Laplace transforms: R series with RC parallel circuit: Difference between revisions
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Laplace Transform S-domain |
Laplace Transform S-domain |
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<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>I_2(12+10/S)=6/S</math> |
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:<math>I_2=6/(12S+10)</math> |
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:<math>I_2=(1/2)(1/(S+(5/6))</math> |
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⚫ | |||
Inverse Laplace transform T-domain |
Inverse Laplace transform T-domain |
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:<math>i_2=(1/2)(e^{-(5/6)t})</math> |
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:substitute this equation back into equation (1) |
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:<math>i_1=(10-20(.5e^{-(5/6)t}))/50</math> |
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:<math>i_1=(1/5)(1-e^{-(5/6)t})</math> |
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Revision as of 00:25, 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)
Solving equations (1) and (2) simultaneously
- Substituting equation (1) into equation (2) gives...
- simplifies to...
Laplace Transform S-domain
Inverse Laplace transform T-domain
- substitute this equation back into equation (1)
Voltage on Capacitor