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CSunderman
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- TL;DR Summary
- 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
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