The current at resonance in a series RLC circuit is described in the bottom derivation, along with it being plugged into the formula for finding the voltage across the inductor in the circuitw:


I have two questions regarding this derivation.

  1. Isn't the current at resonance supposed to include the phase shift φ? Why is it left out of the equation? Example: I*sin(ωt+φ)
  2. When the equation for the current at resonance is plugged into the equation for finding the voltage across the inductor, why is the trigonometric part of the differentiated current equation neglected in the next step?
  • \$\begingroup\$ Sometimes, it helps if you do your own derivation for the series RLC circuit, rather than quoting someone else's results. Usually, if you go through that process on your own, you will develop your own answers to your questions. Here's an example of a parallel RLC analysis: Parallel RLC. Go through that process on your own with a series RLC. \$\endgroup\$
    – jonk
    Commented Sep 4, 2018 at 17:17
  • \$\begingroup\$ At series resonance the inductive and reactive impedances are equal and opposite hence the only remaining component is resistance and this introduces no phase shift. Your 2nd question refers to what equations? \$\endgroup\$
    – Andy aka
    Commented Sep 4, 2018 at 19:05

1 Answer 1

  1. You are basically right. But the phase shift would not change anything in the equations (except the last one). The derivative of sin() is still cos() and the relative phase relation stays the same, no matter what phase sin() has. You can do the math yourself.
  2. cos() is removed in the last equation because cos(t=0) is 1. But if you bring your phase shift in play, then it needs to be considered with cos(phi).

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