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Im practicing solving RLC circuits and there is this exercise that I attempt to solve and now I didn't understand something that was made in the exercise's solution. This is the exercise, I need to find the expression for v(t), t > 0.

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The thing I didn't understand was related to how the question is solved: enter image description here

Should not be an expression for i(t) and then solve somehow to v(t)? Because in and RLC series circuit, current is the same for all components. The book I'm using (Sadiku) divides the chapter into source free RLC series and source free RLC parallel circuits, and respectively, gives a way to calculate i(t) in the series and v(t) for the parallel circuits. I can't understand why the expression in RLC series could be in terms of voltage... It doesn't make sense to me. Apreciate any help...

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    \$\begingroup\$ Current of a capacitor is Cdv/dt. Voltage and current have the same form. i(t) = Cv'(t) = C*(-Aexp(-t) -9Bexp(-9t) ). So the only different in the end is the constants. You can start with whichever you like. Moreover, if you had sine signals you'd also need to calculate the phase constants. \$\endgroup\$ – Andrés Jun 14 '17 at 16:13
  • \$\begingroup\$ Oh that makes a lot of sense, had not seen through this point of view! Thanks a lot! \$\endgroup\$ – João Pedro Jun 14 '17 at 16:19
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    \$\begingroup\$ You could have started with the current and then take the integral. Try it it will be the same in the end. It's probably easier to take the derivative than the integral. \$\endgroup\$ – Andrés Jun 14 '17 at 16:21

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