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I have the following RCL circuit

schematic

simulate this circuit – Schematic created using CircuitLab

After using the Laplace transform my lecture notes mention that we get the following circuit

schematic

simulate this circuit

What I don't understand is why the polarity of the voltage source next to the capacitor is like that and it is not reversed. Shouldn't the initial condition V0 match the polarity of the source? Isn't that that reason we add the voltage source in the first place?

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If you look at this question from EESE and wikipedia you can see that:

  • The source below the capacitor is upside-down (inverse of the correct polarity) , and

  • The expression for the inductor source is not \$s I_0\$ but \$ L I_0\$.

So, either your notes are incorrect or you applied them incorrectly.

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  • \$\begingroup\$ Ok, could you who downvote comment/answer? I mean, I pointed to resources that show exactly "... why the polarity of the voltage source next to the capacitor is like that and not reversed", it is because it should have the opposite polarity. \$\endgroup\$ – jDAQ Jan 30 at 1:29
  • \$\begingroup\$ Also, that comes directly from the definition of the Laplace transform, for systems with initial conditions. \$\endgroup\$ – jDAQ Jan 30 at 1:31
  • \$\begingroup\$ I don't know who downvoted but what you gave is not an answer. Obviously the reason I asked is because I already know the reference polarities and so I know that there is a difference from the one in the notes. Also the sources you provided obviously exist in my 2 textbooks on the subject and in my class notes so I don't know what is the reason you provided them when you must have known that I already have access to much better ones. \$\endgroup\$ – Adam Jan 30 at 10:13

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