I know that a simple circuit like this in an ideal case will have a frequency response of a single pole LPF, where the cutoff frequency is given by:
cutoffFreq = 1/(twoPi*L*2R)
This is reviewed here.
But what would the frequency response look like if you add a capacitor in parallel to the inductor like this:
Would it still be a low pass filter? What would it look like? How would it change based on the capacitance?
I tried running this on circuitlab for simulation but it is giving me an error and I don't know why:
It gives info on this error here but I don't understand it: https://www.circuitlab.com/docs/faq/#q_bode_opsolve_failed
No matter what I do it won't simulate with the inductor in there. But if I replace it with a capacitor it will simulate. Is there something else I'm supposed to be doing?
If it's not hard, what would be the transfer function of this circuit? Thanks.