User:Egm4313.s12.team13.rossmeissl/R1work

Problem Statement
Using the circuit given, the governing circuit equation (4.0), and the capacitance equation (4.1), derive alternate equations (4.2) and (4.3).

Circuit Equation (4.0),

Capacitance Equation (4.1),

Alternate Equation (4.2),

$$ LCI''+RCI'+\frac{1}{C}I = V' $$

Alternate Equation (4.3),

$$ LCQ''+RCQ'+\frac{1}{C}Q = V $$

Solution
First take the derivative of the governing circuit equation (4.0) to get (4.4)

$$ V' = LC\frac{{d}^{3}v_{c}}{d{t}^{3}}+RC\frac{{d}^{2}v_{c}}{d{t}^{2}}+\frac{{d}^{}v_{c}}{d{t}^{}} $$

Since L, R, and C are constants with respect to time

Then, since

$$ I = C\frac{dv_{c}}{dt} $$

that means

$$ I' = C\frac{{d}^{2}v_{c}}{d{t}^{2}} $$

And

$$ I'' = C\frac{{d}^{3}v_{c}}{d{t}^{3}} $$

Substituting these into the derivative of the circuit equation (4.4) gives the first alternate equation (4.2)

$$ V' = LCI''+RCI'+\frac{1}{C}I $$

Then, since

$$ Q = C{v}_{c} $$

$$ Q' = C\frac{d{v}_{c}}{dt} $$

$$ Q'' = C\frac{{d}^{2}{v}_{c}}{d{t}^{2}} $$

Now substituting these into the appropriate places in the circuit equation (4.0) gives the second alternate equation (4.3)

$$ LCQ''+RCQ'+\frac{1}{C}Q = V $$