User:Egm4313.s12.team6.jagolinzer.r1.4

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

Find:
Derive the below two equations from the given:


 * A)     $$LI''+RI'+\frac{1}{C}I=V'$$


 * and


 * B)     $$LQ''+RQ'+\frac{1}{C}Q=V$$

Part A:
From Eq. 2-2(1): $$Q=CV_{c}=\int Idt$$

$$\therefore \frac{dI}{dt}=C\frac{d^2V_{c}}{dt}$$

substituting $$Q=CV_{c}=\int Idt$$ into $$RC\frac{dV_{c}}{dt}$$ we get:   $$RI$$

substituting $$\frac{dI}{dt}=C\frac{d^2V_{c}}{dt}$$ into $$LC\frac{d^2V_{c}}{dt^2}$$ we get:    $$LI'$$

So we now have:   $$V=LI'+RI+V_{c}$$

Deriving the whole equation we get:


 * $$V'=LI''+RI'+\frac{dV_{c}}{dt}$$

by multiplying $$\frac{dV_{c}}{dt}$$ by $$/frac{C}{C}$$ we get:


 * $$V'=LI''+RI'+\frac{CdV_{c}}{Cdt}$$

we can now sub $$Q=CV_{c}=\int Idt$$ for $$\frac{CdV_{c}}{dt}$$ giving us:

Part B:
From Eq. 2-2(1): $$Q=CV_{c}=\int Idt$$

multiplying  $$V_{c}$$   by $$\frac{C}{C}$$ we get $$\frac{CV_{c}}{C}$$

Subbing $$Q=CV_{c}=\int Idt$$ into $$\frac{CV_{c}}{C}$$ we get:


 * $$V=LC\frac{d^2V_{c}}{dt^2}+RC\frac{dV_{c}}{dt}+\frac{Q}{C}$$

taking the first and second derivative of  $$Q=CV_{c}$$   we get:


 * $$Q'=C\frac{dV_{c}}{dt}$$ and $$Q''=C\frac{d^2V_{c}}{dt^2}$$

substitute $$Q'=C\frac{dV_{c}}{dt}$$ into $$RC\frac{dV_{c}}{dt}$$ we get:   $$RQ'$$

substitute $$Q=C\frac{d^2V_{c}}{dt^2}$$ into $$LC\frac{d^2V_{c}}{dt^2}$$ we get:   $$LQ$$