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

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

Given:

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

$V^{\prime }=L{I}^{″}+R{I}^{\prime }+\frac{1}{C}I$ (3)

$V=L{Q}^{″}+R{Q}^{\prime }+\frac{1}{C}Q$ (4)

Solution:

First, let's solve for (3).

Recall that:

$Q=C{V}_c=\int Idt$,

and

$I=C\frac{d{V}_c}{dt}=\frac{dQ}{dt}$.

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

After doing so, we get:

$V=L{I}^{\prime }+RI+V_c$

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:

$V=L{I}^{\prime }+RI+\frac{1}{C}\int Idt$.

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

$V^{\prime }=L{I}^{″}+R{I}^{\prime }+\frac{1}{C}I$.

Now, let's solve for (4).

Once again considering that $Q=C{V}_c$, we can solve for (4) by differentiating

twice and then plugging it into (2).

Deriving twice, we find that:

$Q=C{V}_c$

$Q^{\prime }=C\frac{d{V}_c}{dt}$

$Q^{″}=C\frac{d{V}_c^2}{d{t}^2}$

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

$V=L{Q}^{″}+R{Q}^{\prime }+V_c$.

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

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

$V=L{Q}^{″}+R{Q}^{\prime }+\frac{1}{C}Q$.