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

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

Given:

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

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

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

Solution:

First, let's solve for (3).

Recall that:

${\displaystyle Q=CV_{c}=\int Idt}$,

and

${\displaystyle I=C{\frac {dV_{c}}{dt}}={\frac {dQ}{dt}}}$.

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

After doing so, we get:

${\displaystyle V=L{I}'+RI+V_{c}\!}$

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

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

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

Finally, after differentiating ${\displaystyle V}$ with respect to ${\displaystyle t}$, we get:

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

Now, let's solve for (4).

Once again considering that ${\displaystyle Q=CV_{c}\!}$, we can solve for (4) by differentiating

twice and then plugging it into (2).

Deriving twice, we find that:

${\displaystyle Q=CV_{c}\!}$

${\displaystyle Q'=C{\frac {dV_{c}}{dt}}}$

${\displaystyle Q''=C{\frac {dV_{c}^{2}}{dt^{2}}}}$

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

${\displaystyle V=LQ''+RQ'+V_{c}\!}$.

Once we rearrange ${\displaystyle Q=CV_{c}\!}$, we find that ${\displaystyle V_{c}={\frac {Q}{C}}}$.

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

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