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

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

Given:

$V=LC{\frac {dV_{c}^{2}}{dt^{2}}}+RC{\frac {dV_{c}}{dt}}+V_{c}$ (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:

$Q=CV_{c}\!$ $Q'=C{\frac {dV_{c}}{dt}}$ $Q''=C{\frac {dV_{c}^{2}}{dt^{2}}}$ 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:

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