Laplace transforms: R series with RC parallel circuit

Problem Statement

Find the Voltage across the capacitor for t>=0:

Voltage across capacitor at t({0-})=0

Fig (1)

$v(t0-)=0Volts$

$v(t)=10Volts$

$R_{1}=20{\boldsymbol {\Omega }}$

$R_{2}=30{\boldsymbol {\Omega }}$

$C=.1Farad$

Use Loop Equations to solve for the currents in $i_{1}$ and $i_{2}$

Loop 1

$v(t)=R1(i_{1}+i_{2})+R2(i_{1})$

$10=20(i_{1}+i_{2})+30(i_{1})$

$i_{1}=(10-20i_{2})/50$_______________________________________equation (1)

Loop 2

$v(t)=R1(i_{1}+i_{2})+{\dfrac {1}{C}}\int {i_{2}dt}$

$10=20(i_{1}+i_{2})+{\dfrac {1}{.1}}\int {i_{2}dt}$_______________________equation (2)

Solve equations (1) and (2) simultaneously

Substituting equation (1) into equation (2) gives...

$20((10-20i_{2})/50+i_{2})+{\dfrac {1}{.1}}\int {i_{2}dt}=10$

simplifies to...

$12i_{2}+10\int {i_{2}dt}=6$

Take the Laplace Transform to move to the S-domain

${\mathcal {L}}\left\{\int _{0-}^{\infty }{(i_{2})dt}\right\}=I_{2}/s$

${\mathcal {L}}\left\{(1)\right\}=1/s$

$12I_{2}+10I_{2}/S=6/S$

$I_{2}(12+10/S)=6/S$

$I_{2}=6/(12S+10)$

$I_{2}=(1/2)(1/(S+(5/6))$

Take the inverse Laplace transform to move back into the t-domain

$i_{2}=(1/2)(e^{-(5/6)t})$

substitute this equation back into equation (1)

$i_{1}=(10-20(.5e^{-(5t/6)}))/50$

$i_{1}=(1/5)(1-e^{-(5t/6)})$

Voltage on Capacitor

$v_{capacitor}=10/20(i_{1}+i_{2})$

$v_{capacitor}=10-20((1/5)(1-e^{-(5t/6)})+(1/2)(e^{-(5t/6)}))$

$v_{capacitor}=10-4+4e^{-(5t/6)}-10e{-(5t/6)}$

Answer:

$v_{capacitor}=6-6e^{-(5t/6)}$ Volts

Applying the Initial and Final Theorems:

Initial Value Theorem

$\lim _{s\rightarrow \infty }sF(s)=f(0)$

Final Value Theorem

$\lim _{s\rightarrow 0}sF(s)=f(\infty )$

$v(t0)=0$ Volts

$v(t{\infty })=6$ Volts

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Written by: Andrew Hellie

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