{dV_c}{dt} = \frac{dQ}{dt} $$.

We can use this information to replace the differentiating terms accordingly.

After doing so, we get:

$$ V = L{I}' + RI + V_c \! $$

but knowing that $$ I = \frac{dQ}{dt} $$, we can rearrange the terms to get $$ V_c = \frac{1}{C}\int Idt $$.

Using this information in the previously derived equation, we find that:

$$ V = LI' + RI + \frac{1}{C}\int Idt $$.

Finally, after differentiating $$ V $$ with respect to $$ t $$, we get:

$$ V' = LI'' + RI' + \frac{1}{C}I $$.

Now, let's solve for (4).

Once again considering that $$ Q = CV_c \! $$, we can solve for (4) by differentiating

twice and then plugging it into (2).

Deriving twice, we find that:

$$ Q = CV_c \! $$

$$ Q' = C\frac{dV_c}{dt} $$

$$ Q'' = C\frac{dV^2_c}{dt^2} $$

After plugging this into (2), we see that:

$$ V = LQ'' + RQ' + V_c \! $$.

Once we rearrange $$ Q = CV_c \! $$, we find that $$ V_c = \frac{Q}{C} $$.

We can plug this in to the above equation to get the solution:

$$ V = LQ'' + RQ' + \frac{1}{C}Q $$.