Laplace transforms: R series with RC parallel circuit: Difference between revisions
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Line 30: | Line 30: | ||
: <math>v(t0-)=0 Volts</math> |
: <math>v(t0-)=0 Volts\,</math> |
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: <math>v(t)=10 Volts</math> |
: <math>v(t)=10 Volts\,</math> |
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: <math>R_1=20 \boldsymbol{\Omega}</math> |
: <math>R_1=20 \boldsymbol{\Omega}\,</math> |
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: <math>R_2=30 \boldsymbol{\Omega}</math> |
: <math>R_2=30 \boldsymbol{\Omega}\,</math> |
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: <math>C=.1 Farad</math> |
: <math>C=.1 Farad\,</math> |
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Line 48: | Line 48: | ||
:Loop 1 |
:Loop 1 |
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:<math>v(t)=R1(i_1+i_2)+R2(i_1)</math> |
:<math>v(t)=R1(i_1+i_2)+R2(i_1)\,</math> |
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:<math>10=20(i_1+i_2)+30(i_1)</math> |
:<math>10=20(i_1+i_2)+30(i_1)\,</math> |
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:<math>i_1=(10-20i_2)/50</math>_______________________________________equation (1) |
:<math>i_1=(10-20i_2)/50\,</math>_______________________________________equation (1) |
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:Loop 2 |
:Loop 2 |
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:<math>v(t)=R1(i_1+i_2)+\dfrac{1}{C}\int{i_2 dt}</math> |
:<math>v(t)=R1(i_1+i_2)+\dfrac{1}{C}\int{i_2 dt}\,</math> |
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:<math>10=20(i_1+i_2)+\dfrac{1}{.1}\int{i_2 dt}</math>_______________________equation (2) |
:<math>10=20(i_1+i_2)+\dfrac{1}{.1}\int{i_2 dt}\,</math>_______________________equation (2) |
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Line 68: | Line 68: | ||
:Substituting equation (1) into equation (2) gives... |
:Substituting equation (1) into equation (2) gives... |
||
:<math>20((10-20i_2)/50+i_2)+\dfrac{1}{.1}\int{i_2 dt}=10</math> |
:<math>20((10-20i_2)/50+i_2)+\dfrac{1}{.1}\int{i_2 dt}=10\,</math> |
||
:simplifies to... |
:simplifies to... |
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:<math>12i_2+10\int{i_2 dt}=6</math> |
:<math>12i_2+10\int{i_2 dt}=6\,</math> |
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Line 79: | Line 79: | ||
:<math>\mathcal{L}\left\{\int_{0-}^\infty{(i_2) dt}\right\}=I_2/s</math> |
:<math>\mathcal{L}\left\{\int_{0-}^\infty{(i_2) dt}\right\}=I_2/s\,</math> |
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:<math>\mathcal{L}\left\{(1)\right\}=1/s</math> |
:<math>\mathcal{L}\left\{(1)\right\}=1/s\,</math> |
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:<math>12I_2+10I_2/S=6/S</math> |
:<math>12I_2+10I_2/S=6/S\,</math> |
||
:<math>I_2(12+10/S)=6/S</math> |
:<math>I_2(12+10/S)=6/S\,</math> |
||
:<math>I_2=6/(12S+10)</math> |
:<math>I_2=6/(12S+10)\,</math> |
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:<math>I_2=(1/2)(1/(S+(5/6))</math> |
:<math>I_2=(1/2)(1/(S+(5/6))\,</math> |
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Line 98: | Line 98: | ||
:<math>i_2=(1/2)(e^{-(5/6)t})</math> |
:<math>i_2=(1/2)(e^{-(5/6)t})\,</math> |
||
:substitute this equation back into equation (1) |
:substitute this equation back into equation (1) |
||
:<math>i_1=(10-20(.5e^{-(5t/6)}))/50</math> |
:<math>i_1=(10-20(.5e^{-(5t/6)}))/50\,</math> |
||
:<math>i_1=(1/5)(1-e^{-(5t/6)})</math> |
:<math>i_1=(1/5)(1-e^{-(5t/6)})\,</math> |
||
Voltage on Capacitor |
Voltage on Capacitor |
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:<math>v_{capacitor}=10/20(i_1+i_2)</math> |
:<math>v_{capacitor}=10/20(i_1+i_2)\,</math> |
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:<math>v_{capacitor}=10-20((1/5)(1-e^{-(5t/6)})+(1/2)(e^{-(5t/6)}))</math> |
:<math>v_{capacitor}=10-20((1/5)(1-e^{-(5t/6)})+(1/2)(e^{-(5t/6)}))\,</math> |
||
:<math>v_{capacitor}=10-4+4e^{-(5t/6)}-10e{-(5t/6)}</math> |
:<math>v_{capacitor}=10-4+4e^{-(5t/6)}-10e{-(5t/6)}\,</math> |
||
Answer: |
Answer: |
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::<math>v_{capacitor}=6-6e^{-(5t/6)}</math> Volts |
::<math>v_{capacitor}=6-6e^{-(5t/6)}\,</math> Volts |
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Line 124: | Line 124: | ||
Initial Value Theorem |
Initial Value Theorem |
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:<math>\lim_{s\rightarrow \infty} sF(s)=f(0)</math> |
:<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math> |
||
Final Value Theorem |
Final Value Theorem |
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:<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)</math> |
:<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math> |
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::<math>v(t0)=0</math> Volts |
::<math>v(t0)=0\,</math> Volts |
||
::<math>v(t{\infty})=6</math> Volts |
::<math>v(t{\infty})=6\,</math> Volts |
||
---- |
---- |
Revision as of 14:54, 22 October 2009
Problem Statement
- Find the Voltage across the capacitor for t>=0:
- Voltage across capacitor at t({0-})=0
Use Loop Equations to solve for the currents in and
- Loop 1
- _______________________________________equation (1)
- Loop 2
- _______________________equation (2)
Solve equations (1) and (2) simultaneously
- Substituting equation (1) into equation (2) gives...
- simplifies to...
Take the Laplace Transform to move to the S-domain
Take the inverse Laplace transform to move back into the t-domain
- substitute this equation back into equation (1)
Voltage on Capacitor
Answer:
- Volts
Applying the Initial and Final Theorems:
Initial Value Theorem
Final Value Theorem
- Volts
- Volts
Written by: Andrew Hellie
Checked by: