How do I calculate the expected Gain?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
CSunderman
Messages
1
Reaction score
1
TL;DR
I am doing an experiment in electronics with AC RLC circuits, But there are a few things about what I need to do that I don't understand.
What is The gain equation for an inductor with internal resistance .
Why is the internal resistance in series.
I am doing an experiment in electronics with AC RLC circuits, But there are a few things about what I need to do that I don't understand.
First: While I know in the circuit diagram we include the internal resistance of the Inductor in series, but I don't know why we include it in series rather than in parallel.
Second: The Magnitude of the impedance should be Sqrt(R^2+(Internal R)^2+(wL-(1/wc))^2,
but what should I be using for the gain over the inductor, since it has two components to its gain should I add them when calculating the dBnof the inductor, or is there a more complicated combination .
I.E should the equation be 20Log((XL+(internal Resistance))/Z)
Thanks for help.
for reference: Capacitative Reactance=Xc. Inductive reactance=XL, resistance=R, Total impedance=Z
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
We use the internal resistance in series with the (ideal) inductor because the current through the inductor is the same as the current through its internal resistance. When the currents are the same the components are in series.